math.sjtu.edu.cnmath.sjtu.edu.cn/upload/teachers/11087/Mathematical_analysis_5.pdf · Contents Part...

Analysis

Yi Li

DEPARTMENT OF MATHEMATICS, SHANGHAI JIAO TONG UNIVERSITY, DONGCHUANROAD, SHANGHAI, CHINA 200240

E-mail address: [email protected]

Contents

Part 1. Mathematical analysis 1

Introduction 3

Chapter 1. 5

Chapter 2. 7

Chapter 3. 9

Chapter 4. 11

Chapter 5. 13

Chapter 6. 15

Chapter 7. 17

Chapter 8. 19

Chapter 9. 21

Chapter 10. 23

Chapter 11. 25

Chapter 12. 27

Chapter 13. Multiple integrals 2913.1. Multiple integrals 2913.2. Basic properties 4013.3. Stieltjes integrals 4413.4. Change of variables 6013.5. Improper integrals 6913.6. Differential forms 73Problems 75

Chapter 14. Line and surface integrals 7714.1. Line integrals 7714.2. Surface integrals 8014.3. Green, Gauss, and Stokes formulas 8214.4. Differential forms: revisited 8814.5. Different operators 9114.6. Stokes’ theorem on manifolds 93

v

vi CONTENTS

Problems 93

Chapter 15. Integrals depending on a parameter 9515.1. Proper integrals depending on a parameter 9515.2. Improper integrals depending on a parameter 10015.3. Eulerian integrals 10515.4. Γ, Ψ, Φ functions and Mellin transform 11915.5. Modular forms 15815.6. Problems 158

Chapter 16. Fourier analysis 16316.1. Basic properties 16316.2. Convergence 17616.3. The Fourier transform on R 18716.4. Isoperimetric inequality 207Problems 208

Part 1

Mathematical analysis

Introduction

3

CHAPTER 1

5

CHAPTER 2

7

CHAPTER 3

9

CHAPTER 4

11

CHAPTER 5

13

CHAPTER 6

15

CHAPTER 7

17

CHAPTER 8

19

CHAPTER 9

21

CHAPTER 10

23

CHAPTER 11

25

CHAPTER 12

27

CHAPTER 13

Multiple integrals

13.1. Multiple integrals

We have a canonical metric defined on the Euclidean metric space En :=(Rn, d), where

(13.1.1) d(x, y) :=

(∑

1≤i≤n(xi − yi)

2

)1/2

for any x = (x1, · · · , xn), y = (y1, · · · , yn) ∈ Rn. It is clear that for any x, y, z ∈ Rn

we have(i) d(x, y) ≥ 0 with equality if and only if x = y.

(ii) d(x, y) = d(y, x).(iii) d(x, y) ≤ d(x, z) + d(z, y).En is a basic model for metric space. Besides the metric d given in (13.1.1),

there are lots of metrics on Rn. For example

(13.1.2) d∞(x, y) := max1≤i≤n

|xi − yi|

and also

(13.1.3) d1(x, y) := ∑1≤i≤n

|xi − yi|.

Exercise 13.1.1. Verify that d∞, d1 also satisfy the above properties (i)–(iii). Pleasefind some relations among d, d∞, and d1.

For any p > 0 define

(13.1.4) dp(x, y) :=

(∑

1≤i≤n|xi − yi|p

)1/p

.

Exercise 13.1.2. Show that dp satisfies the above properties (i)–(iii) when p ≥ 1.What happens for 0 < p < 1?

13.1.1. Metric spaces. Let X be an arbitrary nonempty set. A function d :X × X → R := R ∪ ∞ is a metric on X if the following properties hold for anyx, y, z ∈ X:

(i) (Positiveness): d(x, y) ≥ 0 with equality if and only if x = y.(ii) (Symmetry): d(x, y) = d(y, x).

(iii) (Triangle inequality): d(x, y) ≤ d(x, z) + d(z, y).A metric space is a pair (X, d) where d is a metric on X. A function d : X× X → Ris a semi-metric if it satisfies all properties (i)–(iii) except the requirement thatd(x, y) = 0 if and only if x = y. A semi-metric space is a pair (X, d) where d is a

29

30 13. MULTIPLE INTEGRALS

semi-metric on X.

Suppose (X, dX) is a metric space. For any x ∈ X define

(13.1.5) [x] := y ∈ X : d(x, y) 6= ∞.

We introduce a function d[x] : [x]× [x]→ R by

(13.1.6) d[x](y, z) := dX(y, z).

According to (13.1.5) we have

d[x](y, z) ≤ dX(y, x) + dX(x, z) < ∞

so that d[x] is finite.

Exercise 13.1.3. Show that ([x], d[x]) is a finite metric space. Thus a metric spacewith possibly infinite distances splits canonically into subspaces that carry finitemetrics.

Let (X, dX) be a semi-metric space. We say x ∼ y if and only if dX(x, y) = 0.Set

(13.1.7) [x] := y ∈ X : y ∼ x, X/ ∼:= [x] : x ∈ Xand define dX/∼ by

(13.1.8) dx/∼([x], [y]) := dX(x, y)

for any [x], [y] ∈ x/ ∼.

Exercise 13.1.4. Show that (X/ ∼, dX/∼) is a metric space. We will often abusenotation, writing (X/d, d) instead of (X/ ∼, dX/∼).

A map f : (X, dX) → (Y, dY) between two metric spaces is called distance-preserving if

(13.1.9) dY( f (x), f (y)) = dX(x, y)

for any two points x, y ∈ X. A bijective distance-preserving map is called anisometry. Two metric spaces are isometric if there exist an isometry from one tothe other.

Example 13.1.5. Define the distance d between two points x = (x1, x2) and y = (y1, y2)in R2 by

(13.1.10) d(x, y) := |(x1 − y1) + (x2 − y2)|.

Clearly d(x, x) = 0 for all x ∈ R2 and d(x, y) = d(y, x). The triangle inequality followsfrom

d(x, z) = |(x1 − z1) + (x2 − z2)|= |(x1 − y1) + (x2 − y2) + (y1 − z1) + (y2 − z2)|≤ |(x1 − y1) + (x2 − y2)|+ |(y1 − z1) + (y2 − z2)|= d(x, y) + d(y, z).

Hence (R2, d) is a semi-metric space. Given a point x = (x1, x2) ∈ R2, we have

[x]d = y = (y1, x1 + x2 − y1) ∈ R2 : y1 ∈ R.

13.1. MULTIPLE INTEGRALS 31

The real line R together with the metric dE(x, y) := |x− y| is a metric space. Define

(13.1.11) f : (R2/d, d) −→ (R, dE), f ([x]d) := x1 + x2.

If [y]d = [x]d, then x1 + x2 = y1 + y2. Hence the map (13.1.11) is well-defined. Moreover

dE( f ([x]d), f ([y]d)) = dE(x1 + x2, y1 + y2) = |x1 + x2 − y1 − y2| = d([x]d, [y]d).

Thus f is distance-preserving. Define

(13.1.12) g : (R, dE) −→ (R2/d, d), g(x) := [(x, 0)]d.

Hence

f g(x) = f ([(x, 0)]d) = x + 0 = x,g f ([x]d) = g(x1 + x2) = [(x1 + x2, 0)]d = ([x]d);

thus f is an isometry and (R2/d, d) is isometric to the real line.

Various examples of metric spaces will appear everywhere in this part. In thissection we only describe several important ones to begin with.

Example 13.1.6. One can define a (finite) metric on an arbitrary set X by

d(x, y) :=

0, x = y,1, x 6= y.

This example is not particularly interesting but it can serve as the initial pointfor many constructions.

Example 13.1.7. The real line R can be put variant metrics. For instance(1) d(x, y) := |x− y| as in 13.1.4.(2) dln(x, y) := ln(|x− y|+ 1). We can check three conditions in the definition of

metrics: dln(x, y) ≥ ln 1 = 0 for all x, y ∈ R; dln(x, y) = 0 if and only if |x−y| = 0; the triangle inequality dln(x, z) ≤ dln(x, y) + dln(y, z) is equivalent to

|x− z|+ 1 ≤ (|x− y|+ 1)(|y− z|+ 1)

or|x− z| ≤ |x− y||y− z|+ |x− y|+ |y− z|.

However, |x− z| ≤ |x− y|+ |y− z| by the triangle inequality for d defined in(1), we immediately get the triangle inequality for dln.

The metric dln naturally appears in the Fubini-Study metric on projective spaces.

Example 13.1.8. (Direct products) Let (X, dX) and (Y, dY) be two metric spaces. Wedefine a metric on their direct product X×Y by

dX×Y((x1, y1), (x2, y2)) := (dX(x1, x2) + dY(y1, y2))1/2 .

To verify that (X×Y, dX×Y) is a metric space, we suffice to prove the triangle inequality.Using the inequality

√ab ≤

√a +√

b for any a, b ≥ 0, we arrive at

dX×Y((x1, y1), (x3, y3)) =√

dX(x1, x3) + dY(y1, y3)

≤√

dX(x1, x2) + dX(x2, x3) + dY(y1, y2) + dY(y2, y3)

≤√

dX(x1, x2) + dY(y1, y2) +√

dX(x2, x3) + dY(y2, y3)

= dX×Y((x1, y1), (x2, y2)) + dX×Y((x2, y2), (x3, y3)).


In particular, R× R = R2.

Example 13.1.9. (Dilated spaces) Let (X, d) be a metric space. For any λ > 0, defineλX the same set X and

dλ(x, y) := λd(x, y), x, y ∈ X.

Then (λX, dλ) is also a metric space, called the dilation of X by λ.

Example 13.1.10. (Subspaces) If (X, d) is a metric space and Y is a subset of X, thenthe map

dY := d|Yis a metric on Y.

Restricting the distance is the simplest but not the only way to define a metricon a subset. In many cases it is more natural to consider an intrinsic metric, whichis generally not equal to the one restricted from the ambient space.

Example 13.1.11. As a subset of R2, the unit circle S1 carries the restricted Euclideanmetric from R. We can define another (intrinsic) metric dint by

(13.1.13) dint(x, y) := the length of the shorter arc between them.

For example, the arc length distance between two opposite points of the circle is equal to π;the distance between adjacent vertices of a regular n-gon (inscribed into the circle) is equalto 2π/n.

(i) Let θ ∈ [0, π] be the angle between the vectors Ox and Oy, where O is theoriginal point. Then

(13.1.14) dint(x, y) := θ

so that any arc length is less than or equal to π.(ii) By (13.1.14), we see that any circle arc of length less than or equal to π, equipped

with the metric (13.1.14), is isometric to a straight line segment.(iii) However, the whole circle with the intrinsic metric is not isometric to any sub-

set of R2. The reason is that we choose two distinct points x = (0, 1) andy = (0,−1) such that dint(x, (1, 0)) = dint(y, (1, 0)) = π/2. In fact, sup-pose we have an isometry f : (S1, dint) → (X, dE), where X is a subset of R2

with the induced metric dE. f ((1, 0)) and f ((−1, 0)) are middle points on theline segment f (x) f (y). Since we consider the Euclidean metric, it follows thatf ((1, 0)) = f ((−1, 0)) and hence (1, 0) = (−1, 0) as points on S1. This isimpossible.

Let V be a vector space over R.

Definition 13.1.12. A function | · | : V → R is a norm on V if the following condi-tions are satisfied:

(i) (Positiveness) |v| > 0 if v 6= 0 and |0| = 0;(ii) (Homogeneity) |kv| = |k||v| for any k ∈ R;

(iii) (Triangle inequality) |v + w| ≤ |v|+ |w|.A normed space (V, | · |) is a vector space V together with a norm | · |. Finite-dimensional normed spaces are also called Minikowski spaces.


A normed space (V, | · |) induces a natural metric d given by

(13.1.15) d(v, w) := |v− w|.

Hence (V, d) is a metric space.

Example 13.1.13. The Euclidean space Rm is a normed space whose norm is

(13.1.16) |x| :=

(∑

1≤i≤mx2

i

)1/2

.

This norm is called the standard norm on Rm. There are other important norms on Rm.(i) Define

(13.1.17) ||x||1 := ∑1≤i≤m

|xi|, x = (x1, · · · , xm).

Then Rm1 := (Rm, || · ||1) is a normed space.

(ii) Define

(13.1.18) ||x||∞ := max|x1|, · · · , |xm|, x = (x1, · · · , xm).

Then Rm∞ := (Rm, || · ||∞) is a normed space.

(iii) For p ≥ 1, define

(13.1.19) ||x||p :=

(∑

1≤i≤m|xi|p

)1/p

, x = (x1, · · · , xm).

Then Rmp := (Rm, || · ||p) is a normed space. In particular, Rm

2 = (Rm, | · |).

Proposition 13.1.14. Rm1 and Rm

∞ are isometric only if m = 1, 2.

PROOF. Since R11 = R1

∞, we may assume that m ≥ 2.(1) For m = 2, define

f : R21 −→ R2

∞

byf (x1, x2) = (|x2|+ |x1|, |x2| − |x1|).

Then|| f (x)||∞ = |x1|+ |x2| = ||x||1.

Note that f is bijective. Hence f is isometric.(3) For m > 2, we shall show that Rm

1 and Rm∞ are not isometric. Suppose that

we have an isometry f : Rm1 → Rm

∞. Then

max1≤i≤m

| fi(x1, · · · , xm)| = ∑1≤i≤m

|xi|.

The left-hand side has 2m possibilities while the right-hand side has 2m possibili-ties. Hence we have the identity

2m = 2m.

This identity only holds for m = 1 or m = 2.


Example 13.1.15. Let X be an arbitrary set. Denote by `∞(X) the set of all boundedfunctions f : X → R. Then `∞(X) is a vector space and is equipped with a norm givenby

(13.1.20) || f ||∞ := supx∈X| f (x)|.

If X = 1, · · · , m, we write `∞(m) as `∞(X).

Proposition 13.1.16. For any m ≥ 1, we have Rm∞ = `∞(m).

PROOF. For any point x = (x1, · · · , xm) define

fx : 1, · · · , m −→ R, i 7−→ xi.

Then fx is bounded, and

|| fx||∞ = supi∈1,··· ,m

| fx(i)| = sup1≤i≤m

|xi| = ||x||∞.

Thus the map F : Rm∞ → `∞(m), x 7→ fx, is distance-preserving. The inverse map

of F is given by F( f ) = ( f (1), · · · , f (m)). Therefore F is an isometry.

13.1.2. Volume. For n-dimensional rectangle

(13.1.21) = (x1, · · · , xn) ∈ Rn : ai ≤ xi ≤ bi, 1 ≤ i ≤ n = ∏1≤i≤n

[ai, bi],

we define its volume by

(13.1.22) Vol() := ∏1≤i≤n

(bi − ai).

Consider a nonempty bounded domain E ⊂ Rn. We define a family of rectan-gles of E

(13.1.23) AE := (k)1≤k≤K, K ∈ N,

where k are rectangles satisfying

E ⊂⋃

1≤k≤K

k

and Int(k) ∩ Int(j) = ∅ for any j 6= k. Introduce variant notions:

A−E := k ∈ AE : k ⊂ E,A×E := k ∈ AE : k ∩ E 6= ∅ ⊃ A−E ,

A+E := k ∈ AE : k ⊂ Rn \ E.

ThenA−E ∪A

×E ∪A

+E = AE.

Let

(13.1.24) m(AE) := ∑k∈A−E

Vol(k), M(Ak) := ∑k∈A×E

Vol(k).

Observe that 0 ≤ m(AE) ≤ M(AE).


For any two families of rectangles AE = (k)1≤k≤K and AE = (j)1≤j≤J , wesay that AE is a refinement of AE, denote by AE ≺ AE, if for any j, there existsan element k ∈ AE such that j ⊂ k. If AE ≺ AE, then

(13.1.25) m(AE) ≤ m(AE) ≤ M(AE) ≤ M(AE).

Indeed,

m(AE) = ∑j∈A−E

Vol(j)

= ∑j∈A−E , j⊂k∈A−E

Vol(j) + ∑j∈A−E , j⊂k∈A×E

Vol(j)

≥ ∑k∈A−E

∑j⊂k , j∈A−E

Vol(j)

= ∑k∈A−E

Vol(k) = m(AE).

Similarly, we can show that M(AE) ≤ M(AE). We use the notion AE the collectionof all families of rectangles of E. Define

(13.1.26) m(E) := supAE∈AE

m(AE), M(E) := infAE∈AE

M(AE).

Definition 13.1.17. A nonempty bounded domain E ⊂ Rn is said to be admissibleif

m(E) = M(E).

In this case, Vol(E) := m(E) = M(E) is called the volume of E.

Theorem 13.1.18. Suppose that E ⊂ Rn is nonempty and bounded. Then E is admissibleif and only if ∂E is admissible and Vol(∂E) = 0.

PROOF. Assume that E is admissible. For any given ε > 0, we have 0 <M(AE) − m(AE) < ε for some AE ∈ AE. Let A∂E := k ∈ AE : k ∩∂E 6= ∅. It is clear that M(A∂E) = M(AE) − m(AE) < ε and consequentlyinfA∂E∈A∂E M(A∂E) ≤ ε. Hence Vol(∂E) = 0.

Conversely, we assume that ∂E is admissible and Vol(∂E) = 0. For any ε > 0,we have M(A∂E) < ε for some A∂E ∈ A∂E. Choose any AE ∈ AE so that A∂E ⊂A×E \ A

−E . Then M(AE)−m(AE) ≤ M(A∂E) < ε. Thus E is admissible.

Let E ⊂ Rn be a bounded domain.

(1) E is a set of volume zero if E is admissible and Vol(E) = 0.(2) E is admissible if and only if ∂E is a set of volume zero.(3) If E1, E2 are sets of volume zero, then so are E1 ∩ E2, E1 ∪ E2 and E1 \ E2.(4) When E ⊂ R2 is bounded and admissible, we sometimes write

σ(E) = Area(E) = Vol(E).

(5) Vol(E) = 0 if and only if for any ε > 0 there are 1, · · · ,K such that

E ⊂⋃

1≤k≤K

Int(k), Vol

(∑

1≤k≤Kk

)< ε.


Corollary 13.1.19. (1) Let A, B are bounded and admissible subsets of Rn. Then A ∪B, A ∩ B, and A \ B are all admissible.

(2) Let D be a bounded and admissible subset of Rn. Then D is admissible andVol(D) = Vol(D).

Example 13.1.20. Let y = f (x), a ≤ x ≤ b, be a continuous function. Then

D := (x, y) ∈ R2 : y = f (x)is admissible. Choose any partition a = x0 < x1 < · · · < xn = b of [a, b]. Let

M = maxa≤x≤b

f (x), U = [a, b]× [0, M], Mi = maxxi−1≤x≤xi

f (x), mi = minxi−1≤x≤xi

f (x),

where we may assume that f ≥ 0 (otherwise we can replace f by f −mina≤x≤b f (x)).Define

−i := [xi−1, xi]× [0, mi], ×i := [xi−1, xi]× [mi, Mi], +i := [xi−1, xi]× [Mi, M]

which induces an element AD ∈ AD. Then

m(AD) = ∑1≤i≤n

mi(xi − xi−1), M(AD) = ∑1≤i≤n

Mi(xi − xi−1).

Since f is integrable, it follows that

m(D) = M(D) = σ(D) =∫ b

af (x)dx.

Hence the curve Γ = y = f (x) : x ∈ [a, b] is a set of area zero.

Example 13.1.21. Let D ⊂ R2 be a bounded domain. (1) If ∂D is the union of finitelymany admissible curves, then D is admissible. If ∂D = ∪1≤j≤JΓj where Γj are admissi-ble. To prove σ(∂D) = 0 we suffice to show that σ(Γj) = 0 for every j. However thisis obvious. (2) If ∂D is the union of finitely many piecewise smooth curves, then D isadmissible.

Example 13.1.22. Consider the set

D = (x, y) ∈ R2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ D(x)where

D(x) =

1, x ∈ Q,0, x /∈ Q.

Then D is not admissible. This directly follows from

∂D = [0, 1]× [0, 1].

To prove the above fact it suffices to verify that [0, 1] × [0, 1] ⊂ ∂D which is the conse-quence of the definition of D.

13.1.3. Multiple integrals. Let D ⊂ Rn be a bounded, admissible, and closeddomain. A partition of D is a set

(13.1.27) ∆ := (∆Dk)1≤k≤K

where ∆Dk are all admissible and closed, Vol(∆Dj ∩ ∆Dk) = 0 for any j 6= k, andD = ∪1≤k≤K∆Dk. Let

∆Vk := Vol(∆Dk), dk := diam(∆Dk), λ(∆) := max1≤k≤K

dk.


Let f = f (x) be a function defined on D. We say that f is integrable on D,if there is a constant I, such that for any ε > 0 there exists δ > 0 such that theRiemann sum S(∆) := ∑1≤k≤K f (ξk)∆Vk satisfies

(13.1.28) |S(∆)− I| < ε

for any partition ∆ with λ(∆) < δ and any ξk ∈ ∆Dk. In this case, I is called theintegral of f on D and written as

(13.1.29) I =∫

Df dV =

∫D

f (x)dx.

There are some special cases:(a) n = 2: ∫∫

Df (x, y)dxdy

(b) n = 3: ∫∫∫D

f (x, y, z)dxdydz

(c) f ≡ 1:

Vol(D) =∫

Df dV.

13.1.4. Darboux theory. As in the one-dimensional case, we shall show thatany integrable function over a bounded, admissible, and closed domain must bebounded.

Theorem 13.1.23. Let D ⊂ Rn be a bounded, admissible, and closed domain. Then anyintegrable function f on D must be bounded.

PROOF. Otherwise, f is unbounded on D. There exists a point x0 ∈ D suchthat any neighborhood U(x0, δ) of x0 satisfies that f is unbounded on U(x0, δ)∩D.Let

∆Dδ := U(x0, δ) ∩ D.Then Vol(∆Dδ) > 0. Choose ∆ = (∆Dk)1≤k≤K ∪ ∆Dδ with λ(∆) ≤ 2δ. As in theone-dimensional case we can choose some ξk ∈ ∆Dk and ξδ ∈ ∆Dδ such that

∑1≤k≤K

f (ξk)∆Vk + f (ξδ)Vol(∆Dδ) = ∞.

This contradiction shows that f must be bounded.

Exercise 13.1.24. Verify that Vol(∆Dδ) > 0 in the proof of Theorem 13.1.23.

Let f be a bounded function over a bounded, admissible, and closed domainD ⊂ Rn. Set

M := supD

f , m := infD

f .

If ∆ = (∆Dk)1≤k≤K is a partition of D, we define

∆Vk = Vol(∆Dk), Mk = sup∆Dk

f , mk = inf∆Dk

f ,

and the Darboux upper sum and Darboux lower sum

(13.1.30) S(∆) := ∑1≤k≤K

Mk∆Vk, S(∆) := ∑1≤k≤K

mk∆Vk,


respectively. We say that ∆ = (∆Dj)1≤j≤J is a refinement of ∆ = (∆Dk)1≤k≤K if forany j, we have ∆Dj ⊂ ∆Dk for some k. In this case we write ∆ ≺ ∆. The followingare some basic properties:

(a) For any partitions ∆1, ∆2, we have ∆3 ≺ ∆1 and ∆3 ≺ ∆2 for some parti-tion ∆3.

(b) S(∆) ≤ S(∆) ≤ S(∆) for any partition ∆.(c) If ∆1 ≺ ∆2, then

S(∆2) ≤ S(∆1) ≤ S(∆1) ≤ S(∆2).

Let

(13.1.31) I := inf(∆,≺)

S(∆), I := sup(∆,≺)

S(∆).

Theorem 13.1.25. (Darboux) We have

(13.1.32) limλ(∆)→0

S(∆) = I, limλ(∆)→0

S(∆) = I.

In particular, f is integrable if and only if I = I.

Introduce

(13.1.33) ωk := Mk −mk.

Theorem 13.1.26. Let f be a bounded function on a bounded, admissible, and closeddomain D ⊂ Rn. Then f is integrable on D if and only if for any ε > 0 we have

(13.1.34) ∑1≤k≤K

ωk∆Vk < ε

for some partition ∆ = (∆Dk)1≤k≤K.

Corollary 13.1.27. If D ⊂ Rn is a bounded, admissible, and closed domain, and f ∈C(D), then f is integrable on D.

PROOF. Since f is continuous over a compact subset D ⊂ Rn, it follows that fis uniformly continuous on D. For any ε > 0 there exists δ > 0 such that

ω(D) <ε

Vol(D)

whenever D ⊂ D with diam(D) < δ. Choose any partition ∆ = (∆Dk)1≤k≤K withλ(∆) < δ, we arrive at

∑1≤k≤K

ωk∆Vk <ε

Vol(D) ∑1≤k≤K

∆Vk ≤ ε.

Now the result follows from Theorem 13.1.26.

Example 13.1.28. Let (xk)k∈N = [0, 1] ∩Q and define f on D = [0, 1]× [0, 1] by

f (x, y) = 1

k , x = xk, 0 ≤ y ≤ 1,0, otherwise

Then f is integrable and ∫∫D

f (x, y)dxdy = 0.


For any given k ∈ N, we divide [0, 1] into k equality subinterval and then get apartition ∆ = (∆Di)1≤i≤k2 . For any ε > 0, there exists K > 0 such that 1/K < ε/2.Choose k > K with

2kK ≥ |∆Di ∩ (x, y) : x = x1, · · · , xk, 0 ≤ y ≤ 1| .Hence

∑1≤i≤k2

ωi1k2 ≤

2kKK2 +

1k ∑

1≤i≤k2

1k2 ≤

2Kk

+ε

2 ∑1≤i≤k2

1k2 =

2Kk

+ε

2< ε

for k > 4K/ε. Thus f is integrable on D, but is not continuous.For any partition ∆ = (∆Dk)1≤k≤K, there exists (ξk, ηk) ∈ ∆Dk such that f (ξk, ηk) =

0 (unless σ(∆Dk) = 0). Hence

∑1≤k≤K

f (ξk, ηk)∆σk = ∑1≤k≤K, σk>0

f (ξk, ηk)∆σk = 0

implying that the integral of f over D is zero.

Theorem 13.1.29. Let D ⊂ Rn be a bounded, admissible, closed domain. If f is boundedon D and f ∈ C(D \ (D1 ∪ · · · ∪ DK)), where Vol(Dk) = 0, 1 ≤ k ≤ K, then f isintegrable on D.

PROOF. Without loss of generality, we may assume that K = 1, D1 = Γ, andD = is a rectangle. Since Vol(Γ) = 0, it follows that for any ε > 0 there exists afamily AΓ = (′k)1≤k≤K′ such that

Γ ⊂⋃

1≤k≤K′′k, ∑

1≤k≤K′Vol(′k) < ε.

LetQ :=

⋃1≤k≤K′

′k.

Translating ∂Q along x1, · · · , xn-axes by δ1 > 0 so that

Q ⊂ Q′ := x ∈ Rn : d∞(x, ∂Q) < δ1 and Vol(Q′) < ε.

Similarly, we can choose δ2 > 0 so that

Q′ ⊂ Q′′ := x ∈ Rn : d∞(x, ∂Q′) < δ2 and Vol(Q′′) < ε.

Since \ Int(Q′) is compact and f ∈ C( \ Int(Q′)), it follows that f is uniformlycontinuous on \ Int(Q′). Consequently, there exists δ3 > 0 such that | f (x) −f (y)| < ε whenever x, y ∈ \ Int(Q′) and |x− y| < δ3. Let M := sup | f | < ∞and consider any partition (k)1≤k≤K of . Whene Vol(k) < δ := min(δ1, δ2, δ3),one has

∑1≤k≤K

(supk

f − infk

f

)Vol(k)

=

∑k∩Int(Q′)=∅

+ ∑k∩Int(Q′) 6=∅

(supk

f − infk

f

)Vol(k)

≤ ε ∑k∩Int(Q′)=∅

Vol(k) + 2M ∑k∩Int(Q′) 6=∅

Vol(k)

≤ εVol() + 2M ∑k⊂Q′′

Vol(k) ≤ εVol() + 2Mε.


Hence the function f is integrable on .

13.2. Basic properties

Let D ⊂ Rn be a bounded, admissible, and closed domain. Moreover weconsider the set D of all bounded,m admissible, and closed domain in Rn.

Theorem 13.2.1. (Elementary properties) (1) If f , g are integrable on D and α, β ∈ R,then α f + βg is integrable on D and

(13.2.1)∫

D(α f + βg)dV = α

∫D

f dV + β infD

gdV.

(2) If f is integrable on D, then | f | is integrable on D and

(13.2.2)∣∣∣∣∫D

f dV∣∣∣∣ ≤ ∫D

| f |dV.

(3) Suppose D1, D2 ∈ D with Int(D1) ∩ Int(D2) = ∅ and D1 ∪ D2 ∈ D . Then fis integrable on D1 ∪ D2 if and only if f |D1 and f |D2 are integrable. In this case

(13.2.3)∫

D1∪D2

f dV =∫

D1

f dV +∫

D2

f dV.

(4) If f , g are integrable on D ∈ D and f ≤ g on D, then

(13.2.4)∫

Df dV ≤

∫D

gdV.

(4) If f , g are integrable on D ∈ D , then f g is integrable on D.(5) If f , g are integrable on D ∈ D , and g ≥ 0 on D or g ≤ 0 on D, then

(13.2.5)∫

Df gdV = µ

∫D

gdV

for some µ ∈ [m f , M f ], where m f := minD f and M f := maxD f . In particular, whenf ∈ C(D), we have

(13.2.6)∫

Df gdV = f (ξ)

∫D

gdV

for some ξ ∈ D.

13.2.1. Iterated integrals. A basic tool in multiple integrals is the famous Fu-bini’s theorem.

Theorem 13.2.2. (Fubini) Let f is integrable on D = [a, b]× [c, d].(1) If for any x ∈ [a, b] the integral

I(x) :=∫ d

cf (x, y)dy

exists, then I is integrable on [a, b] and

(13.2.7)∫∫

Df (x, y)dxdy =

∫ b

aI(x)dx =

∫ b

a

[∫ d

cf (x, y)dy

]dx.

(2) If for any y ∈ [c, d] the integrable

J(y) :=∫ b

af (x, y)dx

13.2. BASIC PROPERTIES 41

exists, then J is integrable on [c, d] and

(13.2.8)∫∫

Df (x, y)dxdy =

∫ d

cJ(y)dy =

∫ d

c

[∫ b

af (x, y)dx

]dy.

When f ≡ 1, Theorem 13.2.2 is the classical formula for the area of rectangles.The above theorem is a special case of Fubini’s theorem in measure theory.

PROOF. Consider the following partitions

a = x0 < x1 < · · · < xn = b, c = y0 < y1 < · · · < ym = d

with ∆xi := xi − xi−1, ∆yj := yj − yj−1, and

∆Dij := [xi−1, xi]× [yj−1, yj], ∆ = (∆Dij)1≤i≤n,1≤j≤m, mij := infDij

f , Mij := supDij

f .

We shall prove that

limλx→0

∑1≤i≤n

I(ξi)∆xi =∫∫

Df (x, y)dxdy

where λx := max1≤i≤n ∆xi and any ξi ∈ [xi−1, xi]. Since

∑1≤j≤m

mij∆yj ≤ ∑1≤j≤m

∫ yj

yj−1

f (ξi, y)dy =∫ d

cf (ξi, y)dy = I(ξi) ≤ ∑

1≤j≤mMij∆yj,

it follows that

∑1≤i≤n, 1≤j≤m

mij∆xi∆yj ≤ ∑1≤i≤n

I(ξi)∆xi ≤ ∑1≤i≤n, 1≤j≤m

Mij∆xi∆yj.

Because f is integrable on D, letting λ(∆)→ 0 (which is equivalent to both λx andλy → 0) yields (13.2.7).

Remark 13.2.3. We can find an integrable function f over [a, b]× [c, d], but both iteratedintegrals ∫ b

a

[∫ d

cf (x, y)dy

]dx and

∫ d

c

[∫ b

af (x, y)dx

]dy

do not exist. For example,

f (x, y) =

1px

+ 1py

, x = qxpx

, y =qypy∈ Q,

0, otherwise,(x, y) ∈ [0, 1]× [0, 1].

and

f (x, y) = 1

k , x = 1k (k ∈ N), y ∈ Q,

0, otherwise,(x, y) ∈ [0, 1]× [0, 1].

Exercise 13.2.4. Verify the statement in Remark 13.2.3.

Exercise 13.2.5. Compute

I =∫∫

Dx(x− y)2 dxdy, D = [0, 1]× [0, 1].


Theorem 13.2.6. Let f (x1, · · · , xn) be integrable on D = [a1, b1]× · · · × [an, bn]. If

I(x1) :=∫

D∗f (x1, x2, · · · , xn)dx2 · · · dxn

exists for any x1 ∈ [a1, b1], then I is integrable on [a1, b1] and

(13.2.9)∫

Df (x1, · · · , xn)dx1 · · · dxn =

∫ b1

a1

I(x1)dx1

where D∗ = [a2, b2 × · · · × [an, bn] ⊂ Rn−1.

13.2.2. Computations. Let us first consider a X-type domain given by

(13.2.10) D =(x, y) ∈ R2 : a ≤ x ≤ b, ϕ1(x) ≤ y ≤ ϕ2(x)

where

(13.2.11) ϕ1, ϕ2 ∈ C[a, b], ϕ1(x) ≤ ϕ2(x) for any x ∈ (a, b).

Similarly, a Y-type domain is given by

(13.2.12) D =(x, y) ∈ R2 : c ≤ y ≤ d, ψ1(y) ≤ x ≤ ψ2(y)

where

(13.2.13) ψ1, ψ2 ∈ C[c, d], ψ1(y) ≤ ψ2(y) for any y ∈ (c, d).

Theorem 13.2.7. (1) Let D be a X-type domain and f integrable on D. If for any x ∈[a, b], the integral ∫ ϕ2(x)

ϕ1(x)f (x, y)dy

exists, then the iterated integral∫ b

a

[∫ ϕ2(x)

ϕ1(x)f (x, y)dy

]dx

exists and

(13.2.14)∫∫

Df (x, y)dxdy =

∫ b

a

[∫ ϕ2(x)

ϕ1(x)f (x, y)dy

]dx.

(2) Let D be a Y-type domain and f integrable on D. If for any y ∈ [c, d], the integral∫ ψ2(y)

ψ1(y)f (x, y)dx

exists, then the iterated integral∫ d

c

[∫ ψ2(y)

ψ1(y)f (x, y)dx

]dy

exists and

(13.2.15)∫∫

Df (x, y)dxdy =

∫ d

c

[∫ ψ2(y)

ψ1(y)f (x, y)dx

]dy.


PROOF. We only prove (1). Let c = mina≤x≤b ϕ1(x) and d = maxa≤x≤b ϕ2(x).Then D ⊂ [a, b]× [c, d] = D. Define the integrable over D by

f (x, y) :=

f (x, y), (x, y) ∈ D,0, (x, y) ∈ D \ D.

For any x ∈ [a, b], one has∫ d

cf (x, y)du =

(∫ ϕ1(x)

c+∫ ϕ2(x)

ϕ1(x)+∫ d

ϕ2(x)

)f (x, y)dy =

∫ ϕ2(x)

ϕ1(x)f (x, y)dy

which is integrable. According to Theorem 13.2.2, we arrive at∫ b

c

[∫ ϕ2(x)

ϕ1(x)f (x, y)dy

]dx =

∫ b

a

[∫ d

cf (x, y)dy

]dx

=∫∫

Df (x, y)dxdy =

∫∫D

f (x, y)dxdy.

The same argument can be applied to prove (2).

Example 13.2.8. Compute

I =∫ 1

0

(∫ √x

x

sin yy

dy

)dx.

By Theorem 13.2.7 we have

I =∫ 1

0

(∫ y

y2

sin yy

dx)

dy =∫ 1

0(1− y) sin ydy.

Define

Ik :=∫ 1

0yk sin ydy, k = 0, 1, · · · .

By integrations by parts, we have

Ik =∫ 1

0−yk d cos y = −

(yk cos y

∣∣∣∣10−∫ 1

0kyk−1 cos ydy

)

= − cos 1 + k∫ 1

0yk−1 d sin y

= − cos 1 + k

(yk−1 sin y

∣∣∣∣10−∫ 1

0(k− 1)yk−2 sin ydy

)= − cos 1 + k sin 1− k(k− 1)Ik−2

for k ≥ 2. We also have

I0 = 1− cos 1, I1 = sin 1− cos 1.

Hence I = I0 − I1 = 1− sin 1.

Theorem 13.2.9. Let f is continuous on D ⊂ R3 with D ∈ D .(1) If

D =(x, y, z) ∈ R3 : a ≤ x ≤ b, y1(x) ≤ y ≤ y2(x), z1(x, y) ≤ z ≤ z2(x, y)


where y1, y2, z1, z2 are continuous, then

(13.2.16)∫∫∫

Df (x, y, z)dxdydz =

∫ b

adx∫ y2(x)

y1(x)dy∫ z2(x,y)

z1(x,y)f (x, y, z)dz.

(2) If

D =(x, y, z) ∈ R3 : (x, y) ∈ Ω, ϕ(x, y) ≤ z ≤ ψ(x, y)

where Ω ⊂ R2 with D ∈ D , and ϕ, ψ are continuous, then

(13.2.17)∫∫∫


∫∫Ω

dxdy∫ ψ(x,y)

ϕ(x,y)f (x, y, z)dz.

(3) Suppose that ∂D consists of finitely many piecewise smooth surfaces, and for anypoint (x, y, z) ∈ D we have a ≤ z ≤ b and Dz ⊂ R2 with Dz ∈ D . Then

(13.2.18)∫∫∫


∫ b

adz∫∫

Dzf (x, y, z)dxdy.


I =∫∫∫

D(x + y + z)2 dxdydz

where

D =

(x, y, z) ∈ R3 :

x2

a2 +y2

b2 +z2

c2 ≤ 1, a, b, c > 0

.

By symmetry

I =∫∫∫

D

(x2 + y2 + z2 + 2xy + 2yz + 2zx

)dxdydz

=∫∫∫

D

(x2 + y2 + z2

)dxdydz.

For any (x, y, z) ∈ D, we have

−c ≤ z ≤ c, Dz =

x2

a2(1− z2

c2 )+

y2

b2(1− z2

c2 )≤ 1

so that ∫∫∫

Dz2 dxdydz =

∫ c

−c

∫∫Dz

z2 dxdy =∫ c

−cz2Area(Dz)dz

=∫ c

−cz2

(π · a

√1− z2

c2 · b√

1− z2

c2

)dz

=∫ c

−cπabz2

(2− z2

c2

)dz =

415

πabc3.

Hence I = 415 πab(a2 + b2 + c2).

13.3. Stieltjes integrals

In this section we introduce the functions of bounded variation and Stieltjesintegrals.

13.3. STIELTJES INTEGRALS 45

13.3.1. Functions of bounded variation. Let f (x) be a function defined on aclosed interval [a, b]. For any partition ∆ = (xi)0≤i≤n with a = −x0 < x1 < · · · <xn = b consider

(13.3.1)∨∆

f := ∑0≤i≤n−1

| f (xi+1)− f (xi)|.

We say f is a function of bounded variation, or f ∈ BV[a, b], if

sup∆

∨∆

f < ∞.

In this case we define

(13.3.2)b∨a

f := sup∆

∨∆

f .

Let f (x) be a function defined on [a, ∞). We say f ∈ BV[a, ∞), if (∨ba f )b>a is

bounded. In this case we define

(13.3.3)∞∨a

f := supb>a

b∨a

f .

Remark 13.3.1. (1) C[a, b] ; BV[a, b]. Consider the function

f (x) =

x cos π2x , 0 < x ≤ 1,

0, x = 0.

It is clear that f ∈ C[0, 1]. Take ∆ = (xi)0≤i≤2n by

x0 = 0, xi =1

2n + 1− i1 ≤ i ≤ 2n.

Then∨∆

f = ∑0≤i≤2n−1

| f (xi+1)− f (xi)|

= ∑1≤i≤2n−1

∣∣∣∣xi+1 cosπ

2xi+1− xi cos

π

2xi

∣∣∣∣+ ∣∣∣∣ 12n

cos(nπ)

∣∣∣∣=

12n

+ ∑1≤i≤2n−1

∣∣∣∣ 12n− i

cos(2n− i)π

2− 1

2n + 1− icos

(2n + 1− i)π2

∣∣∣∣=

12n

+ ∑1≤i≤2n−1

∣∣∣∣ 12n− i

cos(

nπ − iπ2

)− 1

2n + 1− icos

(nπ +

π

2− iπ

2

)∣∣∣∣=

12n

+ ∑1≤i≤2n−1

∣∣∣∣ 12n− i

cosiπ2− 1

2n + 1− isin

iπ2

∣∣∣∣=

12n

+ ∑1≤k≤n−1

∣∣∣∣ 12n− 2k

cos(kπ)− 12n + 1− 2k

sin(kπ)

∣∣∣∣+ ∑

1≤k≤n

∣∣∣∣ 12n− 2k + 1

cos(

kπ − π

2

)− 1

2n− 2k + 2sin(

kπ − π

2

)∣∣∣∣=

12n

+ ∑1≤k≤n−1

12n− 2k

+ ∑1≤k≤n

12n− 2k + 2


so that ∨∆

f = 1 +12+ · · ·+ 1

n→ ∞.

Hence f /∈ BV[0, 1].

(2) BV[a, b] ; C[a, b]. Consider

f (x) =

0, −1 ≤ x < 0,1, 0 ≤ x ≤ 1.

Then f /∈ C[0, 1] but ∨10 f = 1.

(3) If f is bounded and monotone on [a, b] or [a, ∞), then f ∈ BV[a, b] or BV[a,+∞).Because

b∨a

f = | f (b)− f (a)| or∞∨a

f = | f (∞)− f (a)|.

(4) Lip[a, b] ⇒ BV[a, b]. Here f ∈ Lip[a, b] means that | f (x)− f (y)| ≤ L|x− y|for some L > 0 and any x, y ∈ [a, b]. Since∨

∆

f = ∑0≤i≤n−1

| f (xi+1)− f (xi)| ≤ ∑0≤i≤n−1

L|xi+1 − xi| = L(b− a),

it follows that ∨ba f ≤ L(b− a).

(5) If f ′ is bounded on [a, b], then f ∈ BV[a, b]. But the converse is not true, forexample

f (x) =

x2 sin πx , x ∈ (0, 1],

0, x = 0.

(6) If for any x ∈ [a, b],

f (x) = c +∫ x

aϕ(t)dt

where |ϕ| is integrable, then f ∈ BV[a, b] and

b∨a

f ≤∫ b

a|ϕ(t)|dt.

Indeed,∨∆

f = ∑0≤i≤n−1

| f (xi+1)− f (xi)| = ∑0≤i≤n+1

∣∣∣∣∫ xi+1

xi

ϕ(t)dt∣∣∣∣ ≤ ∫ b

a‖ϕ(t)|dt.

Exercise 13.3.2. Let

f (x) =

x2 sin πx2 , 0 < x ≤ 2,

0, x = 0.

Show that

f (x) =∫ x

0ϕ(t)dt

for some non-integrable |ϕ|, and f /∈ BV[0, 2].


Theorem 13.3.3. Let a, b ∈ R with a < b.(1) BV[a, b] ⊂ B[a, b].(2) If f , g ∈ BV[a, b], then f ± g ∈ BV[a, b].(3) If f , g ∈ BV[a, b] and |g| ≥ σ > 0, then f /g ∈ BV[a, b].(4) Given c ∈ (a, b). Then f ∈ BV[a, b] if and only if f ∈ BV[a, c] and f ∈ BV[c, b].

In this case,b∨a

f =c∨a

f +b∨c

f .

(5) If f ∈ BV[a, b], then g(x) := ∨xa f is increasing and bounded.

(6) f ∈ BV[a, b] if and only if there exists a bounded and increasing function F on[a, b] such that

| f (x)− f (y)| ≤ F(x)− F(y)

whenever y < x.(7) f ∈ BV[a, b] if and only if f = g− h for some bounded and increasing functions

f and g.(8) If f ∈ BV[a, b] and f is continuous at x0 ∈ [a, b], then ∨x

a is continuous at x0.(9) If f ∈ BV[a, b] ∩ C[a, b], then f = g − h for some continuous and increasing

functions f and g.(10) If f ∈ BV[a, b] ∩ C[a, b], then for any partition ∆ = (xi)0≤i≤n of [a, b] we have

b∨a

f = limλ(∆)→0

∑0≤i≤n−1

| f (xi+1)− f (xi)|.

PROOF. (1)–(3) and (10) are obvious.(4) Let f ∈ BV[a, b] and c ∈ (a, b). Consider two partitions ∆y = (yi)0≤i≤m

with a = y0 < y1 < · · · < ym = c and ∆z = (zj)0≤j≤n with c = z0 < z1 < · · · <zn = b. Then ∆ = ∆y ∪ ∆z forms a partition of [a, b]. Then∨

∆

f = ∑0≤i≤m−1

| f (yi+1)− f (yi)|+ ∑0≤j≤n−1

| f (yj+1)− f (yj)| =∨∆y

f +∨∆z

f

so that ∨ca f + ∨b

c f ≤ ∨ba f .

Conversely, suppose f ∈ BV[a, c] and f ∈ BV[c, b]. Consider a partition ∆ =(xi)0≤i≤n with a = x0 < x1 < · · · < xn = b. If c = xi0 for some i0, then

∨∆

f = ∑0≤i≤i0−1

| f (xi+1)− f (xi)|+ ∑i0≤i≤n−1

| f (xi+1)− f (xi)| ≤c∨a

f +b∨c

f .

If c 6= xi for any 0 ≤ i ≤ n, we define ∆ := ∆ ∪ c. Then∨∆

f = ∑0≤i≤i0−1

| f (xi+1)− f (xi)|+ | f (c)− f (xi0)|

+ | f (xi0+1)− f (c)|+ ∑i0+1≤i≤n−1

| f (xi+1)− f (xi)|

≤c∨a

f +b∨c

f .

Hence ∨ba f = ∨c

a f + ∨bc f .


(5) For any a ≤ x < y ≤ b, we have

y∨a

f =x∨a

f +y∨x

f

so that g(y)− g(x) = ∨yx f ≥ 0. Thus g(x) ≤ ∨b

a f < ∞.(6) If f ∈ BV[a, b], we take F(x) := ∨x

a f . By (5), F is increasing, bounded, and

F(y)− F(x) =y∨x

f ≥ | f (y)− f (x)|

for any x < y. Conversely, for any partition ∆ = (xi)0≤i≤n with a = x0 < x1 <· · · < xn = b, we have∨

∆

f = ∑0≤i≤n−1

| f (xi+1)− f (xi)|

≤ ∑0≤i≤n−1

[F(xi+1)− F(xi)] = F(b)− F(a) < ∞.

(7) If f ∈ BV[a, b], we take g(x) := ∨xa f . By (5), the function g is increasing

and bounded. Define h := g− f . For any x < y, one has

h(y)− h(x) = [g(y)− g(x)]− [ f (y)− f (x)] ≥ [g(y)− g(x)]−y∨x

f ≥ 0.

If f = g− h with g, h being increasing and bounded, then F := g + h is increasingand

| f (x)− f (y)| ≤ [g(x)− g(y)] + [h(x)− h(y)] = F(x)− F(y)

whenever y < x.(8) Without loss of generality, we may assume that a < x0 < b. For any ε > 0

there exists a partition ∆ = (xi)0≤i≤n with x0 < x1 < · · · < xn = b such that

∑0≤i≤n−1

| f (xi+1)− f (xi)| =∨∆

f ≥b∨x0

f − ε.

In particular, | f (x1)− f (x0)| < ε. Hence

b∨x0

< ε + ∑0≤i≤n−1

| f (xi+1)− f (xi)|

< 2ε + ∑1≤i≤n−1

| f (xi+1)− f (xi)| ≤ 2ε +b∨x1

f

and then g(x1) − g(x0) < 2ε for any x1 x0. Thus limε→0 g(x0 + ε) = g(x0).Similarly, g(x0 − 0) = g(x0).

(9) If f ∈ BV[a, b] ∩ C[a, b], then by (7), f = g− h with g, h being increasingand bounded. By (8), g ∈ C[a, b] so that h ∈ C[a, b].


13.3.2. Stieltjes integrals. Suppose f , g are bounded over [a, b]. Consider apartition ∆ = (xi)0≤i≤n with ∆xi = xi − xi−1 and λ = max1≤i≤n ∆xi. For anyξi ∈ [xi−1, xi], define the Stieltjes sum by

(13.3.4) S := ∑1≤i≤n

f (ξi)∆g(xi), ∆g(xi) := g(xi)− g(xi−1).

The Stieltjes integral of f with respect to g is defined to be

(13.3.5) (S)∫ b

af (x)dg(x) =

∫ b

af (x)dg(x) := lim

λ→0S = lim

λ→0∑

1≤i≤nf (ξi)∆g(xi).

When g(x) = x, we have

(S)∫ b

af (x)dg(x) = (R)

∫ b

af (x)dx.

If g is increasing, then

(S)∫ b

af (x)dg(x) exists⇐⇒ lim

λ→0∑

1≤i≤nωi∆g(xi) = 0

where ωi = Mi −mi, Mi = sup[xi−1,xi ]f , and mi = inf[xi−1,xi ]

f .

Theorem 13.3.4. If f ∈ C[a, b] and g ∈ BV[a, b], then

(S)∫ b

af (x)dg(x)

exists.

PROOF. We first assume that g is increasing. For any ε > 0, there exists δ > 0such that | f (x) − f (y)| < ε whenever |x − y| < δ. Consider any partition ∆ =(xi)0≤i≤n with a = x0 < x1 < · · · < xn = b with λ < δ. Then

∑1≤i≤n

ωi∆g(xi) ≤ ε ∑1≤i≤n

∆g(xi) = [g(b)− g(a)]ε.

Thus the Stieltjes integral exists.In general, g = g1 − g2 with g1, g2 being increasing and bounded. Then the

Stieltjes integral also exists.

Corollary 13.3.5. (1) If f is integrable over [a, b] and g ∈ Lip[a, b], then

(S)∫ b

af (x)dg(x)

exists.(2) If f is integrable over [a, b] and

g(x) = c +∫ x

aϕ(t)dt

with |ϕ| ≤ L, then

(S)∫ b

af (x)dg(x)

exists.


Theorem 13.3.6. The following statements hold, where the Stieltjes integrals exist:∫ b

adg(x) = g(b)− g(a),∫ b

a[ f1(x)± f2(x)]dg(x) =

∫ b

af1(x)dg(x)±

∫ b

af2(x)dg(x),∫ b

af (x)d[g1(x)± g2(x)] =

∫ b

af (x)dg1(x)±

∫ b

af (x)dg2(x),∫ b

ak f (x)d[`g(x)] = k`

∫ b

af (x)dg(x),∫ b

af (x)dg(x) =

∫ c

af (x)dg(x) +

∫ b

cf (x)dg(x).

Theorem 13.3.7. (Integral by parts) If f , g are bounded over [a, b] and the Stieltjesintegral

(S)∫ b

af (x)dg(x)

exists, then

(13.3.6)∫ b

af (x)dg(x) = f (x)g(x)

∣∣∣∣ba−∫ b

ag(x)d f (x)

exists.

PROOF. Consider a partition ∆ = (xi)0≤i≤n with a = x0 < x1 < · · · < xn = band choose ξi ∈ [xi−1, xi]. Then

S = ∑1≤i≤n

f (ξi)∆g(xi) = ∑1≤i≤n

f (ξi)[g(xi)− g(xi−1)]

= f (ξn)g(b)− f (ξ1)g(a)− ∑1≤i≤n−1

g(xi)[ f (ξi+1)− f (ξi)]

so that

S− f (x)g(x)∣∣∣∣ba

= g(b)[ f (ξn)− f (b)]− g(a)[ f (ξ1)− f (a)]

− ∑1≤i≤n−1

g(xi)[ f (ξi+1)− f (ξi)].

Hence we obtain (13.3.6).

Let us consider the function

ρ(x) =

0, x ≤ 0,1, x > 0.

Note that 0 is the first discontinuity point. Let c ∈ [a, b), f is continuous at c, andf ∈ BV[a, b]. Define

I = (S)∫ b

af (x)dρc(x)

where

ρc(x) := ρ(x− c) =

0, x ≤ c,1, x > c.


Then I exists and

S = ∑1≤n

f (ξi)∆ρc(xi) =

(∑i 6=k

+ ∑i=k

)f (ξi)[ρc(xi)− ρc(xi−1)]

= f (ξk)[ρc(xk)− ρc(xk−1)] = f (ξk)

where ∆ = (xi)0≤i≤n is a partition of [a, b] with a = x0 < x1 < · · · < xn = b, andxk−1 < c < xk. Letting λ→ 0 yields

(S)∫ b

af (x)dρc(x) = f (c), a ≤ c < b.

Similarly

(S)∫ b

af (x)dρ(c− x) = − f (c), a < c ≤ b.

Theorem 13.3.8. Suppose that f ∈ C[a, b], g ∈ C([a, b] \ c0, c1, · · · , cm), wherea = c0 < c1 < · · · < cm = b are first discontinuity points, g′(x) exists (except possiblyfor finitely many points), and |g′| is integrable over [a, b]. Then∫ b

af (x)dg(x) =

∫ b

af (x)g′(x)dx + f (a)[g(a + 0)− g(a)]

+ ∑1≤k≤m−1

f (ck)[g(ck + 0)− g(ck − 0)] + f (b)[g(b)− g(b− 0)].


I =∫ 2

−2xdg(x), g(x) =

x + 2, −2 ≤ x ≤ −1,

2, −1 < x < 1,x2 − 1, 1 ≤ x ≤ 2.

According to Theorem 13.3.8 we have

I =∫ 2

−2xg′(x)dx + (−1)(2− 1) + 1(0− 2)

=∫ −1

−2xdx +

∫ 2

12x2 dx− 1− 2 =

12

x2∣∣∣∣−1

−2+

23

x3∣∣∣∣21− 3 =

16

.

Theorem 13.3.10. (1) If f is bounded over [a, b], g is increasing, and the Stieltjes integral

(S)∫ b

af (x)dg(x)

exists, then ∫ b

af (x)dg(x) = µ

∫ b

adg(x) = µ[g(b)]− g(a)]

where µ ∈ [m f , M f ]. When f ∈ C[a, b], we can take µ = f (ξ) for some ξ ∈ [a, b].(2) If f ∈ C[a, b] and g ∈ BV[a, b], then∣∣∣∣∫ b

af (x)dg(x)

∣∣∣∣ ≤ Mb∨a

g

where M = supa≤x≤b | f (x)|.(3) If ( fn)n∈N ⊂ C[a, b], fn ⇒ f , and g ∈ BV[a, b], then

limn→∞

∫ b

afn(x)dg(x) =

∫ b

af (x)dg(x).


(4) If f ∈ C[a, b], (gn)n∈N ⊂ BV[a, b] with gn → g and ∨bagn ≤ V for any n ∈ N,

then g ∈ BV[a, b] and

limn→∞

∫ b

af (x)dgn(x) =

∫ b

af (x)dg(x).

PROOF. (1) LetS = ∑

1≤i≤nf (ξi)[g(xi)− g(xi−1)]

for a partition ∆ = (xi)0≤i≤n with a = x0 < x1 < · · · < xn = b of [a, b] andξi ∈ [xi−1, xi]. Since g is increasing, it follows that

m f ≤1

g(b)− g(a)(S)

∫ b

af (x)dg(x) ≤ M f .

(2) One has|S| ≤ M ∑

1≤i≤n|g(xi)− g(xi−1)| ≤ M

∨∆

f .

(3) For any ε > 0, there exists N ∈ N such that

maxa≤x≤b

| fn(x)− f (x)| < ε

whenever n > N. Hence∣∣∣∣∫ b

afn(x)dg(x)−

∫ b

af (x)dg(x)

∣∣∣∣ ≤ maxa≤x≤b

| fn(x)− f (x)|b∨a

g ≤ εb∨a

g

for any n > N.(4) Consider any partition ∆ = (xi)0≤i≤m of [a, b]. Then

∨∆

gn = ∑1≤i≤m

|gn(xi)− gn(xi−1)| ≤b∨a

gn ≤ V.

Letting n→ ∞ yields∑

1≤i≤m|g(xi)− g(xi−1)| ≤ V

so that g ∈ BV[a, b]. Consider

S := ∑1≤i≤m

f (ξi)∆g(xi), Sn = ∑1≤i≤m

f (ξi)∆gn(xi).

For any ε > 0 there exists δ > 0 so that | f (x)− f (y)| < ε whenever |x − y| < δ.Hence ∣∣∣∣Sn −

∫ b

af (x)dgn(x)

∣∣∣∣ =

∣∣∣∣∣Sn − ∑1≤i≤m

∫ xi

xi−1

f (x)dgn(x)

∣∣∣∣∣≤ ∑

1≤i≤m

∫ xi

xi−1

| f (ξi)− f (x)|dgn(x)

≤ ε ∑1≤i≤m

∫ xi

xi−1

dgn(x) ≤ εb∨a

gn ≤ εV

where ∆ = (xi)0≤i≤m is a partition with max1≤i≤m ∆xi < δ. Similarly we have∣∣∣∣S− ∫ b

af (x)dg(x)

∣∣∣∣ ≤ εV.


We can find an integer N ∈ N such that |Sn− S| < ε for any n ≥ N. Consequently,∣∣∣∣∫ b

af (x)dgn(x)−

∫ b

af (x)dg(x)

∣∣∣∣≤

∣∣∣∣∫ b

af (x)dgn(x)− Sn

∣∣∣∣+ |Sn − S|+∣∣∣∣∫ b

af (x)dg(x)− S

∣∣∣∣≤ εV + ε + εV = (1 + 2V)ε.

Thus we get the desired result.

13.3.3. Applications. The first application is the connection between Stieltjesintegrals and Riemann integrals.

Theorem 13.3.11. (Catalan) Suppose f , g ∈ C(D), D ⊂ R2 is a bounded, admissibleand closed domain. Let m = minD g, M = maxD g, ϕ ∈ C[m, M], and

ψ(u) :=∫∫

(x,y)∈D, m≤g(x,y)≤uf (x, y)dxdy.

Then

(13.3.7)∫∫

Df (x, y)ϕ(g(x, y))dxdy =

∫ M

mϕ(u)dψ(u).

PROOF. Without loss of generality, we may assume that f > 0 on D. Choosea partition ∆ = (∆ui)0≤i≤m with m = u0 < u1 < · · · < un−1 < un = M. Then∫∫

Df (x, y)ϕ(g(x, y))dxdy = ∑

1≤i≤n

∫∫ui−1≤g≤ui

f (x, y)ϕ(g(x, y))dxdy

= ∑1≤i≤n

ϕ (g(ξ∗i , η∗i ))∫∫

ui−1≤g≤ui

f (x, y)dxdy

= ∑1≤i≤n

ϕ (g(ξ∗i , η∗i )) [ψ(ui)− ψ(ui−1)]

→∫ M

mϕ(u)dψ(u)

as λ→ 0.

By bxc we mean the largest integer smaller or equal to x. Tied to this functionis the fractional part of x defined by

〈x〉 := x− bxc.Given two functions f and g defined on [a, ∞), where a ≥ 0 and g(x) > 0. We

shall writef (x) = O(g(x)) or f (x) g(x),

if there exist two constants M > 0 and x0 ≥ a such that | f (x)| ≤ Mg(x) for allx ≥ x0. We shall write

f (x) = o(g(x))if for each ε > 0, there exists a constant x0 := x0(ε) such that | f (x)| ≤ εg(x) forall x ≥ x0.

Given two functions f and g defined on [a, ∞), where a ≥ 0, we write

f (x) ∼ g(x)


to mean that limx→∞ f (x)/g(x) = 1.

For any set A ⊂ N and x > 1, we define

(13.3.8) A(x) := #a ≤ x : a ∈ A.Compute∫ x

1f (t)dA(t) = ∑

n≤x

∫ n+0

n−0f (t)dA(t) = ∑

n∈x,n∈A

∫ n+0

n−0f (t)dA(t) = ∑

n∈x,n∈Af (n).

Proposition 13.3.12. (1) If f ∈ C[1, x], then

(13.3.9) ∑a≤x,a∈A

f (x) =∫ x

1f (t)dA(t).

(2) (Abel summation formula) Let(an)n∈N be a sequence of complex numbers andlet f : [1,+∞)→ C. For each real number x ≥ 1, let

A(x) = ∑n≤x

an

and assume that f (x) has a continuous derivative for x ≥ 1. Then

(13.3.10) ∑n≤x

an f (n) = A(x) f (x)−∫ x

1A(t) f ′(t)dt.

PROOF. First assume that x = N an integer. Then

∑n≤x

an f (n) = A(1) f (1) + ∑2≤n≤N

[A(n)− A(n− 1)] f (n)

= A(1) f (1) + ∑2≤n≤N

A(n) f (n)− ∑1≤n≤N−1

A(n) f (n + 1)

= ∑1≤n≤N−1

A(n)[ f (n)− f (n + 1)] + A(N) f (N)

= A(N) f (N)− ∑1≤i≤N−1

A(i)∫ i+1

if ′(t)dt

= A(N) f (N)− ∑1≤i≤N−1

∫ i+1

iA(t) f ′(t)dt

= A(N) f (N)−∫ N

1A(t) f ′(t)dt.

Let us now assume that x is not an integer. Set N := bxc. Since A9t) is constanton the interval [N, x], it follows that

A(x) f (x)−∫ x

1A(t) f ′(t)dt = A(x) f (x)−

∫ x

NA(t) f ′(t)dt−

∫ N

1A(t) f ′(t)dt

= A(x) f (x)− A(N)∫ x

Nf ′(t)dt−

∫ N

1A(t) f ′(t)dt

= A(N) f (N)−∫ N

1A(t) f ′(t)dt = ∑

n≤Nan f (n).

Because ∑n≤x an f (n) = ∑n≤N an f (n) the proof is complete.


The Euler’s constant γ is defined by

(13.3.11) γ := 1−∫ ∞

1

t− btct2 dt = 0.57721 · · · .

A long-standing conjecture is that Euler’s constant γ is irrational and moreovertranscendental. We shall show that this definition of γ coincides with the usualone, i.e.,

(13.3.12) γ = limN→∞

(∑

1≤n≤N

1n− ln N

).

Theorem 13.3.13. For all x ≥ 1,

(13.3.13) ∑n≤x

1n= ln x + γ + O

(1x

).

PROOF. Setting an ≡ 1 and f (t) = 1/t in Proposition 13.3.12, we obtainA(x) = ∑n≤x 1 = bxc and

∑n≤x

1n

=bxc

x+∫ x

1

btct2 dt = 1− x− bxc

x+∫ x

1

dtt−∫ x

1

t− btct2 dt

= 1 + O(

1x

)+ ln x−

∫ ∞

0

t− btct2 dt +

∫ ∞

x

t− btct2 dt

= ln x + γ + O(

1x

)+∫ ∞

x

t− bxct2 dt

= ln x + γ + O(

1x

)+ O

(∫ ∞

x

dtt2

)= ln x + γ + O

(1x

)as x → ∞.

Let f : N → C be a function for which there exists a positive constant A suchthat limx→∞

1x ∑n≤x f (n) = A. By (13.3.10) we have

∑n≤x

f (n) ln n = F(x)x ln x−∫ x

1F(t)dt

where F(x) := 1x ∑n≤x f (n). Since

limx→∞

1Ax

∫ x

1F(t)dt = lim

x→∞

F(x)A

= 1

it follows that

(13.3.14) ∑n≤x

f (n) ln n = A(1+ o(1))x ln x+ Ax ln x1 + o(1)

ln x= A(1+ o(1))x ln x

as x → ∞.

Proposition 13.3.14. Let a, b ∈ N with a < b and let f : [a, b] → R be a monotonicfunction on [a, b]. There exists a real number θ := θ(a, b) ∈ [0, 1] such that

(13.3.15) ∑a<n≤b

f (n) =∫ b

af (t)dt + θ[ f (b)− f (a)].


PROOF. Consider the function g(t) := btc ∈ BV[a, b]. Then

∑a<n≤b

f (n)−∫ b

af (t)dt =

∫ b

af (t)dbtc −

∫ b

af (t)dt = −

∫ b

af (t)d〈t〉

= [− f (t)〈t〉]∣∣∣∣ba+∫ b

a〈t〉d f (t) =

∫ b

a〈t〉d f (t).

Without loss of generality, we may assume that f is increasing. Then we can finda real number θ ∈ [0, 1] such that∫ b

a〈t〉d f (t) = θ

∫ b

ad f (t) = θ[ f (b)− f (a)]

because 〈t〉 ∈ [0, 1].

As the direct consequence we have

ln n! = (n− 1) ln n + O(ln n),

∑n≤x

nα =xα+1

α + 1+ O(xα), α ≥ 0,

∑n≤x

nα ln n =xα+1

α + 1

(ln x− 1

α + 1

)+ O (xα ln x) , α > 0.

One can also obtain a more accurate asymptotic expression in the case wherethe function is decreasing.

Proposition 13.3.15. If f : [1, ∞)→ R is continuous, decreasing and such that limx→∞ f (x) =0, then there exists a constant A such that

(13.3.16) ∑1≤n≤x

f (n) =∫ x

1f (t)dt + A + O( f (x)).

PROOF. Given a positive integer N, consider the expression

D(N) :=∫ N

1f (t)dt− ∑

2≤n≤Nf (n) = ∑

2≤n≤N

[∫ n

n−1f (t)dt− f (n)

].

Since D(N) ≥ 0, to prove the convergence of the sequence (D(N))N∈N, it sufficesto show that

R(N) := ∑n≥N+1

[∫ n


]= O( f (N)).

For each pair of positive integers M and N with M ≥ N + 3 we have

∑N+1≤n≤M−1

f (n) + f (M) ≤∫ M

Nf (t)dt ≤ f (N) + ∑

N+1≤n≤M−1f (n),

so that

0 ≤ ∑N+1≤n≤M

[∫ n


]≤ f (N)− f (M) ≤ f (N)

and then 0 ≤ R(N) ≤ f (N), which implies R(N) = O( f (N)).


Consider a family of functions (bi(x))i∈N∪0 ⊂ C[0, 1] satisfying

b0(x) ≡ 1,(13.3.17)b′i(x) = ibi−1(x), i ≥ 1,(13.3.18) ∫ 1

0bi(x)dx = 0, i ≥ 1.(13.3.19)

Observe that

(13.3.20) ∑i≥0

bi(x)yi

i!=

yexy

ey − 1=

yey − 1

· ∑n≥0

(xy)n

n!.

Exercise 13.3.16. Prove (13.3.20).

The Bernoulli numbers Bi are defined by

(13.3.21) ∑i≥0

Biyi

i!=

yey − 1

.

Using the Taylor’s series ey − 1 = ∑i∈N yi/i! we get

u = (ey − 1) ∑i≥0

Biyi

i!= ∑

i≥1

[∑

0≤j≤i−1

Bj

(i− j)!

]yi

which gives us

(13.3.22) B0 = 1, bi(x) = ∑0≤j≤i

(ij

)Bjxi−j.

In particular, B1 = − 12 , B2 = 1

6 , and B3 = 0. Moreover,

b0(x) = 1,

b1(x) = x− 12

,

b2(x) = x2 − x +16

,

B3(x) = x3 − 32

x2 +12

x.

Exercise 13.3.17. Show that Bi = 0 for i is odd and i ≥ 3.

We define the i-th Bernoulli function Bi(x) as the periodic function of period1 which coincides with bi(x) on [0, 1). Note that

Bi = Bi(0).

Theorem 13.3.18. (Euler-Maclaurin summation formula) For any integer k ≥ 0 andfor any function f ∈ Ck+1[a, b], a, b ∈ Z, we have

∑a<n≤b

f (n) =∫ b

af (x)dx + ∑

0≤i≤k

(−1)i+1Bi+1

(i + 1)!

[f (i)(b)− f (i)(a)

]+

(−1)k

(k + 1)!

∫ b

aBk+1(x) f (k+1)(x)dx.(13.3.23)


PROOF. We give a proof for k = 1. Recall that

∑a<n≤b

f (n) =∫ b

af (t)dbtc =

∫ b

af (t)dt−

∫ b

af (t)d〈t〉.

Since b1(x) = x− 12 , it follows that B1(x) = 〈x〉 − 1

2 and B1(n) = − 12 = B1 for any

n ∈ Z. Moreover, B′i = iBi−1, B2(t) ∈ C(R), B′2(t) exists on R \ Z, and B′i(t) existson R for i ≥ 3. Hence

∑a<n≤b

f (n) =∫ b

af (x)dx−

∫ b

af (t)dB1(t)

=∫ b

af (t)dt− B1(t) f (t)

∣∣∣∣ba+∫ b

aB1(t) f ′(t)dt

=∫ b

af (t)dt− B1[ f (b)− f (a)] +

12

∫ b

af ′(t)dB2(t)

=∫ b

af (t)dt− B1[ f (b)− f (a)]

+12

B2[ f ′(b)− f ′(a)]−

∫ b

aB2(t) f ′′(t)dt

which is exact the formula (13.3.23) when k = 1.

Another version of Euler-MacLaurin formula is the following

Proposition 13.3.19. Let 0 < y < x be two real numbers and assume that f : [y, x]→ Rhas a continuous derivative on [y, x]. Then

(13.3.24) ∑y<n≤x

f (n) =∫ x

yf (t)dt +

∫ x

y〈t〉 f ′(t)dt + 〈t〉 f (t)

∣∣∣∣xy.

PROOF. Set a := byc and b = bxc, and write

I :=∫ x

u〈t〉 f ′(t)dt =

(∫ b

a+1+∫ x

b+∫ a+1

y

)〈t〉 f ′(t)dt = I1 + I2 + I3.

Compute

I1 =∫ b

a+1(t− btc) f ′(t)dt = ∑

a+1≤k≤b−1

∫ k+1

k(t− k) f ′(t)dt

= ∑a+1≤k≤b−1

[t f (t)

∣∣∣∣k+1

k−∫ k+1

kf (t)dt

]− ∑

a+1≤k≤b−1k[ f (k + 1)− f (k)]

= ∑a+1≤k≤b−1

f (k + 1)−∫ b

a+1f (t)dt = ∑

a+2≤k≤bf (k)−

∫ b

a+1f (t)dt.


Similarly,

I2 :=∫ x

b(t− btc) f ′(t)dt =

∫ x

b(t− b) f ′(t)dt

= t f (t)∣∣∣∣xb−∫ x

bf (t)dt− b[ f (x)− f (b)] = 〈x〉 f (x)−

∫ x

bf (t)dt,

I3 = −〈y〉 f (y)−∫ a+1

yf (t)dt + f (a + 1).

Combining together Ii’s we obtain

I = ∑a+2≤k≤b

f (k) + 〈t〉 f (t)∣∣∣∣xy−∫ x

yf (t)dt + f (a + 1)

which proves (13.3.24).

Taking f (t) = −1/t in (13.3.24) we get

∑1<n≤x

1n=∫ x

1

dtt−∫ x

1

〈t〉t2 dt +

〈x〉x

which implies that

∑n≤x

1n− ln x = 1−

∫ x

1

〈t〉t2 dt +

〈x〉x

.

In particular,

(13.3.25) 1−∫ ∞

1

〈t〉t2 dt = γ = lim

N→∞

(∑

1≤n≤N

1n− ln N

).

Corollary 13.3.20. For any n ≥ 1, we have

(13.3.26) ∑m≤n

1m

= ln n + γ +1

2n− 1

12n2 +θ

60n4

for some θ = θn ∈ [0, 1] and γ is the Euler constant.

PROOF. Choose f (t) = 1t , a = 1, b = n, and k = 3 in (13.3.23). We have

∑2≤m≤n

1n

=∫ n

1

dtt+−B1

1!

(1n− 1)+

B2

2!

(− 1

n2 + 1)+−B3

3!

(2n3 −

21

)+

B4

4!

(−6n4 + 6

)+−14!

∫ n

1B4(t)

24t5 dt

= ln n +12

(1n− 1)− 1

12

(1n2 − 1

)+

1120

(1n4 − 1

)−∫ n

1

B4(t)t5 dt.

Thus

∑m≤n

1n− ln n =

12+

112− 1

120+

12n− 1

12n2 +1

120n4 −∫ n

1

B4(t)t5 dt.

Letting n→ ∞ yields

(13.3.27) γ =12+

112− 1

120−∫ ∞

1

B4(t)t5 dt


and therefore

∑m≤n

1m

= γ + ln n +1

2n− 1

12n2 +1

120n4 +∫ ∞

n

B4(t)t5 dt.

Since |B4(t)| ≤ 130 , it follows that∣∣∣∣∣ ∑

m≤n

1m− γ− ln n− 1

2n+

112n2

∣∣∣∣∣ ≤ 1120n4 +

∫ ∞

n

|B4(t)|t5 dt

≤ 1120n4 +

130

∫ ∞

n

dtt5 =

160n4 .

Consequently

∑m≤n

1m

= γ + ln n +1

2n− 1

12n2 +θ

60n4

for some θ = θn ∈ [0, 1].

In general, one can prove

(13.3.28) γ =12+ ∑

2≤i≤k

Bii−∫ ∞

1

Bk(t)tk+1 dt

for any k ≥ 0.

13.4. Change of variables

Recall that ∫ b

af (x)dx =

∫ d

cf (ϕ(u))ϕ′(u)du

where ϕ(u) is differentiable and monotone. In particular, we have∫ d

cϕ′(u)du = b− a = length of [a, b],

where a = ϕ(c) and b = ϕ(d).

How about the two-dimensional case:∫∫D

f (x, y)dxdy =∫∫

Df (x(u, v), y(u, v))

[?]

dudv.

If it was true, then, in particular, by taking f ≡ 1, we obtain∫∫D

[?]

dudv =∫∫

Ddxdy = Area(D).

13.4.1. Geometric insight. Assume that(i) D ⊂ R2 is a bounded, admissible, and closed domain, T : D → Ω is a

C1-diffeomorphisms with Ω = T(D). Write T(u, v) := (x(u, v), y(u, v)).(ii) The determinant of T

det(T) :=∂(x, y)∂(u, v)

6= 0

for any (u, v) ∈ D. Hence the Jacobian of T

JT := |det(T)|

13.4. CHANGE OF VARIABLES 61

is always positive on D.For any point (u0, v0) ∈ D with du, dv 1, consider the four points

P1 = (x(u0, v0), y(u0, v0)) = (x0, y0),P2 = (x(u0 + du, v0), y(u0 + du, v0)) ≈ (x0 + xu(u0, v0)du, y0 + xv(u0, v0)du) ,P3 = (x(u0 + du, v0 + dv), y(u0 + du, v0 + dv)),P4 = (x(u0, v0 + dv), y(u0, v0 + dv))

in Ω. The area of the “rectangle” P1P2P3P4 is approximated as∣∣∣∣ xu(u0, v0) yu(u0, v0)xv(u0, v0), yv(u0, v0)

∣∣∣∣ dudv = JT(u0, v0)dudv.

Therefore∫∫D

f (x(u, v), y(u, v))JT(u, v)dudv “ = ”∫∫

Ωf (x, y)dxdy.

Equivalently, one has∫∫T(D)

f (x, y)dxdy “ = ”∫∫

D( f T)JT(u, v)dudv

or ∫∫T(D)

f “ = ”∫

D( f T)JT.

13.4.2. Change of variables. We first state a theorem for 2D.

Theorem 13.4.1. (Change of variables for 2D) Let D ⊂ R2 be a bounded and closeddomain with piecewise smooth boundary ∂D, and D ⊂ D for some open domain D in R2.Suppose T : D → Ω = T(D) is a C1-diffeomorphism with JT > 0 on D. Then

(13.4.1)∫∫

Ωf (x, y)dxdy =

∫∫D

f (x(u, v), y(u, v))JT(u, v)dudv

or

(13.4.2)∫∫

T(D)f =

∫∫D( f T)JT.

PROOF. Divide the intervals [a, b] and [c, d] into 2n subintervals with the samelengths, respectively,

a = x0 < x1 < · · · < xM = b, c = y0 < y1 < · · · < yM = d, M := 2n.

Writeij := [xi−1, xi]× [yj−1, yj], 1 ≤ i, j ≤ M

and let

An := ij : ij ⊂ D, Bn := ij : ij ∩ D 6= ∅, Cn := Bn \ Int(An).

It is clear that

· · · ⊂ An ⊂ An+1 ⊂ · · · ⊂ D ⊂ · · · ⊂ Bn+1 ⊂ Bn ⊂ · · ·and

limn→∞

Vol(An) = Vol(D) = limn→∞

Vol(Bn)

since D is admissible. We then mat choose a sufficiently large enough n 1 sothat Bn ⊂ D. Consider two special cases of T:

(i) Case 1: T(u, v) = (u, y(u, v)),


(ii) Case 2: T(u, v) := (x(u, v), v).Step 1. Suppose T is either of Case 1 or Case 2. For any ∈ Bn, we have

Vol(T()) = JT(u, v)Vol()

for some point (u, v) ∈ . Suppose T is of Case 1, i.e., T(u, v) = (u, y(u, v)) forsome C1-function y. On U, we have JT > 0 so that

0 < JT =

∣∣∣∣det[

1 0∂y∂u

∂y∂v

]∣∣∣∣ = ∣∣∣∣∂y∂v

∣∣∣∣ .

We may assume that ∂y/∂v > 0 on U. If = [a, b] × [c, d], then y(u, ·) isincreasing for any given u ∈ [a, b]. Consequently

T() =(x, y) ∈ R2 : a ≤ x ≤ b, y(x, c) ≤ y ≤ y(x, d)

which is a X-type domain. Compute

Vol(T()) =∫ b

adx∫ y(x,d)

y(x,c)dy =

∫ b

a[y(x, d)− y(x, c)] dx

= [y(x, d)− y(x, c)] (b − a)

for some x ∈ [a, b]. By the mean value theorem, we obtain

Vol(T()) =∂y∂v

(u, v)(d − c)(b − a) = JT(u, v)Vol()

for some v ∈ [c, d].

Step 2. Suppose that T is either of Case 1 or Case 2. If f ∈ C(Ω), then∫∫Ω

f (x, y)dxdy =∫∫

Df T(u, v)JT(u, v)dudv.

Set H := maxΩ | f |. Since there exists an integer B ∈ N such that Bn ⊂ D forn ≥ N, it follows that maxBn JT ≤ K for some K > 0 (independent of n) andany n ≥ N. Step 1 implies that Vol(T(Cn)) ≤ KVol(Cn) for any n ≥ N, so thatlimn→∞ Vol(T(Cn)) = 0. Write Dij := D ∩ ij. For any (uij, vij) ∈ ij writeξij = x(uij, vij) and ηij = y(uij, vij). Then

∑Dij

f (ξij, ηij)Vol(T(Dij)) =

∑Dij⊂An

+ ∑Dij⊂D\Int(An)

f (ξij, ηij)Vol(T(Dij)) = I + J.

For Dij ⊂ An we have Dij = ij and

Vol(T(ij)) = JT(uij, vij)Vol(ij)

for some (uij, vij) ∈ ij. Then

I = ∑ij⊂An

f (ξij, ηij)Vol(T(ij))

= ∑ij⊂An

f (ξij, ηij)JT(uij, vij)Vol(ij)

= ∑ij⊂An

f T(uij, vij)JT(uij, vij)Vol(ij).


For Dij ⊂ D \ Int(An) we take any (uij, vij) ∈ Dij so that

J = ∑Dij⊂D\Int(An)

f (ξij, ηij)Vol(T(Dij))

= ∑Dij⊂D\Int(An)

f T(uij, vij)JT(uij, vij)Vol(Dij)

+ ∑Dij⊂D\Int(An)

[f (ξij, ηij)Vol(T(Dij))− f T(uij, vij)JT(uij, vij)Vol(Dij)

].

Since D \ Int(An) ⊂ Cn, it follows that∣∣∣∣∣∣ ∑Dij⊂D\Int(An)

f (ξij, ηij)Vol(T(Dij))

∣∣∣∣∣∣ ≤ H ∑Dij⊂D\Int(An)

Vol(T(Dij))

≤ HVol(T(Cn))

≤ HKVol(Cn),∣∣∣∣∣∣ ∑Dij⊂D\Int(An)


∣∣∣∣∣∣ ≤ HKVol(Cn).

Thus ∣∣∣∣∣∣ ∑Dij⊂D\Int(An)

[f (ξij, ηij)Vol(T(Dij))− f T(uij, vij)JT(uij, vij)Vol(Dij)

]∣∣∣∣∣∣≤ 2HKVol(Cn) → 0

as n→ ∞. Because f is integrable on Ω, we conclude that

∑Dij

f (ξij, ηij)Vol(T(Dij)) → ∑Dij


→∫∫

Df T(u, v)JT(u, v)dudv.

Here the left-hand side tends to∫∫

Ω f (x, y)dxdy.

Step 3. For any (u0, v0) ∈ U we have T = T2 T1 in a neighborhood of (u0, v0),where T1, T2 are C1-diffeomorphisms and Ti is of Case 1 or Case 2.

Since JT(u0, v0) = |det(T)(u0, v0)| > 0, it follows that at least one of xu, yu, xv, yvis nonzero at (u0, v0). Without loss of generality, we may assume that ∂x

∂u (u0, v0) 6=0. Consider T(u, v) = (x(u, v), y(u, v)), (u, v) ∈ U, and

T1(u, v) := (x(u, v), v)

which is of Case 2. Since

det(T1) = det([

∂x∂u

∂x∂v

0 1

])=

∂x∂u6= 0

at (u0, v0). According to implicit function theorem, we can find neighborhoodsU0, Ω0 satisfying U ⊃ U0 3 (u0, v0) and Ω ⊃ Ω0 3 (x(u0, v0), v0), and g ∈ C1(Ω0)such that

T1 : D0 −→ Ω0, (u, v) 7−→ (ξ, η)


is a C1-diffeomorphism, and furthermore, T−11 (ξ, η) = (g(ξ, η), η). In particular,

we have

g(x(u, v), v) = u

for any (u, v) ∈ U0. Define

T2(ξ, η) := (ζ, y(g(ζ, η), η))

which is of Case 1. Then

T2 T1(u, v) = T2(x(u, v), v) = (x(u, v), y(g(x(u, v), v), v)) = (x(u, v), y(u, v))

for any (u, v) ∈ U0.

Step 4. Proof of (13.4.1). Step 3 implies that for any (u, v) ∈ D, there ex-ists a neighborhood Dδ(u, v) (here δ depends on (u, v)) such that T = T2 T1 inDδ(u, v), where T1, T2 are defined as above. Since (Dδ/2(u, v))(u,v)∈D is an opencover of D and D is compact (because D is bounded and compact in R2), it fol-lows that (Dδi/2(ui, vi))1≤i≤N is an open cover of D for some N ∈ N. Let δ :=min1≤i≤N δi/2. Choose n sufficiently large so that any rectangle ij := [xi−1, xi]×[yj−1, yj] has diameter

√(xi − xi−1)2 + (yj − yj−1)2 < δ. For any ij ∩ Int(D) 6=

∅, we can find an integer 1 ≤ k ≤ N such that Dij := ij ∩ D ⊂ ij ⊂ Dδk (uk, vk).

Therefore, in Dij, we have T = Tij2 Tij

1 . Write

Tij1 (u, v) := (ζ(u, v), η(u, v)) , Tij

2 (u, v) := (x(ζ, η), y(ζ, η)) .

Step 2 then implies∫∫Ω

f (x, y)dxdy = ∑1≤i,j≤M

∫∫T(Dij)

f (x, y)dxdy

= ∑1≤i,j≤M

∫∫Tij

2 (Tij1 (Dij))

f (x, y)dxdy

= ∑1≤i,j≤M

∫∫Tij

1 (Dij)f Tij

2 (ζ, η)JTij

2(ζ, η)dζdη

= ∑1≤i,j≤M

∫∫Dij

f Tij2 Tij

1 (u, v)JTij

2 Tij

1 (u, v)JTij

1(u, v)dudv.

Since

JT(u, v) = JTij

2 Tij1(u, v) =

(J

Tij2 Tij

1 (u, v))

JTij

1(u, v)

by the chain rule, it follows that∫∫Ω

f (x, y)dxdy = ∑1≤i,j≤M

∫∫Dij

f T(u, v)JT(u, v)dudv

=∫∫

Tf T(u, v)JT(u, v)dudv.

Thus we have proved (13.4.1).


Theorem 13.4.2. (Change of variables for nD) Let U, V ⊂ Rn be open subsets andT : U → V be a C1-diffeomorphic with T(x) = y. For any bounded and closed domainD ⊂ U with piecewise smooth boundary ∂D, we have

(13.4.3)∫

T(D)f (y)dy =

∫D

f T(x)JT(x)dx

for any f ∈ C(T(D)), where dx = dx1 · · · dxn.

13.4.3. Examples. Recall first that 2D polar coordinates:

(13.4.4) x = r cos θ, y = r sin θ, 0 ≤ r < ∞, 0 ≤ θ ≤ 2π.

Define a map T by

(13.4.5) T(r, θ) = (r cos θ, r sin θ)

whose Jacobian is given by JT(r, θ) = r.


I =∫∫

Ωdxdy, Ω = (x, y) ∈ R2 : x2 + y2 ≤ 1.

Let D = (r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π and T(r, θ) be given by (13.4.5). Then

I =∫∫

D1 T(r, θ)JT(r, θ)drdθ =

∫ 1

0

∫ 2π

0rdrdθ = 2π

∫ 1

0rdr = π.

The 3D cylinder coordinates are given by

(13.4.6) x = r cos θ, y = r sin θ, z = z,

where 0 ≤ r < ∞, 0 ≤ θ ≤ 2π, and z ∈ R. Define a map T by

(13.4.7) T(r, θ, z) = (r cos θ, r sin θ, z)

whose Jacobian is given by JT(r, θ, z) = r.


I =∫∫

Ωdxdydz, Ω = (x, y, z) ∈ R3 : 0 ≤ z ≤ 25− x2 − y2.

Let x = r cos θ, y = r sin θ, z = z, where 0 ≤ r ≤ 5 and 0 ≤ θ ≤ 2π, and D =(r, θ, z) : 0 ≤ r ≤ 5, 0 ≤ θ ≤ 5, 0 ≤ z ≤ 25− r2. Then

I =∫∫∫

Drdrdθdz =

∫ 5

0rdr

∫ 2π

0dθ∫ 25−r2

0dz = 2π

∫ 5

0r(25− r2)dr

= π∫ 5

0(25− r2)dr2 = −π

2(r2 − 25)2

∣∣∣∣50

=6252

π.

The 3D spherical coordinates are given by

(13.4.8) x = r sin ϕ cos θ, y = r sin ϕ sin θ, z = r cos ϕ

where 0 ≤ r < ∞, 0 ≤ ϕ ≤ π, and 0 ≤ θ ≤ 2π. Define a map T by

(13.4.9) T(r, ϕ, θ) = (r sin ϕ cos θ, r sin ϕ sin θ, r cos ϕ)

whose Jacobian is given by JT(r, ϕ, θ) = r2 sin ϕ.



I =∫∫∫

Ωdxdydz, Ω =

(x, y, z) ∈ R3 :

x2

a2 +y2

b2 +z2

c2 ≤ 1

where a, b, c > 0. Let x = ar sin ϕ cos θ, y = br sin ϕ sin θ, and z = cr cos ϕ. Then

I =∫ 1

0dr∫ π

0dϕ∫ 2π

0abcr2 sin ϕdθ =

abc3

2π∫ π

0sin ϕdϕ =

4π

3abc.

In general, the nD spherical coordinates are given by

x1 = r cos ϕ1,x2 = r sin ϕ1 cos ϕ2,x3 = r sin ϕ1 sin ϕ2 cos ϕ3,

· · · · · · · · ·(13.4.10)xn−1 = r sin ϕ1 sin ϕ2 · · · sin ϕn−2 cos ϕn−1,

xn = r sin ϕ1 sin ϕ2 · · · sin ϕn−2 sin ϕn−1,

where 0 ≤ r < ∞, 0 ≤ ϕ1, ϕ2, · · · , ϕn−2 ≤ π, and 0 ≤ ϕn−1 ≤ 2π. Define a map Tby

(13.4.11) T(r, ϕ1, ϕ2, · · · , ϕn−1) = (x1, x2, · · · , xn)

according to (13.4.10) whose Jacobian is

JT(r, ϕ1, ϕ2, · · · , ϕn−1) = rn−1 sinn−2 ϕ1 sinn−3 ϕ2 · · · sin ϕn−2.


I =∫

BnR(0)

dx1 · · · dxn, R > 0.

Then

I =∫ R

0dr∫ π

0dϕ1 · · ·

∫ π

0dϕn−2

∫ 2π

0rn−1 sinn−2 ϕ1 sinn−3 ϕ2 · · · sin ϕn−2 dϕn−1

=2πRn

n ∏1≤i≤n−2

∫ π

0sinn−1−i ϕi dϕi.

For the integral

Jk :=∫ π

0sink ϕdϕ = 2

∫ π/2

0sink ϕdϕ

we have

Jk =

(2m−1)!!(2m)!! π, k = 2m,(2m)!!

(2m+1)!! 2, k = 2m + 1.

Hence I = (2π)kR2k/(2k)!! for n = 2k and I = 2(2π)kR2k+1/(2k + 1)!! for n =2k + 1.

Example 13.4.7. Prove

2π(√

17− 4) ≤ I :=∫∫

x2+y2≤1

dxdy√16 + sin2 x + sin2 y

≤ π

4.

Letx = r cos θ, y = r sin θ, 0 ≤ θ ≤ 1, 0 ≤ θ ≤ 2π.


Then

I ≤∫∫

x2+y2≤1

14

dxdy =π

4,

I ≥∫

x2+y2≤1

dxdy16 + x2 + y2 =

∫ 1

0dr∫ 2π

0

rdθ√16 + r2

= 2π∫ 1

0

rdr√16 + r2

= 2π√

16 + r2∣∣∣∣10

= 2π(√

17− 4).

Exercise 13.4.8. If f ∈ C[−1, 1], then∫∫|x|+|y|≤1

f (x + y)dxdy =∫ 1

−1f (u)du.

Example 13.4.9. (1) Compute

(13.4.12) IA :=∫∫

x2+y2≤A2e−(x2+y2)dxdy, A > 0.

Let x = r cos θ, y = r sin θ, 0 ≤ r ≤ A, and 0 ≤ θ ≤ 2π. We get

IA =∫ A

0dr∫ 2π

0e−r2

rdθ = 2π∫ A

0re−r2

dr = π(

1− e−A2)

.

Consider (∫ A

−Ae−x2

dx)2

=∫−A≤x,y≤A

e−(x2+y2)dxdy.

Then B2√2A

(0) ⊃ A = [−A, A]× [−A, A] ⊃ B2A(0) and hence

√π(

1− e−A2)1/2

= I1/2A ≤

∫ A

−Ae−x2

dx ≤ I1/2√2A

=√

π(

1− e−2A2)1/2

.

Taking A→ ∞ implies

(13.4.13)∫ ∞

−∞

1√π

e−x2dx = 1.

In general, we have

(13.4.14)∫

Rn

1πn/2 e−|x|

2dx = 1.

Let us consider∑

1≤i,j≤naijxixj = xAxT ,

where the matrix A = (aij)1≤i,j≤n is positive definite.(2) Compute

(13.4.15) IA =∫

Rne−xAxT

dx

where A is a positive definite n× n matrix. There exists a matrix Q ∈ O(n) such that

A = QΛQT , Λ = diag(λ1, · · · , λn)

where λ1, · · · , λn > 0 are eigenvalues of A. Therefore

IA =∫

Rne−xQΛ(xQ)T

dx =∫

Rne−yΛyT

dy = ∏1≤i≤n

(1√λi

√π

)=

√πn

det(A).


(3) Compute

(13.4.16) Ia,b,c =∫

Re−(ax2+2bx+c)dx, a > 0, b, c ∈ R.

Then

Ia,b,c =∫

Re−a[(x+ b

a )2+ ac−b2

a2 ]dx = eb2−ac

a

∫R

e−a(x+ ba )

2dx

= eb2−ac

a

√π

a= e−

det(A∗)a

√π

a

where A∗ is the matrix (a bb c).

More general, consider

(13.4.17) Iaij ,bi ,c =∫

Rne−(∑1≤i,j≤n aijxixj+2 ∑1≤i≤n bixi+c)dx1 · · · dxn

where A = (aij)1≤i,j≤n is positive definite, b1, · · · , bn, c ∈ R. Write

x = (x1, · · · , xn), b = (b1, · · · , bn),

so thatIaij ,bi ,c =

∫Rn

e−(xAxT+2b·xT+c)dx.

As before, we have A = QΛQT for some Q ∈ O(m) and Λ = diag(λ1, · · · , λm) withλ1, · · · , λn > 0. Hence

xAxT + 2b · xT + c = (xQ)Λ(xQ)T + 2b · xT + c

= yΛyT + 2b · (Q−1)TyT + c

= ∑1≤i≤n

λiy2i + 2 ∑

1≤i≤nbiyi + c

= ∑1≤i≤n

λi

(yi +

biλi

)2

+

(c− ∑

1≤i≤n

b2i

λi

)where y = xQ = (y1, · · · , yn) and b = b · (Q−1)T = b · (QT)−1 = (b1, · · · , bn).Moreover, for the matrix

A∗ =[

A bT

b c

]its determinant is given by

det(A∗) = det[

A bT

0 c− bA−1bT

]=

(c− bA−1b−1

)det(A)

=

c− b[(QT)−1Λ−1Q−1

]bT

det(A) =(

c− bΛ−1bT)

det(A)

=

(c− ∑

1≤i≤n

b2i

λi

)det(A).

Consequently,

xAxT + 2b · xT + c = ∑1≤i≤n

λi

(yi +

biλi

)2

+det(A∗)det(A)

which implies that Iaij ,bi ,c =√

πn/ det(A) exp[−det(A∗)/ det(A)].

13.5. IMPROPER INTEGRALS 69

13.5. Improper integrals

If we let A→ ∞ in (13.4.12) we get

limA→∞

∫∫x2+y2≤A2

e−(x2+y2)dxdy = π.

Thus we may define ∫∫R2

e−(x2+y2)dxdy

as the limitlim

A→∞

∫∫x2+y2≤A2

e−(x2+y2)dxdy.

This example gives an idea to define improper integrals.

13.5.1. Improper integrals over unbounded domains. Let D ⊂ R2 be a un-bounded domain with boundary ∂D consisting of finitely many smooth curves.Consider a bounded function f in D which is integrable over any admissible sub-domain of D.

Definition 13.5.1. We say that f is integrable over D if the limit

limd(Γ)→∞

∫∫DΓ

f (x, y)dxdy

exists. Here Γ is a curve in D with |Γ| = 0 and cut down a bounded domain DΓ.In this case, we write∫∫

Df (x, y)dxdy = lim

d(Γ)→∞

∫∫DΓ

f (x, y)dxdy

and call it as the improper integral of f over D. Otherwise, we say the integral isdivergent.

Lemma 13.5.2. Suppose that f is nonnegative on D and (Di)i∈N is a increasing sequenceof domains with limn→∞ d(Γn) = ∞, where (Γi)i∈N is a sequence of curves in D with|Γi| = 0 and each Γi cuts down a bounded domain Di. Then∫∫

Df (x, y)dxdy

converges if and only if the sequence∫∫Dn

f (x, y)dxdy

n∈N

converges. In this case

(13.5.1)∫∫

Df (x, y)dxdy = lim

n→∞

∫∫Dn

f (x, y)dxdy.

PROOF. One direction is obvious. We now assume that the sequence con-verges. Given Γ and DΓ as above, there exists N ∈ N such that DΓ ⊂ Dn forsufficiently large number n N. Since f is nonnegative, it follows that∫∫

DΓ

f (x, y)dxdy ≤∫∫

Dnf (x, y)dxdy −→ I := lim

n→∞

∫∫Dn

f (x, y)dxdy

as n→ ∞. Conversely, for any ε > 0 there is an integer N ∈ N such that∫∫DN

f (x, y)dxdy > I − ε.


Choose a admissible curve Γ with DΓ ⊃ DN so that∫∫DΓ

f (x, y)dxdy ≥∫∫

DN

f (x, y)dxdy > I − ε.

Hence we get the desired result.


I =∫∫

x2+y2≥a2

dxdyrp , r :=

√x2 + y2, a > 0.

According to Lemma 13.5.2 we have

I = limA→∞

∫ 2π

0dθ∫ A

a

rdrrp = lim

A→∞

2π

2− pr2−p

∣∣∣∣A0

.

Thus the integral is convergent for p > 2 and divergent for p ≤ 2.

There are some basic properties about improper integrals.

Proposition 13.5.4. Let D ⊂ R2 be a unbounded domain with piecewise smooth bound-ary ∂D.

(1) Let 0 ≤ f ≤ g in D. Then∫∫D

g(x, y)dxdy is convergent =⇒∫∫

Df (x, y)dxdy is convergent,∫∫

Df (x, y)dxdy is divergent =⇒

∫∫D

g(x, y)dxdy is divergent.

(2) f is integrable over D if and only if | f | is integrable over D.

Theorem 13.5.5. (1) Consider the domain D = (r, θ) : a ≤ r < ∞, α ≤ θ ≤ β,where 0 ≤ α, β ≤ 2π and a > 0.

| f (x, y)| ≤ Mrp (p > 2) =⇒

∫∫D

f (x, y)dxdy is convergent,

| f (x, y)| ≥ mrp (p ≤ 2) =⇒

∫∫D

f (x, y)dxdy is divergent.

(2) If f ∈ C([a, ∞)× [c, ∞)) and∫ ∞

adx∫ ∞

cf (x, y)dy and

∫ ∞

adx∫ ∞

c| f (x, y)|dy

exist, then f is integrable over D and∫∫[a,∞)×[c,∞)

f (x, y)dxdy =∫ ∞

adx∫ ∞

cf (x, y)dy.

(3) If T : D → T(D) is a C1-diffeomorphism with JT over D, where D is a domainin R2, then ∫∫

T(D)d(x, y)dxdy =

∫∫D

f T(u, v)JT(u, v) dudv

provided one of integrals exists.

13.5. IMPROPER INTEGRALS 71


I =∫∫

R2e−(x2+y2)dxdy.

This integral has been computed in (13.4.12). By Theorem 13.5.5, we have

I =∫ 2π

0dθ∫ ∞

0e−r2

dr = 2π∫ ∞

0e−r2

dr = π.

Consequently,

(13.5.2)∫ ∞

0e−x2

dx =

√π

2.

Letting u = x2 in (13.5.2) yields√

π

2=

12

∫ ∞

0u−1/2e−udu =

12

∫ ∞

0u

12−1e−udu.

Define the Gamma function by

(13.5.3) Γ(s) :=∫ ∞

0us−1e−udu, s > 0.

Note that Γ(s) is well-defined for any s > 0. Actually, for s ∈ C with positive real part,the associated Gamma function Γ(s) is also well-defined. Compute

Γ(n+ 1) =∫ ∞

0une−udu =

∫ ∞

0−unde−u = −

[un

eu

∣∣∣∣∞0−∫ ∞

0nun−1e−udu

]= nΓ(n).

HenceΓ(n + 1) = n!Γ(1) = n!

since Γ(1) = 1. Moreover,

Γ(

12

)=√

π,(13.5.4)

Γ(s + 1) = sΓ(s) for s > 0.(13.5.5)

13.5.2. Improper integrals of unbounded functions. Let D ⊂ R2 be a boundeddomain and P0 ∈ D. Consider a function f defined on D \ P with the propertythat f is unbounded on any neighborhood of P0. Choose a Jordan curve γ aboundP0 with |γ| = 0 and write σ the domain enclosed by γ in D.We assume that∫∫

D\σf (x, y)dxdy

exists for any such curves γ.

Definition 13.5.7. We say that f is integrable over D if the limit

limρ(γ)→0

∫∫D\σ

f (x, y)dxdy

exists. Here ρ(γ) := supP∈γ |P− P0|. In this case, we write∫∫D

f (x, y)dxdy = limρ(γ)→0

∫∫D\σ

f (x, y)dxdy.

Otherwise, we say that the integrable is divergent.



I =∫∫

x2+y2≤a2

dxdyrp , r =

√x2 + y2, a > 0.

By definition one has

I = limρ→0

∫∫ρ2≤x2+y2≤a2

dxdyrp = lim

ρ→02π∫ a

ρ

rdrrp .

Hence I is convergent only for p < 2.

13.5.3. Applications. Introduce the Beta function by

(13.5.6) B(p, q) :=∫ 1

0xp−1(1− x)q−1dx, p, q > 0.

Note that B(p, q) is well-defined for p, q > 0.

Proposition 13.5.9. For any p, q > 0, one has

(13.5.7) B(p, q) =Γ(p)Γ(q)Γ(p + q)

.

PROOF. Write

Γ(p) = 2∫ ∞

0s2p−1e−s2

ds, Γ(q) = 2∫ ∞

0r2q−1e−t2

dt.

Therefore

Γ(p)Γ(q) = 4∫∫

[0,∞)×[0,∞)s2p−1t2q−1e−(s

2+t2)dsdt

= 4∫ π/2

0cos2p−1 θ sin2q−1 θdθ

∫ ∞

0e−r2

r2p+2q−2rdr

=

(2∫ π/2

0cos2p−1 θ sin2q−1 θ dθ

)(2∫ ∞

0r2(p+q)−1e−r2

dr)

= B(p, q)Γ(p + q),

where s = r cos θ and t = r sin θ.

A byproduct is

(13.5.8) B(p, q) = 2∫ π/2

0cos2p−1 θ sin2q−1 θ dθ.


(13.5.9) I =∫ π/2

0sinn xdx, J =

∫ π/2

0cosn xdx, n ∈ N.

In fact, using (13.5.7) and (13.5.8) we arrive at

I =12

∫ 1

0u

n+12 −1(1− u)

12−1du =

12

B(

n + 12

,12

)=

12

Γ(

n+12

)Γ(

12

)Γ( n+2

2) .

13.6. DIFFERENTIAL FORMS 73

13.6. Differential forms

For two vectors a = (a1, a2) and b = (b1, b2), we define

a ∧ b :=∣∣∣∣ a1 a2

b1 b2

∣∣∣∣ = a1b2 − a2b1.

The wedge product satisfies(1) a ∧ b = −b ∧ a,(2) a ∧ a = 0,(3) R2 together with wedge product and scalar multiplication forms a vector

space over R.(4) If e1, e2 is a basis of R2, then

a1 ∧ a2 =

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ e1 ∧ e2

where a1 = a11e1 + a12e2 and a2 = a21e1 + a22e2.In this section we shall define wedge product between differential forms.

13.6.1. Differential forms and wedge products. Consider a domain U ⊂ Rn.Recall that

(13.6.1) d f = ∑1≤i≤n

∂ f∂xi

dxi

for any f ∈ C1(U). Define

(13.6.2) Λ1(U) := generated by dx1, · · · , dxn =

∑

1≤i≤nai(x)dxi

where ai(x) are functions of x. It is clear that (Λ1(U),+) is a vector space anddim Λ1(U) = n.

Define the wedge product ∧ formally by satisfying

(13.6.3) dxi ∧ dxj = −dxj ∧ dxi

for any 1 ≤ i, j ≤ n. Actually, it comes from tensor products. It is clear thatdxi ∧ dxi = 0 for any 1 ≤ i ≤ n. Let(13.6.4)

Λ2(U) := generated by dxi ∧ dxj = generated by dxi ∧ dxj (1 ≤ i < j ≤ n).

Note that dim Λ2(U) = (n2) =

n(n−1)2 .

Similarly we can define k-forms by

Λk(U) := generated by dxi1 ∧ · · · ∧ dxik

= generated by dxi1 ∧ · · · ∧ dxik (1 ≤ i1 < · · · < ik ≤ n)(13.6.5)

=

∑

1≤i1<···<ik≤nai1···ik (x)dxi1 ∧ · · · ∧ dxik

.

Note also that dim Λk(U) = (nk) and Λk(U) = 0 for any k > n.


For ω ∈ Λi(U) and η ∈ Λj(U) write

ω = ∑1≤k1<···<ki≤n

ak1···kidxk1 ∧ · · · ∧ dxki

= ∑|I|=i

aIdxI ,

η = ∑1≤`1<···<`j≤n

b`1···`jdx`1 ∧ · · · ∧ dx`i

= ∑|J|=j

bJdxJ .

We define

(13.6.6) ω ∧ η := ∑|I|=i,|J|=j

aIbJdxI ∧ dxJ ∈ Λi+j(U).

For a function f on U define

(13.6.7) f ω := ∑|I|=i

( f aI)dxI .

Theorem 13.6.1. (1) If ω ∈ Λp(U) and η ∈ Λq(U) with p + q > n, then ω ∧ η = 0.(2) If ω ∈ Λp(U) and η ∈ Λq(U), then ω ∧ η = (−1)pqη ∧ω.(3) If ω ∈ Λp(U) and p is odd, then ω ∧ω = 0.(4) For any differential forms ω, η, and σ, one has

(ω+ η)∧σ = ω∧σ+ η∧σ, σ∧ (ω+ η) = σ∧ω+σ∧ η, (ω∧ η)∧σ = ω∧ (η∧σ).

As notion in Theorem 13.4.1. Write

T(u, v) = (x(u, v), y(u, v)) .

Since dx = xudu + xvdv and dy = yudu + yvdc, it follows that

dx ∧ dy =

∣∣∣∣ xu xvyu yv

∣∣∣∣ du ∧ dv =∂(x, y)∂(u, v)

du ∧ dv.

If we think of dx ∧ dy as “positive” area element, then ∂(x, y)/∂(u, v) = JT and(13.4.1) is equivalent to

(13.6.8)∫∫

T(D)f (x, y)dx ∧ dy =

∫∫D

f T(u, v)∂(x, y)∂(u, v)

du ∧ dv.

13.6.2. Cohomology groups. Consider a subset of Λk(U):

(13.6.9) Λk∞(U) :=

ω ∈ Λk(U) with ai1···ik ∈ C∞(U)

.

Define the exterior differential

(13.6.10) d : Λk∞(U) −→ Λk+1

∞ (U)

by

d

(∑

1≤i1<···<ik≤nai1···ik (x)dxi1 ∧ · · · ∧ dxik

):= ∑

1≤i1<···<ik≤ndai1···ik ∧ dxi1 ∧ · · · ∧ dxik .

In particular,

d

(∑

1≤i≤naidxi

)= ∑

1≤i<j≤n

(∂aj

∂xi− ∂ai

∂xj

)dxi ∧ dxj.

Since d2 = 0 that is obvious, we get a complex (Λ•(U), d) as

· · · d−−−−→ Λk∞(U)

d−−−−→ Λk+1∞ (U)

d−−−−→ · · ·

PROBLEMS 75

Define the k-th cohomology group of U by

(13.6.11) Hk(U) :=Ker(d|Λk

∞(U))

Im(d|Λk−1∞ (U)

).

We say ω ∈ Λk∞(U) is closed if dω = 0, and is exact if ω = dη for some η ∈

Λk−1∞ (U). Note that any exact form must be closed.

Example 13.6.2. Consider 1-form

α = [2x + y cos(xy)] dx + [x cos(xy)] dy

on R2. If α = d f , then

fx = 2x + y cos(xy), fy = x cos(xy)

which impliesf (x, y) = x2 + sin(xy)

up to a additive constant. Hence α is exact.

Example 13.6.3. Consider 1-form

α =1

2π

xdy− ydxx2 + y2

on R2 \ (0, 0). It is clear that dα = 0 but α is not exact. Otherwise, if α = d f for somesmooth function, then

1 =∫

S1α =

∫S1

d f =∫

∂S1f = 0

by Stokes’ formula, which is a contradiction.

A domain U ⊂ Rn is called star-shaped with respect to 0, if 0 ∈ U and 0x ⊂ Ufor any x ∈ U. Observe that a convex domain must be star-shaped, however, theconverse may not be true.

For any k-form ω = ∑1≤i1<···<ik≤n ai1···ik dxi1 ∧ · · · ∧ dxik define a map

(13.6.12) P : Λk∞(U) −→ Λk−1

∞ (U)

by

Pω := ∑1≤i1<···<ik≤n

∑1≤j≤k

(−1)j−1[∫ 1

0tk−1ai1···ik (tx)dt

]dxi1 ∧ · · · ∧ dxij ∧ · · · ∧ dxik .

Then

(13.6.13) (d P + P d)ω = ω.

If ωis closed, then (13.6.13) shows that ω = d(Pω) and hence ω is exact. Conse-quently,

(13.6.14) Hk(U) = 0 (1 ≤ k ≤ n) for any star-shaped domain U.

Exercise 13.6.4. Verify (13.6.13).

Problems

CHAPTER 14

Line and surface integrals

14.1. Line integrals

Consider an admissible, continuous, and simple curve L ⊂ R3 with startingpoint A and ending point B. Choose a partition T : A = P0, P1, · · · , Pn−1, Pn = Bof L, and write

∆si :=∣∣∣Pi−1Pi

∣∣∣ , 1 ≤ i ≤ n and λ(T) := max1≤i≤n

∆si.

14.1.1. Type I line integral. Given a function f defined on L. For any point(ξi, ηi, ζi) ∈ Pi−1Pi consider the Riemann sum

ST( f ) := ∑1≤i≤n

f (ξi, ηi, ζi)∆si.

Define

(14.1.1)∫

Lf (x, y, z)ds := lim

λ(T)→0ST( f ).

If the limit (14.1.1) exists, we say that the integral in (14.1.1) is a type I line integral.The following are basic properties

(1) The integral (14.1.1) is independent of the orientation of L.(2) If both integrals∫

Lf (x, y, z)ds and

∫L

g(x, y, z)ds

exist, then ∫L(α f + βg)ds = α

∫L

f ds + β∫

Lgds.

(3) For any curves L1 and L2, one has∫L1+L2

f ds =∫

L1

f ds +∫

L2

f ds.

Theorem 14.1.1. If L is a smooth curve with representation

L = (x, y, z) ∈ R3 : x = x(t), y = y(t), z = z(t), α ≤ t ≤ βand f ∈ C(L), then

(14.1.2)∫

Lf (x, y, z)ds =

∫ β

αf (x(t), y(t), z(t))

√x(t)2 + y(t)2 + z(t)2 dt.

PROOF. By definition∫L

f (x, y, z)ds =∫ β

αf (x(t), y(t), z(t))

dsdt

dt.

77

78 14. LINE AND SURFACE INTEGRALS

Here ds stands for arc length which is determined by dsdt =

√x(t)2 + y(t)2 + z(t)2.

Indeed,

s(t) =∫ t

α

√x(t)2 + β(t)2 + z(t)2 dt

for any t ∈ [α, β].


I =∫

L

√1− x2 − y2 ds, L : x2 + y2 = x.

Let x = 12 (1 + cos t) and y = 1

2 sin t with 0 ≤ t ≤ 2π. Then

I =12

∫ 2π

0

√1− cos t

2dt =

12

∫ 2π

0

∣∣∣∣sint2

∣∣∣∣ dt =∫ π

0sin

t2

dt = 2.


I =∫

Γ(x + y)2ds, Γ = (x, y) ∈ R2 : x2 + y2 = 1, y ≥ 0 ∪ (x, 0) : −1 ≤ x ≤ 1.

Write Γ1 = (x, y) ∈ R2 : x2 + y2 = 1, y ≥ 0 and Γ2 = (x, 0) : −1 ≤ x ≤ 1.Then

I1 :=∫

Γ1

(x + y)2ds =∫

Γ1

(x2 + y2 + 2xy)ds =∫

Γ1

ds + 2∫

Γ1

xyds = π,

I2 :=∫

Γ2

(x + y)2ds =∫ −1

1x2 dx =

23

.

Hence I = I1 + I2 = 23 + π.

14.1.2. Type II line integral. Consider a vector-valued function

F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z)) .

Consider the Riemann sum

ST(F) := ∑1≤i≤n

[P(ξi, ηi, zi)∆xi + Q(ξi, ηi, ζi)∆yi + R(ξi, ηi, ζi)∆zi] .

Define ∫L

F · ds =∫

LP(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz

= limλ(T)→0

ST(F).(14.1.3)

If the limit (14.1.3) exists, we say that the integral in (14.1.3) is a type II line inte-gral. The following are basic properties:

(1) If −L denotes the revised curve of L, then∫L

Pdx + Qdy + Rdz = −∫−L

Pdx + Qdy + Rdz.

(2) If integrals ∫L

Pi dx + Qi dy + Ri dz, i = 1, 2,

14.1. LINE INTEGRALS 79

exist, then ∫L(αP1 + βP2)dx + (αQ1 + βQ2)dy + (αR1 + βR2)dz

= α∫

LP1 dx + Q1 dy + R1 dz + β

∫L

P2 dx + Q2 dy + R2 dz.

(3) For any curves L1 and L2, one has∫L1+L2

Pdx + Qdy + Rdz =∫

L1

Pdx + Qdy + T dz +∫

L2

Pdx + Qdy + Rdz.

(4) If L = (x, y, z) ∈ R3 : x = x(t), y = y(t), z = z(t), α ≤ t ≤ β, then theunit tangent vector τ of L at (x, y, z) is

τ =(x(t), y(t), z(t))√

x(t)2 + y(t)2 + z(t)2= (cos α, cos β, cos γ) ∈ S2.

Then∫:Pdx + Qdy + Rdz =

∫L

F τ ds

=∫ β

αP(x(t), y(t), z(t))x(t)dt + Q(x(t), y(t), z(t))y(t)dt(14.1.4)

+ R(x(t), y(t), z(t))z(t)dt.


I =∫

L(−y)dx + xdy.

Let x = cos t and y = sin t with 0 ≤ t ≤ π, then

I =∫ π

0[− sin t(− sin t) + cos t · cos t]dt = π.


I =∫

L(x2 − y2)dx− 2xydy, L = (x, y) ∈ R2 : y = xα, α > 0, 0 ≤ x ≤ 1.

Let x = t and y = tα with 0 ≤ t ≤ 1. Then

I =∫ 1

0(t2 − t2α)dt−

∫ 1

02αt1+αtα−1dt =

∫ 1

0

(t2 − t2α − 2αt2α

)dt

=∫ 1

0

[t2 − (1 + 2α)t2α

]dt =

13− t2α+1

∣∣∣∣10

= −23

.


I =∫

Lxdx + ydy + zdz, Γ = x2 + y2 + z2 = 1 ∩ x + y + z = 0.

Let n denote the normal vector on S2. Then

I =∫

L(x, y, z) · τ ds =

∫L

n · τ ds = 0.


14.2. Surface integrals

Consider a domain Σ ⊂ R3 with representation

x = x(u, v), y = y(u, v), z = z(u, v)

with (u, v) ∈ D ⊂ R2, where D is a closed and bounded domain with smooth orpiecewise smooth boundary ∂D.

Assume that(1) x(u, v), y(u, v), z(u, v) ∈ C2, and write r(u, v) = (x(u, v), y(u, v), z(u, v),(2) the Jacobian matrix [

xu yu zuxv yv zv

]has rank 2. This implies that ru, rv, ru × rv form a basis at each point of D.

14.2.1. Type I surface integrals. Fix a point P1 = (u0, v0) and consider

P2 = (u0 + ∆u, v0), P3 = (u0 + ∆u, v0 + ∆v), P4 = (u0, v0 + ∆v),

and the corresponding points

Q1 = (x(u0, v0), y(u0, v0), z(u0, v0)) ,Q2 = (x(u0 + ∆, v0), y(u0 + ∆u, v0), z(u0 + ∆u, v0)) ,Q3 = (x(u0 + ∆u, v0 + ∆v), y(u0 + ∆u, v0 + ∆v), z(u0 + ∆u, v0 + ∆v)) ,Q4 = (x(u0, v0 + ∆v), y(u0, v0 + ∆v), z(u0, v0 + ∆v))

on Σ. As ∆u, ∆v → 0, we have−−−→Q1Q2 ≈ ru(u0, v0)∆u and

−−−→Q1Q4 ≈ rv(u0, v0)∆v.

Hence the area of Q1Q2Q3Q4 is approximated as−−−→Q1Q2 ×

−−−→Q1Q4 ≈ |ru × rv|(u0, v0)∆u∆v.

As a consequence, we get

(14.2.1) Area(Σ) =∫∫

D|ru × rv|dudv.

Theorem 14.2.1. The area of Σ is equal to

(14.2.2) Area(Σ) =∫∫

D

√EG− F2 dudv

where E = ru · ru, F = ru · rv, and G = rv · rv.

PROOF. Since

ru × rv =

∣∣∣∣∣∣i j k

xu yu zuxv yv zv

∣∣∣∣∣∣ =(

∂(y, z)∂(u, v)

,∂(z, x)∂(u, v)

,∂(x, y)∂(u, v)

)it follows that

|ru × rv|2 = (yuzv − yvzu)2 + (zuxv − zvxu)

2 + (xuyu − xvyu)2 = EG− F2

Thus we get (14.2.2).

Consider the following surface

Σ = (x, y, f (x, y)) : (x, y) ∈ Dand f ∈ C1(D). Then r = (x, y, f (x, y)) and

rx = (1, 0, fx), ry = (0, 1, fy).

14.2. SURFACE INTEGRALS 81

Since EG− F2 = (1 + f 2x )(1 + f 2

y )− f 2x f 2

y = 1 + f 2x + f 2

y = 1 + |∇ f |2 where ∇ f =

( fx, fy), it follows that

(14.2.3) Area f = Area(Σ) =∫∫

D

√1 + |∇ f |2dxdy

by (14.2.2). For any h ∈ C∞0 (D) consider Area f+th for any t > 0. Then Σ is called a

minimal surface if 0 = ddt |t=0Area f+th for any h ∈ C∞

0 (D). It is clear that

(14.2.4) Σ is minimal⇐⇒ div

(∇ f√

1 + |∇ f |2

)= 0

where div(F) := Px + Qy + Rz for F = (P, Q, R) is the divergence of F. Theequation in (14.2.4) is called the minimal surface equation for f .


Assume the surface Σ is given by

Σ = (x, y, z) ∈ R3 : H(x, y, z) = 0

where H ∈ C1, Hz 6= 0 on Σ, and the projection of Σ on xy-plane is a closed andbounded domain D with piecewise smooth boundary ∂D. By the implicit functiontheorem we have

H(x, y, z) = 0⇐⇒ z = f (x, y), (x, y) ∈ D

for some f ∈ C1. Then fx = −Hx/Hz and fy = −Hy/Hz. Hence

Area(Σ) =∫∫

D

√1 + |∇ f |2dxdy =

∫∫D

√1 +

H2x + H2

y

H2z

dxdy =∫∫

D

|∇H||Hz|

dxdy.

For any function we define the type I surface integral of f by

(14.2.5)∫∫

Σf (x, y, z)dS := lim

λ→0∑

1≤i≤nf (ξi, ηi, ζi)∆Si

where Σ = t1≤i≤n∆Σi, (ξi, ηi, ζi) ∈ ∆Σi, |∆Σi| = ∆Si, and λ = max1≤i≤n diam(∆Σi).If Σ is given by

x = x(u, v), y = y(u, v), z = z(u, v), (u, v) ∈ D,

then

(14.2.6)∫∫

Σf (x, y, z)dS =

∫∫D

f (x(u, v), y(u, v), z(u, v))√

EG− F2 dudv.


I =∫∫

Σx2zdS, Σ = (x, y, z) ∈ R3 : z =

√1− x2 − y2.

Then

I =∫∫

x2+y2≤1x2√

1− x2 − y2

√1 +

x2

1− x2 − y2 +y2

1− x2 − y2 dxdy

=∫∫

x2+y2≤1x2 dxdy =

12

∫∫x2+y2≤1

(x2 + y2)dxdy =π

4.


14.2.2. Type II surface integrals. Given a point P0 in the domain Σ ⊂ R3 andchoose a normal vector n0. Σ is called oriented if for any closed curve starting atP0, the normal vector n at any point P on this closed curve is continuously variousand finally returns to n0.

Assume furthermore that Σ is oriented. The unit normal vector is

(14.2.7) n = (cos α, cos β, cos γ) =±1√

EG− F2

(∂(y, z)∂(u, v)

,∂(z, x)∂(u, v)

,∂(x, y)∂(u, v)

)where +1 or −1 depends on the given orientation. Define∫∫

ΣF · ndS =

∫∫Σ[P(x, y, z) cos α + Q(x, y, z) cos β + R(x, y, z) cos γ] dS

=∫∫

Σ[Pdydz + Qdzdx + Rdxdy] ,

where F = (P, Q, R). The following are basic properties∫∫Σ

F · ndS = −∫∫

ΣF · ndS,(14.2.8) ∫∫

Σ(αF + βG) · ndS = α

∫∫Σ

F · ndS + β∫∫

ΣG · ndS,(14.2.9) ∫∫

Σ1+Σ2

F · ndS =∫∫

Σ1

F · ndS +∫∫

Σ2

F · ndS.(14.2.10)


I =∫∫

Σx3dydx + y3dzdx + z3dxdy, Σ =

x2

a2 +y2

b2 +z2

c2 = 1, z ≥ 0

.

Here a, b, c > 0. Let

x = a sin ϕ cos θ, y = b sin ϕ sin θ, z = c cos ϕ, 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π

2.

Then

I = abc∫ π/2

0dϕ∫ 2π

0

(a2 sin5 ϕ cos4 θ + b2 sin5 ϕ sin4 θ + c2 sin ϕ cos4 ϕ

)dθ.

Using the formula ∫ π/2

0cos2p θ sin2q θ dθ =

12

B(

p +12

, q +12

),

we obtain

I = abc[

a2 · B(

3,12

)· B(

12

,52

)+ b2 · B

(3,

12

)· B(

12

,52

)+ c2πB

(52

, 1)]

= abc(

a2 + b2 + c2) πΓ(5/2)

Γ(7/2)=

2π

5abc(

a2 + b2 + c2)

.

14.3. Green, Gauss, and Stokes formulas

In this section we generalize the Netwon-Leibniz formula∫ b

af ′(x)dx = f (b)− f (a)

to higher-dimensional cases.

14.3. GREEN, GAUSS, AND STOKES FORMULAS 83

14.3.1. Green’s formula. A Jordan curve is a curve L ⊂ R2 with parametriza-tion r(t) := (x(t), y(t)) for α ≤ t ≤ β and r(α) = r(β) and r(t1) 6= r(t2) for anyt1 6= t2 with t1, t2 ∈ (α, β).

Given a domain D ⊂ R2 let

Ω(D, p) := f ∈ C(I, D) : f (0) = f (1) = pwhere p is a point in D and the notion f ∈ C(I, D) means that f : I → D is acontinuous function from I := [0, 1] into D. The constant map cp : I → D withcp(s) := p is an element of Ω(D, p). We say that f is homotopic to g, denoted byf ∼ g, if there exists a continuous map H : I × I → D satisfying

H(s, 0) = f (s), G(s, 1) = g(s), f (0) = g(0) = H(0, t) = H(1, t) = f (1) = g(1).

Exercise 14.3.1. Show that ∼ is an equivalence relation in Ω(D, p).

Let [ f ] := g ∈ Ω(D, p) : g ∼ f be the equivalence class of f and define

π1(D, p) := [ f ] : f ∈ Ω(D, p).For f , g ∈ Ω(D, p) define f ∗ g by

( f ∗ g)(s) :=

f (2s), 0 ≤ s ≤ 12 ,

g(1− 2s), 12 ≤ s ≤ 1.

Define[ f ] · [g] := [ f ∗ g].

Exercise 14.3.2. Show that the above product · does not depend on the choice ofrepresents. Hence (π1(D, p), ·) is a group.

We say that D is simply-connected if π1(D, p) = [cp], or equivalently, iff ∼ cp for any f ∈ Ω(D, p).

Theorem 14.3.3. (Green) Suppose that D ⊂ R2 is a simply-connected domain enclosedby a smooth or piecewise smooth closed Jordan curve. If P, Q ∈ C1(D), then

(14.3.1)∫

∂DPdx + Qdy =

∫∫D

(∂Q∂x− ∂P

∂y

)dxdy.

Where we have given an orientation on the boundary ∂D.

PROOF. We give a proof for the domain

D =(x, y) ∈ R2 : ϕ(y) ≤ x ≤ ψ(y), c ≤ y ≤ d

.

Compute∫∂D

Qdy =∫

ABQ(x, y)dy +

∫BC

Q(x, y)dy +∫

CEQ(x, y)dy +

∫EA

Q(x, y)dy

=∫ d

cQ(ψ(y), y)dy−

∫ d

cQ(ϕ(y), y)dy =

∫ d

cdy∫ ψ(y)

ϕ(y)

∂Q∂x

dx

=∫∫

D

∂Q∂x

dxdy,

where

A = (ϕ(c), c), B = (ψ(c), c), C = (ψ(d), d), D = (ϕ(d), d).


Similarly, for the domain

D =(x, y) ∈ R2 : a ≤ x ≤ b, ϕ(x) ≤ y ≤ ψ(x)

one has ∫

∂DPdx = −

∫∫D

∂P∂y

dxdy.

If the domain D can be written as

D =(x, y) ∈ R2 : ϕ(y) ≤ x ≤ ψ(y), c ≤ y ≤ d

=

(x, y) ∈ R2 : a ≤ x ≤ b, ϕ(x) ≤ y ≤ ψ(x)

then ∫

∂DPdx + Qdy =

∫∫D

(∂Q∂x− ∂P

∂y

)dxdy.

Let us assume that the domain D can be cut into a finite number of domainsof type Dy = (x, y) ∈ R2 : ϕ(y) ≤ x ≤ ψ(y), c ≤ y ≤ d. By additivity,the double integral over the domain D is the sum of the integrals over the piecesof type Dy into which D is divided. Formula (14.3.1) holds for each such piece.But the adjacent pieces induce opposite orientations on their common boundary,so that when the integrals over the boundaries are added, all that remains aftercancellatyion is the integral over the boundary ∂D.

Similarly if D admits a partition into domains of type Dx = (x, y) ∈ R2 : a ≤x ≤ b, ϕ(x) ≤ y ≤ ψ(x) the formula (14.3.1) still holds.

If the domain D ⊂ R2 contains a subdomain D′, then D \D′ induces a naturalorientation for ∂(D \ D′). By cutting D \ D′ along a line contained in D we easilysee that the formula (14.3.1) still holds in this case.

Assume that the boundary ∂D is a smooth curve parametrized by

x = x(t), y = y(t), α ≤ t ≤ β.

If τ denotes the unit tangent vector, then

dx = x(t)dt =x(t)√

x(t)2 + y(t)2

√x(t)2 + y(t)2dt = cos(τ, x)ds

where (τ, x) stands for the angle between τ and the positive x-axis. Similarly, wehave

dy = y(t)dt = cos(τ, y)dswhere (τ, y) stands for the angle between τ and the positive y-axis. If n is the(outward) unit normal vector, then

cos(τ, x) = − cos(n, y), sin(τ, y) = cos(n, x).

If F, G ∈ C1(D), then∫∫D

(∂F∂x

+∂G∂y

)dxdy =

∫∂D

Fdy− Gdx

=∫

∂D[F sin(τ, y)− G cos(τ, x)] ds

=∫

∂D[F cos(n, x) + G cos(n, y)] ds.


Remark 14.3.4. If d ∈ C1[a, b] and D := [a, b]× [0, 1], then Green’s formula (14.3.1)implies∫ b

af ′(x)dx =

∫ 1

0dy∫ b

af ′(x)dx =

∫∫D

f ′(x)dxdy

=∫

∂Df (x)dy =

∫ 1

0f (b)dy +

∫ 0

1f (a)dy = f (b)− f (a).

If D is a simply-connected domain with smooth boundary ∂D, then

(14.3.2) Area(D) =∫∫

Ddxdy =

∫∂D

xdy = −∫

∂Dydx =

12

∫∂D

xdy− ydx.


I =∫

L2xydx + (x2 − y2)dy

where L is any smooth curve from A = (0, 0) to B = (0, 1). Choosing a smooth curveabove x-axis, we get

−I +∫

AB2xydx + (x2 − y2)dy =

∫∫D

(∂Q∂x− ∂P

∂y

)dxdy = 0

where D is the domain enclosed by AB and −L, and P = 2xy and Q = x2 − y2.

14.3.2. Gauss-Ostrogradskii formula. The Gauss-Ostrogradskii formula con-nects the integral over the boundary of a three-dimensional domain with an inte-gral over the domain itself.

Theorem 14.3.6. (Gauss-Ostrogradskii) Let D ⊂ R3 be a closed and bounded domainwith smooth or piecewise smooth boundary ∂D. If P, Q, R ∈ C1(D), then

(14.3.3)∫∫

∂DPdydz + Qdzdx + Rdxdy =

∫∫∫D

(∂P∂x

+∂Q∂y

+∂R∂z

)dxdydx.

PROOF. We give a proof for the domain

D =(x, y, z) ∈ R3 : (y, z) ∈ Ω, ϕ(y, z) ≤ z ≤ ψ(y, z)

where Ω ⊂ R2 is a closed and bounded domain with smooth or piecewise smoothboundary ∂Ω and ϕ, ψ ∈ C(Ω). We claim that∫∫

∂DPdydz =

∫∫∫D

∂P∂x

dxdydz.

Let

S1 := (x, y, z) ∈ R3 : x = ψ(y, z), (y, z) ∈ Ω,S2 := (x, y, z) ∈ R3 : x = ϕ(y, z), (y, z) ∈ Ω,

with the naturally induced orientations. Then∫∫∫D

∂P∂x

dxdydz =∫∫

Ωdydz

∫ ψ(y,z)

ϕ(y,z)

∂P∂x

dx

=∫∫

Ω[P(ψ(y, z), y, z)− P(ϕ(y, z), y, z)] dydz

=∫∫

S1

Pdydz +∫∫

S2

Pdydz =∫∫

∂DPdydz.


Similarly if D = (x, y, z) ∈ R3 : (x, y) ∈ Ω, ϕ(x, y) ≤ z ≤ ψ(x, y), then∫∫∫D

∂R∂z

dxdydz =∫∫

∂DRdxdy;

if D = (x, y, z) ∈ R3 : (z, x) ∈ Ω, ϕ(z, x) ≤ y ≤ ψ(z, x), then∫∫∫D

∂Q∂y

dxdydz =∫∫

∂DQdzdx.

If the domain D is a simple domain, that is, a domain admits each of the threetypes of partitions described above, we obtain (14.3.3).

If D ⊂ R3 is a closed and bounded domain with smooth boundary ∂D, then

Vol(D) =∫∫∫

Ddxdydz =

∫∫∂D

xdydz =∫∫

∂Dydzdx =

∫∫∂D

zdxdy

=∫∫

∂D

xdydz + ydzdx + zdxdy3

.(14.3.4)

Example 14.3.7. Let S ⊂ R3 be a closed and bounded surface and (ξ, η, ζ) ∈ R3 \ S afixed point. Compute the Gauss integral

(14.3.5) I(ξ, η, ζ) :=∫∫

S

cos(r, n)|r|2 dS

where r = (x− ξ, y− η, z− ζ) and n is the unit outward normal vector on S. Since

cos(r, n) =x− ξ

rcos(n, x) +

y− η

rcos(n, y) +

z− ζ

rcos(n, z)

where r := |r|, it follows that

I(ξ, η, ζ) =∫∫

S

[x− ξ

r3 cos(n, x) +y− η

r3 cos(n, y) +z− ζ

r3 cos(n, z)]

dS

=∫∫

S

x− ξ

r3 dydz +y− η

r3 dzdx +z− ζ

r3 dxdy.

If (ξ, η, ζ) /∈ D, the domain enclosed by S, then

I(ξ, η, ζ) =∫∫∫

D

(3r3 −

3|r|2r5

)dxdydz = 0.

If (ξ, η, ζ) ∈ D, we can take a sphere Sε := (x, y, z) ∈ R3 : (x − ξ)2 + (y − η)2 +(z− ζ)2 = ε2 ⊂ Int(D) for some sufficiently small number ε 1. Hence

I(ξ, η, ζ) =∫∫

Sε

cos(r, n)r2 dS =

∫∫Sε

dSε2 =

1ε2 · 4πε2 = 4π.

14.3.3. Stokes’ formula. Let S ⊂ R3 be an oriented smooth compact surfacewith smooth or piecewise smooth boundary ∂S. Choose a unit outward normalvector n on S. We say that the boundary ∂S is inherited an orientation from S if onevery point of ∂S, the unit tangent vector and the normal vector n form a naturalcoordinates system.


Theorem 14.3.8. (Stokes) Let S ⊂ R3 be an oriented smooth compact surface withsmooth or piecewise smooth boundary ∂S inherited an orientation from S. If P, Q, R ∈C1(S), then

(14.3.6)∫

∂SPdx + Qdy + Rdz =

∫∫S

∣∣∣∣∣∣dydz dzdx dxdy

∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣ .

PROOF. We prove (14.3.6) for the surface S given by S = (x, y, z) ∈ R3 : z =f (x, y), (x, y) ∈ D where D ⊂ R2 is the domain enclosed by piecewise smoothJordan curve. Then∫

∂SPdx =

∫∂D

P(x, y, f (x, y)dx = −∫∫

D

(∂P∂y

+∂P∂z

∂ f∂y

)dxdy

by Theorem 14.3.3. On the other hand,∫∫S

∂P∂x

dzdx− ∂P∂y

dxdy = −∫∫

D

(∂P∂y

+∂P∂z· ∂ f

∂y

)dxdy.

Hence ∫∂S

Pdx =∫∫

S

∂P∂z

dzdx− ∂P∂y

dxdy.

The same argument shows∫∂S

Qdy =∫∫

S

(−∂Q

∂z

)dydz +

∂Q∂x

dxdy.

Thus ∫∂S

Pdx + Qdy =∫∫

S

∂P∂z

dzdx− ∂Q∂z

dydz +(

∂Q∂x− ∂P

∂y

)dxdy.

If the surface is given by S = (x, y, z) ∈ R3 : x = g(y, z), (y, z) ∈ D, then∫∂s

Qdy +∫

∂SRdz =

∫∫S

(∂R∂y− ∂Q

∂z

)dydz +

∂Q∂x

dxdy− ∂R∂x

dzdx;

if the surface is given by S = (x, y, z) ∈ R3 : y = h(z, x), (z, x) ∈ D, then∫∂S

Pdx +∫

∂SRdz =

∫∫S

(∂P∂z− ∂R

∂x

)dzdx +

∂R∂y

dydz− ∂P∂y

dxdy.

If the surface is a simple surface, that is, a surface admits each of the three types ofpartitions described above, we obtain (14.3.6).

The formula can also be written as

(14.3.7)∫

∂SPdx + Qdy + Rdz =

∫∫S

∣∣∣∣∣∣cos α cos β cos γ

∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣ dS,

where α = (n, x), β = (n, y), and γ = (n, z).


14.4. Differential forms: revisited

Let U ⊂ Rn be a domain and ω ∈ Λk∞(U). Recall that

dω = ∑1≤i1<···<ik≤n

(dωi1···ik

)∧ dxi1 ∧ · · · ∧ dxik

where ω = ∑1≤i1<···<ik≤n ωi1···ik dxi1 ∧ · · · ∧ dxik .Define the graded algebra

(14.4.1) Λ∞(U) := ∑0≤k≤n

Λk∞(U)

and the linear map

(14.4.2) d : Λ∞(U) −→ Λ∞(U), ω 7−→ dω.

We have proved that d2 = 0.

If ω = Pdx + Qdy ∈ Λ1∞(R2), then

dω =

(∂Q∂x− ∂P

∂y

)dx ∧ dy.

If ω = Pdx + Qdy + Rdz ∈ Λ1∞(R3), then

dω =

(∂R∂y− ∂Q

∂z

)dy ∧ dz +

(∂P∂z− ∂R

∂x

)dz ∧ dx +

(∂Q∂x− ∂P

∂y

)dx ∧ dy.

If ω = Pdy ∧ dz + Qdz ∧ dx + Rdx ∧ dy ∈ Λ2∞(R3), then

dω =

(∂P∂x

+∂Q∂y

+∂R∂z

)dx ∧ dy ∧ dz.

Theorem 14.4.1. (Leibniz rule) If ω ∈ Λk∞(U) and η ∈ Λ`

∞(U), then

(14.4.3) d(ω ∧ η) = dω ∧ η + (−1)kω ∧ dη.

PROOF. Write ω = ωi1···ik dxi1 ∧ · · · ∧ dxik and η = ηj1···j`dxj1 ∧ · · · ∧ dxj` andcompute

d(ω ∧ η) = d(ωi1···ik ηj1···j`dxi1 ∧ · · · ∧ dxik ∧ dxj1 ∧ · · · ∧ dxj`

)=

(∂ωi1···ik

∂xidxiηj1···j` +

∂ηj1···j`∂xi

dxiωi1···ik

)∧dxi1 ∧ · · · ∧ dxik ∧ dxj1 ∧ · · · ∧ dxj`

= dω ∧ η

+ ωi1···ik dxi1 ∧ · · · ∧ dxik ∧(

∂ηj1···j`∂xi

(−1)kdxi1 ∧ dxj1 ∧ · · · ∧ dxj`

)= dω ∧ η + (−1)kω ∧ dη.


14.4. DIFFERENTIAL FORMS: REVISITED 89

14.4.1. Applications. Write Λk`(U) the space of C`-differential k-forms on a

domain U ⊂ Rn.

If we view dx ∧ dy the positive are element dxdy, then Green’s formula statesthat

(14.4.4)∫

∂Dω =

∫D

dω

where ω = Pdx + Qdy ∈ Λ11(D). Similarly, we have the following versions of

Stokes and Gauss formulae:

(14.4.5)∫

∂Sω =

∫S

dω,

where ω = Pdx + Qdy + Rdz ∈ Λ11(S), and

(14.4.6)∫

∂Ω=∫

Ωdω,

where ω = Pdy ∧ dz + Qdz ∧ dx + Rdx ∧ dy ∈ Λ21(Ω), respectively.

14.4.2. Line integrals independent of paths. Let D ⊂ R3 be a domain andu ∈ C1(D). For any smooth curve Γ ⊂ D from A0 to A, we have∫

Γdu =

∫∂Γ

u = u(A)− u(A0)

which is independent of the choice of Γ.

Definition 14.4.2. Let D ⊂ R3 be a domain and P, Q, R ∈ C(D). If for any A0, A ∈D and any piecewise smooth curve Γ ⊂ D from A0 to A, the line integral∫

ΓPdx + Qdy + Rdz

depends only on A0 and A, then we say that this line integral is independent ofthe choice of Γ in D.

Theorem 14.4.3. Let D ⊂ R3 be a domain and P, Q, R ∈ C(D). Then the following areequivalent (TFAE):

(1) The line integral ∫Γ

Pdx + Qdy + Rdz

is independent of the choice of Γ in D.(2) For any piecewise smooth Jordan curve Γ ⊂ D, we have∫

ΓPdx + Qdy + Rdz = 0.

(3) There exists a function u ∈ C1(D) so that Pdx + Qdy + Rdz = du.

PROOF. (1)⇔ (2) is clear.(1)⇒ (3) : Given A0 = (x0, y0, z0) ∈ D. For any A = (x, y, z) ∈ D define

u(x, y, z) :=∫

Γ0

P(ξ, η, ζ)dξ + Q(ξ, η, ζ)dη + R(ξ, η, ζ)dζ,


where Γ0 is a fixed smooth curve from A0 to A. Since A ∈ D, there is a sufficientlysmall ∆x such that A∆x := ()x + ∆x, y, z) ∈ D and hence AA∆x ⊂ D. Let Γ∆x :=Γ ∪ AA∆x. Compute

u(x + ∆x, y, z)− u(x, y, z) =∫

Γ∆x

Pdξ + Qdη + Rdζ −∫

Γ0

Pdξ + Qdη + Rdζ

=∫

AA∆x

Pdξ + Qdη + Rdζ

=∫ x+∆x

xP(ξ, η, ζ)dξ = P(x + θ∆x, y, z)∆x

for some θ ∈ (0, 1). Hence ux = P. Similarly, one has uy = Q and uz = R.(3) ⇒ (1) : Suppose u ∈ C1(D) with du = Pdx + Qdy + Rdz. Let A0 :=

(x0, y0, z0) and A1 = (x1, y1, z1) ∈ D. For any piecewise smooth path Γ from A0 toA1, parametrized by

x = x(t), y = y(t), z = z(t), α ≤ t ≤ β.

Then∫Γ

Pdx + Qdy + Rdz =∫ β

α

[P(x(t), y(t), z(t))x(t) + Q(x(t), y(t), z(t))y(t)

+ R(x(t), y(t), z(t))z(t)]

dt

=∫ β

α

ddt

u(x(t), y(t), z(t))dt

= u(x1, y1, z1)− u(x0, y0, z0).

Hence the line integral is independent of the choice of Γ.

Theorem 14.4.4. Let D ⊂ R3 be a simply-connected domain and P, Q, R ∈ C1(D).Then TFAE:

(1) The line integral ∫Γ

Pdx + Qdy + Rdz

is independent of the choice of Γ in D.(2) Ry = Qz, Pz = Rx, and Qx = Py.

PROOF. (1) ⇒ (2) : In this case we have du = Pdx + Qdy + Rdz for someu ∈ C1(D) by Theorem 14.4.3. Then ux = P, uy = Q, and uz = R. Henceu ∈ C2(D) and

Ry = (uz)y = (uy)z = Qz.

Similarly, Pz = Rx and Qx = Py.(2) ⇒ (1) : Let Γ be any given piecewise smooth Jordan curve in D. The

assumption that D is simply-connected, implies there exists a piecewise smoothoriented surface S so that ∂S = Γ. By Stokes formula we get

∫Γ

Pdx + Qdy + Rdz =∫∫

S

∣∣∣∣∣∣dydz dzdx dxdy

∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣ = 0.

14.5. DIFFERENT OPERATORS 91

By Theorem 14.4.3, we conclude that the line integral is independent of the choiceof Γ in D.

For dimension two, using Green’s formula, we have

Theorem 14.4.5. (1) Let D ⊂ R2 be a domain and P, Q ∈ C(D). Then TFAE(a) The line integral ∫

ΓPdx + Qdy

is independent of the choice of Γ in D.(b) For any piecewise smooth Jordan curve Γ in D, we have∫

ΓPdx + Qdy = 0.

(c) du = Pdx + Qdy for some u ∈ C1(D).(2) Let D ⊂ R2 be a simply-connected domain and P, Q ∈ C1(D). Then TFAE

(a) The line integral ∫Γ

Pdx + Qdy

is independent of the choice of Γ in D.(b) Qx = Py in D.


I =∫

Γ

xdy− ydxx2 + y2

where Γ is a Jordan curve not containing (0, 0). Let

P(x, y) :=−y

x2 + y2 , Q(x, y) :=x

x2 + y2 .

If 0 /∈ D, then

I =∫

ΓPdx + Qdy =

∫∫S

(∂Q∂x− ∂P

∂y

)dxdy = 0.

Here D is the domain enclosed by Γ. If 0 ∈ D, then

I =∫

Sε

Pdx + Qdy =∫ 2π

0dθ = 2π,

where Sε := (x, y) ∈ R2 : x2 + y2 = ε for ε 1.

14.5. Different operators

Let Ω ⊂ R3 be a domain.(1) For f ∈ C1(Ω), define

(14.5.1) grad( f ) := ( fx, fy, fz) = fxi + fy j + fzk.

(2) For F = (P, Q, R) with P, Q, R ∈ C1(Ω), define

(14.5.2) div(F) := Px + Qy + Rz.

Then the Gauss formula can be written as∫∫∫Ω

div(F)dV =∫∫

∂ΩF · ndS.


(3) For F = (P, Q, R) with P, Q, R ∈ C1(Ω), define

(14.5.3) rot(F) :=

∣∣∣∣∣∣i j k∂

∂x∂

∂y∂∂z

P Q R

∣∣∣∣∣∣ .

Then the Stokes’ formula can be written as∫∂S

F · ds =∫∫

Srot(F) · ndS.

14.5.1. Hamilton’s operators. Define

(14.5.4) ∇ :=(

∂

∂x,

∂

∂y,

∂

∂z

)=

∂

∂xi +

∂

∂yj +

∂

∂zk.

For f , P, Q, R ∈ C1(D) and F = (P, Q, R), we have

∇ f := grad( f ),(14.5.5)∇ · F := div(F),(14.5.6)∇× F := rot(F).(14.5.7)

For any f ∈ C2(D) one has

(14.5.8) ∇ ·∇ f = div grad( f ) = div ∇( f ) = ∆ f ,

where ∆ := ∂2

∂x2 +∂2

∂y2 +∂2

∂z2 denotes the Laplacian.

Exercise 14.5.1. Let D ⊂ R3 be a domain, f , P, Q, R, U, V, W ∈ C2(D) with F =(P, Q, R) and G = (U, V, W), and λ, µ be constant. Prove that

(i) div(λF + µG) = λdiv(F) + µdiv(G)(ii) ∇ · (λF + µG) = λ(∇ · F) + µ(∇ ·G)

(iii) rot(λF + µG) = λrot(F) + µrot(G)(iv) ∇× (λF + µG) = λ(∇× F) + µ(∇×G).(v) div( f F) = f div(F) +∇ f · F and ∇ · ( f F) = f (∇ · F) + (∇ f ) · F

(vi) rot( f F) = f rot(F) +∇ f × F and ∇× ( f F) = f (∇× F) +∇ f × F(vii) div(F ×G) = G · rot(F)− F · rot(G)

(viii) ∇ · (F ×G) = G · (∇× F)− F · (∇×G)(x) rot(∇ f ) = ∇× (∇ f ) = div(rot(F)) = ∇ · (∇× F) = 0

14.5.2. Green’s identities. For any f , g ∈ C2(D) one has∫∫∫D(∇ f ·∇g + f ∆g) dV =

∫∫∂D

f∂g∂n

dS,(14.5.9) ∫∫∫D( f ∆g− g∆g) dV =

∫∫∂D

(f

∂g∂n− g

∂ f∂n

)dS.(14.5.10)

The second identity follows from the first one. To prove (14.5.9) we compute∫∫∫D(∇ f ·∇g + f ∆g) dV =

∫∫∫D∇ · ( f∇g)dV

=∫∫

∂Df∇g · ndS =

∫∫∂D

f∂g∂n

dS.

As a consequence of (14.5.9) we obtain the divergence theorem:

(14.5.11)∫∫∫

D∆ f dV =

∫∫∂D

∂g∂n

dS.

PROBLEMS 93

14.6. Stokes’ theorem on manifolds

To be constructed.

Problems

CHAPTER 15

Integrals depending on a parameter

15.1. Proper integrals depending on a parameter

Integral depending on a parameter means that

F(t) :=∫

Etf (x, t)dx, t ∈ T.

There are two cases:(A) proper: For any t ∈ T, the function ϕt(x) := f (x, t) is integrable over Et

in the proper sense.(B) Improper: The function ϕt(x) is integrable over Et in the improper sense

for some or all t ∈ T.

15.1.1. Continuity. Consider a continuous function f on [a, b]× [c, d] and itsintegral

(15.1.1) F(y) :=∫ b

af (x, y)dx

for each y ∈ [c, d].

Theorem 15.1.1. If f ∈ C([a, b]× [c, d]), then F ∈ C[c, d].

PROOF. The continuity of f implies f is uniformly continuous in [a, b]× [c, d].For any ε > 0 there is a positive number δ > 0 such that

| f (x1, y1)− f (x2, y2)| < ε

whenever√(x1 − x2)2 + (y1 − y2)2 < δ. Given y0 ∈ [c, d]. For any y ∈ (y0 −

δ, y0 + δ) we have

|F(y)− F(y0)| ≤∫ b

a| f (x, y)− f (x, y0)|dx < (b− a)ε.

Hence F is continuous at y0.

Corollary 15.1.2. If f ∈ C([a, b]× [c, d]), then

(15.1.2) limy→y0

∫ b

af (x, y)dx =

∫ b

alim

y→y0f (x, y)dx.

Corollary 15.1.3. (Hadamard’s lemma) If f ∈ C1(I), where I ⊂ R is an open intervalcontaining x0 ∈ R, then

f (x) = f (x0) + ϕ(x)(x− x0)

in some neighborhood of x0, where ϕ is continuous and ϕ(x0) = f ′(x0).

95

96 15. INTEGRALS DEPENDING ON A PARAMETER

PROOF. Recall that

f (x0 + h)− f (x0) =

[∫ 1

0f ′(x0 + th)dt

]h

for sufficiently small h. Define

ϕ(x) :=∫ 1

0f ′(x0 + t(x− x0))dt.

By Theorem 15.1.1, ϕ is continuous and ϕ(x0) = f ′(x0).

15.1.2. Integration of an integral depending on a parameter. If f is continu-ous on [a, b]× [c, d]. Then

F(y) =∫ b

af (x, y)dx

is continuous on [c, d] by Theorem 15.1.1, and consequently, F(y) is integrable.Hence, according to Theorem 13.2.2, we obtain

Theorem 15.1.4. If f ∈ C([a, b]× [c, d]), then

(15.1.3)∫ d

cdy∫ b

af (x, y)dx =

∫∫[a,b]×[c,d]

f (x, y)dxdy =∫ b

adx∫ d

cf (x, y)dy.


I =∫ 1

0

xb − xa

ln xdx, b > a > 0.

Sincexb − xa

ln x=∫ b

xy dy,

we get

I =∫ 1

0dx∫ b

axy dy =

∫ b

ady∫ 1

0xy dx =

∫ b

a

dy1 + y

= ln1 + b1 + a

.

15.1.3. Differentiation. Consider the function

F(a) :=∫ ∞

0e−ax dx, a > 0.

Then F(a) = 1/a and F′(a) = −1/a2. However∫ ∞

0

ddx

e−ax dx =∫ ∞

0−ae−ax dx = −1.

Therefore, in general we might not interchange the differentiation and integration.

Theorem 15.1.6. If f ∈ C([a, b] × [c, d]) and fy ∈ C([a, b] × [c, d]), then F(y) ∈C1[c, d] and

(15.1.4) F′(y) =∫ b

afy(x, y)dx.

15.1. PROPER INTEGRALS DEPENDING ON A PARAMETER 97

PROOF. Given y0 ∈ [c, d]. We have∣∣∣∣F(y0 + h)− F(y0)−[∫ b

afy(x, y0)dx

]h∣∣∣∣

=

∣∣∣∣∫ b

a

[f (x, y0 + h)− f (x, y0)− fy(x, y0)h

]dx∣∣∣∣

≤∫ b

amax

0≤θ≤1| fy(x, y0 + θh)− fy(x, y0)||h|dx.

Let

ϕ(h) :=∫ b

amax

0≤θ≤1| fy(x, y0 + θh)− fy(x, y0)|dx.

Since fy is continuous, it follows that fy(x, y) ⇒ fy(x, y0) on [a, b] as y→ y0. Henceϕ(h)→ 0 as h→ 0. Thus we get (15.1.4).

Similarly, if f ∈ C([a, b]× [c, d]) and fx ∈ C([a, b]× [c, d]), then

(15.1.5)d

dx

∫ d

cf (x, y)dy =

∫ d

cfx(x, y)dy.

Theorem 15.1.7. Suppose f , fy ∈ C([a, b]× [c, d]) and a(y), b(y) are differentiable on[c, d] and a ≤ a(y), b(y) ≤ b. Then

F(y) :=∫ b(y)

a(y)f (x, y)dx

is differentiable and moreover

(15.1.6) F′(y) =∫ b(y)

a(y)fy(x, y)dx + f (b(y), y)b′(y)− f (a(y), y)a′(y).

PROOF. Let α = a(y), β = b(y), and

F(y) =∫ β

αf (x, y)dx =: Φ(y, β, α).

Then Φβ = f (β, y), Φα = − f (α, y), and

Φy =∫ β

αfy(x, y)dx.

Therefore

F′(y) =∂

∂yΦ(y, β, α) = Φy + Φββ′ + Φαα′

=∫ β

αfy(x, y)dx + f (β, y)b′(y)− f (α, y)α′(y).


Example 15.1.8. (Bessel’s equation) The function

(15.1.7) u(x) :=∫ π

0cos(nϕ− x sin ϕ)dϕ

satisfies

(15.1.8) x2u′′(x) + xu′(x) + (x2 − n2)u(x) = 0.


Example 15.1.9. (Elliptic integral) Compute the circumference of

x2

a2 +y2

b2 = 1, b > a > 0.

Let x = a cos θ and y = b sin θ. Then

L = 4∫ π/2

0

√a2 sin2 θ + b2 cos2 θ dθ = 4b

∫ π/2

0

√1− k2 sin2 θ dθ

where k =√

b2 − a2/b.The 2nd complete elliptic integral is given by

(15.1.9) E(k) :=∫ π/2

0

√1− k2 sin2 θ dθ =

∫ 1

0

√1− k2x2√

1− x2dx, 0 ≤ k ≤ 1.

Hence

L = 4bE

(√b2 − a2

b

).

Note that we actually have

(15.1.10) E(k) =π

2 ∑n≥0

[(2n)!

22n(n!)2

]2 k2n

1− 2n.

The 1st complete elliptic integral is given by

(15.1.11) K(k) :=∫ π/2

0

dθ√1− k2 sin2 θ

=∫ 1

0

dx√(1− x2)(1− k2x2)

, 0 ≤ k ≤ 1.

It can be shown that

(15.1.12) K(k) =π

2 ∑n≥0

[(2n)!

22n(n!)2

]2

k2n.

The 3rd complete elliptic integral is given by

(15.1.13) Π(n, k) :=∫ π/2

0

dθ√1− k2 sin2 θ(1− n sin2 θ)

, 0 ≤ k ≤ 1.

Exercise 15.1.10. Verify that

E′(k) =R(k)− K(k)

k,

K′(k) =E(k)

k(1− k2)− K(k)

k,

Πn(n, k) =E(k) + k2−n

n K(k) + n2−k2

n Π(n, k)2(k2 − n)(n− 1)

,

Πk(n, k) =k

n− k2

[E(k)

k2 − 1+ Π(n, k)

].

Example 15.1.11. (Gauss) Let a1 > b1 > 0 and define

(15.1.14) an+1 =an + bn

2, bn+1 =

√anbn.

Then Gauss shows that(i) an is decreasing and bn is increasing,

15.1. PROPER INTEGRALS DEPENDING ON A PARAMETER 99

(ii) limn→∞ an = limn→∞ bn =: L(a1, b1) exist,(iii) L(a1, b1) =

π2G(a1,b1)

where

(15.1.15) G(a1, b1) :=∫ π/2

0

dx√a2

1 cos2 x + b21 sin2 x

=1a1

K

√ a21 − b2

1a1

.

It is clear that (an)n∈N is decreasing and bounded and (bn)n∈N is increasing andbounded. Let

α := limn→∞

an, β := limn→∞

bn.

But

an+1 =an + bn

2and so, for n → ∞, we get α = β. We denote this common limit by L = L(a1, b1). LetG := G(a1, b1) be given in (15.1.15). We put

sin x =2a1 sin t

(a1 + b1) + (a1 − b1) sin2 t;

as t change from 0 to π/2, so grows x from 0 to π/2. Differentiation gives

cos xdx = 2a1(a1 + b1)− (a1 − b1) sin2 t

[(a1 + b1) + (a1 − b1) sin2 t]2cos tdt.

Hencedx√

a21 cos2 x + b2

1 sin2 x=

dt√[(a1 + b1)/2]2 cos2 t + a1b1 sin2 t

.

Putting a2 = (a1 + b1)/2 and b2 =√

a1b1 we get

G =∫ π/2

0

dx√a2

1 cos2 x + b21 sin2 x

=∫ π/2

0

dt√a2

2 cos2 t + b22 sin2 t

.

By repeated application of this construction we arrive at

G =∫ π/2

0

dx√a2

n cos2 x + b2n sin2 x

, n ∈ N.

Here an and bn are defined by (15.1.14). Since π/2an < G < π/2bn, letting n → ∞yields

G =π

2L(a1, b1).

We consider an application of this formula and compute the integral

G =∫ π/2

0

dx√1 + cos2 x

=∫ π/2

0

dx√2 cos2 x + sin2 x

.

Here a1 =√

2 and b1 = 1; the numerical sequences (an)n∈N and (bn)n∈N convergerapidly to L(

√2, 1) in this case and a5 and b5 are approximately equal to 1.198154. Hence

we may put L(√

2, 1) ≈ 1.198154 and obtain the approximate value

G =π

2L(√

2, 1)≈ 1.3110138.


15.2. Improper integrals depending on a parameter

Let f (x, y) be a function defined on [a, ∞)× [c, d]. Suppose for any y ∈ [c, d]the improper integral

(15.2.1) F(y) :=∫ ∞

af (x, y)dx

exists. We say the integral (15.2.1) converges uniformly in y ∈ [c, d], if for anyε > 0 there exists A0 > 0 such that

maxc≤y≤d

∣∣∣∣∫ ∞

Af (x, y)dx

∣∣∣∣ < ε

whenever A > A0.

Let f (x, y) be a function defined on [a, b) × [c, d]. Suppose for any y ∈ [c, d]the improper integral

(15.2.2) F(y) :=∫ b

af (x, y)dx

exists. We say that the integral (15.2.2) converges uniformly in y ∈ [c, d], if for anyε > 0 there exists δ > 0 such that

maxc≤y≤d

∣∣∣∣∫ b

b−ηf (x, y)dx

∣∣∣∣ < ε

whenever 0 < η < δ.

Example 15.2.1. (1) Show that ∫ ∞

1

dxx2 + y2

converges uniformly in y ∈ R. In fact,∫ ∞

b

dxx2 + y2 ≤

∫ ∞

b

dxx2 =

1b< ε

for b > 1/ε.(2) Show that ∫ ∞

0e−xy dx

converges uniformly on every set [y0, ∞), where y0 > 0. However it does not uniformlyconverge on (0, ∞). If y ≥ y0 > 0, then∫ ∞

be−xy dx =

1y

e−by ≤ 1y0

e−by0 → 0

as b→ ∞. However ∫ ∞

be−xy dx =

1y

e−by → ∞

as y→ 0+.(3) Show that

Φ(x) =∫ ∞

0xαyα+β+1e−(1+x)y dy, α, β > 0

15.2. IMPROPER INTEGRALS DEPENDING ON A PARAMETER 101

converges uniformly in y ∈ [0, ∞). Compute

Φ(x) =∫ ∞

b(xy)αe−xyyβ+1e−y dy ≤ Mα

∫ ∞

byβ+1e−y dy

where Mα = maxt≥0 tα/et = αα/eα. Since∫ ∞

0yβ+1e−y dy = Γ(β + 2),

it follows that ∫ ∞

byβ+1e−y dy < ε

when b 1.

15.2.1. Cauchy criterion. A fundamental tool is the following

Theorem 15.2.2. The integral (15.2.1) converges uniformly in y ∈ [c, d] if and only if forany ε > 0 there exists A0 > 0 such that

maxc≤y≤b

∣∣∣∣∫ A2

A1

f (x, y)dx∣∣∣∣ < ε

for any A1, A2 > A0.

Example 15.2.3. Show that ∫ ∞

0e−tx2

dx

converges uniformly in every [t0, ∞) with t0 > 0. Since∫ ∞

be−tx2

dx =1√

t

∫ ∞

b√

te−u2

du ≤ 1√t0

∫ ∞

b√

t0

e−u2du→ 0

as b→ ∞.

15.2.2. Sufficient conditions. We consider the following three sufficient con-ditions on uniform convergence.

Theorem 15.2.4. (Weierstrass test) If | f (x, y)| ≤ F(x) on for any (x, y) ∈ [a, ∞)×[c, d] and the integral ∫ ∞

aF(x)dx

converges, then ∫ ∞

af (x, y)dx

converges uniformly in y ∈ [c, d].

Theorem 15.2.5. (Abel-Dirichlet test) Assume either (1) (Abel)(i) the integral ∫ ∞

af (x, y)dx

converges uniformly in y ∈ [c, d],(ii) g(·, y) is monotone for any given y ∈ [c, d],

(iii) g(x, y) is uniformly bounded (i.e., |g(x, y)| ≤ L for any (x, y) ∈ [a, ∞) ×[c, d]),

or (2) (Dirichlet)


(i) the integral ∫ ∞

af (x, y)dx

is uniformly bounded, i.e.,∣∣∣∣∫ A

af (x, y)dx

∣∣∣∣ ≤ L

for any A > a and y ∈ [c, d],(ii) g(·, y) is monotone for any given y ∈ [c, d],

(iii) g(·, y) uniformly converges to 0 as x → ∞ (i.e., for any ε > 0 there existsA0 > 0 such that maxc≤y≤d |g(x, y)| ≤ ε for any x ≥ A0).

Then the integral ∫ ∞

af (x, y)g(x, y)dx


PROOF. We only prove (2). For any A2 > A1 > a there exists ξ ∈ (A1, A2)such that∫ A2

A1

f (x, y)g(x, y)dx = g(A1, y)∫ ξ

A1

f (x, y)dx + g(A2, y)∫ A2

ξf (x, y)dx.

By (iii) for any ε > 0 we can find A0 > a such that |g(x, y)| > ε/4M for anyx > A0 and y ∈ [c, d]. Consequently, whenever A2 > A1 > A0, we arrive at∣∣∣∣∫ A2

A1

f (x, y)g(x, y)dx∣∣∣∣ ≤ ε

4M

∣∣∣∣∫ ξ

af (x, y)dx−

∫ A1

af (x, y)dx

∣∣∣∣+

ε

4M

∣∣∣∣∫ A2

af (x, y)dx−

∫ ξ

af (x, y)dx

∣∣∣∣≤ ε

4M(2M + 2M) = ε.

By Theorem 15.2.2 we prove the result.

When f (x, y) is nonnegative, we have the following

Theorem 15.2.6. (Dini) Suppose that f (x, y) ∈ C([a, b]× [c, d]) and f (x, y) ≥ 0. Ifthe integral ∫ ∞

af (x, y)dx

is continuous in [c, d], then ∫ ∞

af (x, y)dx


PROOF. Otherwise, there exists ε0 > 0 such that for any integer n > a we canfind yn ∈ [c, d] satisfying ∫ ∞

nf (x, yn)dx ≥ ε0.

The compactness of [c, d] allows us to assume that yn → y0 ∈ [c, d] as n → ∞.Since the improper integral ∫ ∞

af (x, y0)dx

15.2. IMPROPER INTEGRALS DEPENDING ON A PARAMETER 103

converges, we can find A > a such that∫ ∞

Af (x, y0)dx <

ε0

2.

When n > A, we have∫ ∞

Af (x, yn)dx ≥

∫ ∞

nf (x, yn)dx ≥ ε0.

The continuity implies that∫ ∞

Af (x, y)dx =

∫ ∞

af (x, y)dx−

∫ A

af (x, y)dx

is also continuous and hence

ε0 ≤ limn→∞

∫ ∞

Af (x, yn)dx =

∫ ∞

Af (x, y0)dx <

ε0

2

a contradiction.

Example 15.2.7. Show that the integral

F(x) =∫ ∞

0yxe−y dy

converges uniformly in every interval [0, M] with M > 0. Since yxe−y ≤ yMe−y and theintegral ∫ ∞

0yMe−y dy

converges, it follows from Theorem 15.2.4 that F(x) converges uniformly in [0, M]. How-ever F does not uniformly converge in [0, ∞), the reason is as follows. For any A0 > 1,take A1 > A0, A2 = A1 + 1, and choose a point x0 ∈ (0, ∞) with Ax0

1 e−A2 ≥ 1. Hence∫ A2

A1

yx0 e−y dy ≥∫ A2

A1

Ax01 e−A2 dy ≥ 1.

15.2.3. Basic properties. Suppose that the integral∫ ∞

0f (x, y)dx

converges uniformly for any y ∈ [c, d]. Let

F(y) :=∫ ∞

af (x, y)dx, y ∈ [c, d].

Take any increasing sequence (an)n≥0 with a0 = a and an → ∞ and set

(15.2.3) un(y) :=∫ an

an−1

f (x, y)dx, n ∈ N.

Then

(15.2.4) F(y) = ∑n∈N

un(y).

As a consequence, we have


Theorem 15.2.8. The integral ∫ ∞

af (x, y)dx

converges uniformly in y ∈ [c, d], if and only if for any increasing sequence (an)n≥0 witha0 = a and an → ∞, the series ∑n∈N un(y) converges uniformly in y ∈ [c, d], whereun(y) is given by (15.2.3).

Theorem 15.2.9. (Continuity) If f ∈ C([a, ∞)× [c, d]) and the integral∫ ∞

af (x, y)dx

converges uniformly in y ∈ [c, d], then

(15.2.5) limy→y0

∫ ∞

af (x, y)dx =

∫ ∞

alim

y→y0f (x, y)dx

for any y0 ∈ [c, d].

PROOF. Uniform continuity implies that the series ∑n∈N un(y) converges uni-formly in y ∈ [c, d]. Since

un(y) =∫ an

an−1

f (x, y)dx

is continuous, it follows that ∑n∈N un(y) = F(y) is continuous. Now the theoremfollows from Corollary 15.1.2.

Theorem 15.2.10. (Differentiation) If f , fy ∈ C([a, b]× [c, d]), the integral∫ ∞

af (x, y)dx

converges for any y ∈ [c, d], and the integral∫ ∞

afy(x, y)dx


(15.2.6)d

dy

∫ ∞

af (x, y)dx =

∫ ∞

afy(x, y)dx.

PROOF. DefineG(y) :=

∫ ∞

afy(x, y)dx.

By Theorem 15.2.9, the function G(y) is continuous. According to Theorem 15.2.11,we have∫ y

cG(z)dz =

∫ y

cdz∫ ∞

afz(x, z)dx =

∫ ∞

adx∫ y

cfz(x, z)dx = F(y)− F(c).

Hence G(y) = F′(y).

Theorem 15.2.11. (Integration) If f ∈ C([a, b]× [c, d]) and the integral∫ ∞

af (x, y)dx


(15.2.7)∫ d

cdy∫ ∞

af (x, y)dx =

∫ ∞

adx∫ d

cf (x, y)dy.

15.3. EULERIAN INTEGRALS 105

PROOF. The left-hand side of (15.2.7) equals∫ d

c∑

n∈Nun(y)dy = ∑

n∈N

∫ d

cun(y)dy = ∑

n∈N

∫ an

an−1

dx∫ d

cf (x, y)dy

=∫ ∞

adx∫ d

cf (x, y)dy

which is the right-hand side of (15.2.7).

Example 15.2.12. (Dirichlet integral) Compute

(15.2.8) I =∫ ∞

0

sin xx

dx.

Consider the function

I(α) :=∫ ∞

0e−αx sin x

xdx, α ≥ 0.

According to Theorem 15.2.4 and Theorem 15.2.5 we get

I′(α) = −∫ ∞

0e−αx sin xdx = − 1

1 + α2 , I = limα→0+

I(α).

Since |I(α)| ≤ 1α , it follows that limα→∞ I(α) = 0 and hence I(α) = − tan−1 α + π

2 . Inparticular, I = π/2.

15.3. Eulerian integrals

Recall that ∫ 1

0(ln x)ndx = (−1)nn!

for any n ∈ N. Letting t = ln(1/x) we obtain Euler’s formula

(15.3.1) n! =∫ 1

0

(ln

1x

)ndx =

∫ ∞

0e−ttn dt.

15.3.1. Gamma function. Define the Gamma function in the sense of Le-gendre

(15.3.2) Γ(s) :=∫ ∞

0e−tts−1 dt, s ∈ R.

Write (15.3.2) as ∫ ∞

0e−tts−1 dt =

∫ 1

0e−tts−1 dt +

∫ ∞

1e−tts−1 dt.

When s ≤ 0, limt→0t1−s

ett1−s = 1 so that (15.3.2) is divergent. Therefore the Gammafunction converges for s > 0:∫ 1

0e−sts−1 dt ≤

∫ 1

0ts−1 dt =

1s

,∫ ∞

1e−tts−1 dt ≤

∫ ∞

1

ts−1

t2+bsc (2 + bsc)!dt ≤ (2 + bsc)!.

Theorem 15.3.1. The Gamma function is convergent for any s > 0. Moreover, Γ(s) ∈C∞(0, ∞) and

(15.3.3) Γ(n)(s) =∫ ∞

0e−tts−1(ln t)sdt.


PROOF. Given [a, b] ⊂ (0, ∞). We have

0 <∫ 1

0e−tts−1 dt <

∫ 1

0ta−1e−t dt < ∞,

0 <∫ ∞

1e−tts−1 dt <

∫ ∞

1tb−1e−t dt < ∞

Hence the integral ∫ ∞

0e−tts−1 dt

is uniformly convergent and hence Γ(s) ∈ C(0, ∞). Similarly, we can prove

Γ′(s) =∫ ∞

0e−tts−1(ln t)dt ∈ C(0, ∞)

and thus Γ(s) ∈ C1(0, ∞), etc.

If γ is the Euler constant given by

(15.3.4) γ := limn→∞

(∑

1≤k≤n

1k− ln n

)then

(15.3.5) Γ′(1) = −∫ 1

0

1− e−u − e−1u

udy = −γ.

In fact,

Γ′(1) =∫ ∞

0e−t ln tdt = lim

n→∞

(∫ 1

1/ne−t ln tdt +

∫ n

1e−t ln tdt

).

Since ∫ 1

1/ne−t ln tdt = −

∫ 1

1/nln tde−t = −e−1/n ln n +

∫ 1

1/n

e−t

tdt,∫ n

1e−t ln tdt = −

∫ n

1ln tde−t = −e−n ln n +

∫ n

1

e−t

tdt

= −e−n ln n +∫ 1

1/n

e−1/u

udu,

it follows that

Γ′(1) = limn→∞

[−(

e−n + e−1/n)

ln n +∫ 1

1/n

e−u + e−1/u

udu

]

= limn→∞

[−(

e−n + e−1/n)

ln n−∫ 1

1/n

1− e−u − e−1/u

udu + ln n

]

= −∫ 1

0

1− e−u − e−1/u

udu + lim

n→∞

(1− e−1/u

)ln n− lim

n→∞e−n ln n

= −∫ 1

0

1− e−u − e−1/u

udu

and

−Γ′(1) =∫ 1

0

1− e−u − e−1/u

udu =

∫ 1

0

1− e−u

udu−

∫ ∞

1

e−u

udu.


Using ∫ ∞

1

e−u

udu =

∫ ∞

1

1u

limn→∞

(1− u

n

)ndu,∫ 1

0

1− e−u

udu =

∫ 1

0

1u

limn→∞

[1−

(1− u

n

)n]du,

we arrive at

−Γ′(1) = limn→∞

[∫ 1

0

1− (1− un )

n

udu−

∫ n

1

(1− un )

n

udu]

= limn→∞

[∫ n

0

1− (1− un )

n

udu−

∫ n

1

duu

]= lim

n→∞

(∫ n

0

1− yn

1− ydy− ln n

)= lim

n→∞

(∑

1≤k≤n

1k− ln n

)= γ.

Theorem 15.3.2. One has(1) Γ(1 + s) = sΓ(s) for any s > 0.(2) Γ(s) can be extended to R \ Z≤0.(3) lims→−n(s + n)Γ(s) = (−1)n/n!.

PROOF. The first result is clear. To prove (2) we first consider the interval−1 < s < 0. In this case we have 1 + s > 0 and hence Γ(1 + s) is well-defined.Hence

Γ(1 + s)s

=∫ ∞

0e−t ts

sdt.

For any δ, ∆ > 0, one has∫ ∆

δe−t ts

sdt = −e−t ts

s

∣∣∣∣∆δ

+∫ ∆

δe−tts−1 dt

=∫ ∆

δ

(e−t − 1

)ts−1 dt +

(1− e−t) ts

s

∣∣∣∣∆δ

.

Letting δ→ 0 and ∆→ ∞ we define the Gamma function for s ∈ (−1, 0) as

Γ(s) :=Γ(s + 1)

s=∫ ∞

0

(e−t − 1

)ts−1 ds, s ∈ (−1, 0).

Similarly, where s ∈ (−n,−n + 1), we can obtain

Γ(s) :=∫ ∞

0

[e−t + ∑

0≤k≤n−1(−1)k+1 tk

k!

]ts−1 dt, t ∈ (−n,−n + 1).

Using (1) we have

lims→−n

(s + n)Γ(s) = lims→−n

Γ(s + n + 1)s(s + 1) · · · (s + n− 1)

=Γ(1)

(−n)(−n + 1) · · · (−n + n− 1)=

(−1)n

n!.

Thus Γ(s) ≈ (−1)n

n!1

s+n as s→ −n.


By Theorem 15.3.2 we observe that Γ(s)Γ(1− s) has poles at all integers. Recallthat

(15.3.6) sin(πs) = πs ∏n∈N

(1− s2

n2

)has zeros at all integer. Consequently, we see that the function

Γ(s)Γ(1− s) sin(πs)

has no poles and zeros.

We can define Γ(z) for a complex number z ∈ C by the same formula:

(15.3.7) Γ(z) :=∫ ∞

0e−ttz−1 dt, z ∈ C.

As before, Γ(z) is convergent for Re(z) > 0, and Theorem 15.3.2 remains true. Thefunction

f (z) := Γ(z)Γ(1− z) sin(πz)is holomorphic in C. Since

limz→0

f (z) = limz→0

sin(πz)z

= π,

f (√−1y) =

√−1y|Γ(

√−1y)|2 sinh(πy)

= −√−1y

2π

y(eπy − e−πy)

e−πy − eπy

2√−1

= π,

a result in complex analysis shows that f (z) ≡ π for any z ∈ C.

Theorem 15.3.3. For any z ∈ C \ Z one has

(15.3.8) Γ(z)Γ(1− z) =π

sin(πz).

We give an alternative proof of (15.3.8) when z = s ∈ R \Z as follows. For anys ∈ (0, 1) we have

Γ(s)Γ(1− s) = Γ(s)∫ ∞

0e−tt−s dt =

∫ ∞

0tse−tys−1 dy ·

∫ ∞

0e−tt−s dt

=∫ ∞

0

[∫ ∞

0e−t(y+1)dt

]ys−1dy =

∫ ∞

0

ys−1

1 + ydy.

If s = 2m+12n with 0 < m < n, then∫ ∞

0

ys−1

1 + ydy = n

∫R

u2m

1 + u2n du.

By Exercise 15.3.4, we have∫ ∞

0

ys−1

1 + ydy =

π

sin(xπ), s =

2m + 12n

.

Hence we get (15.3.8) for s = 2m+12n . Because any real number in (0, 1) can be

approximated by a sequence of rational numbers with form 2m+12n , by taking a

limit we obtain (15.3.8) for any s ∈ (0, 1).


Exercise 15.3.4. (1) For any natural numbers m and n show that

∫R

x2m

1 + x2n dx =π

n1

sin 2m+12n π

.

(2) For any z ∈ C \ Z, we have

Γ(z)Γ(−z) = − π

z sin(πz), Γ(1 + z)Γ(1− z) =

πzsin(πz)

.

(3) For any n ∈ N and z ∈ C \ Z we have

Γ(n + z)Γ(n− z) =πz

sin(πz) ∏1≤k≤n−1

[(k + z)(k− z)].

(4) When z + 12 ∈ C \ Z, we have

Γ(

12+ z)

Γ(

12− z)=

π

cos(πz).

(5) For any n ∈ N and z ∈ C \ Z, we have

Γ(

2n + 12

+ z)

Γ(

2n + 12− z)=

π

cos(πz) ∏1≤k≤n

[(2k− 1

2

)2− z2

].

As an immediate consequence of Theorem 15.3.3, we see that Γ(z) has no ze-ros on C, hence 1/Γ(z) is an entire function.

Let us consider Euler’s constant

(15.3.9) E := ∏1≤k≤n−1

Γ(

kn

), n ≥ 2.

We claim that

(15.3.10) E =(2π)

n−12

√n

,

a result due to Euler. Because

E2 = ∏1≤k≤n−1

[Γ(

kn

)Γ(

n− kn

)]= ∏

1≤k≤n−1

π

sin kπn

by (15.3.8). Observe that

∑0≤k≤n−1

xk =xn − 1x− 1

= ∏1≤k≤n−1

(x− e

2kπ√−1

n

).


Letting x → 1 yields

n = ∏1≤k≤n−1

(1− e

2kπ√−1

n

)= ∏

1≤k≤n−1−e

kπ√−1

n

(e

kπ√−1

n − e−kπ√−1

n

)= ∏

1≤k≤n−1

(−2√−1e

kπ√−1

n sinkπ

n

)= (−2

√−1)n−1e

π√−1

n · n(n−1)2 ∏

1≤k≤n−1sin

kπ

n

= 2n−1(−√−1e

π√−1

2

)n−1

∏1≤k≤n−1

sinkπ

n= 2n−1 ∏

1≤k≤n−1sin

kπ

n.


15.3.2. Beta function. Recall the Bata function given by

(15.3.11) B(a, b) =∫ 1

0ta−1(1− t)b−1dt, a, b > 0.

We have proved the following basic properties

B(a, b) = B(b, a), B(a, b) =Γ(a)Γ(b)Γ(a + b)

.

Theorem 15.3.5. (Legendre) For any s ∈ R \ 12 Z≤0 we have

(15.3.12) Γ(s)Γ(

s +12

)=√

π21−2sΓ(2s).

PROOF. Consider

Γ(

s +12

)=

Γ(s)Γ( 12 )

B(s, 12 )

, Γ(2s) =Γ(s)Γ(s)B(s, s)

.

HenceΓ(s)Γ(s + 1

2 )

Γ(2s)=

Γ( 12 )B(s, s)B(s, 1

2 )=√

πB(s, s)B(s, 1

2 ).

To compute B(s, s) we consider the change of variable 12 − t = 1

2√

u so that

B(s, s) =∫ 1

0ts−1(1− s)s−1dt =

∫ 1

0(t− t2)s−1dt

=∫ 1

0

[14−(

12− t)2]s−1

dt = 2∫ 1/2

0

[14−(

12− t)2]s−1

dt

=1

4s−1

∫ 1

0(1− u)s−1 du

2√

u= 21−2sB

(s,

12

).

Therefore we obtain (15.3.12).

Theorem 15.3.6. If F ∈ C1(0, ∞) satisfies(a) F(1 + x) = xF(x),(b) F(x) is not identically equal to zero,(c) F(x)F(x + 1

2 ) =√

π21−2xF(2x),then F(x) = Γ(x) in (0, ∞).


PROOF. Since Γ(x) is not identically zero by Theorem 15.3.3, we can con-sider the quotient Q(x) := F(x)/Γ(x) and the logarithmic derivative P(x) :=(ln Q(x))′ = Q′(x)/Q(x). Conditions (a) and (c) show that

P(x + 1) = p(x), P(x) + P(

x +12

)= P(2x).

In particular,

P(x) =12

[P( x

2

)+ P

(x + 1

2

)].

Therefore, by replacing x by x/2 and (x = 1)/2, we have

P( x

2

)=

12

[P( x

4

)+ P

(x4+

12

)],

P(

x + 12

)=

12

[P(

x + 14

)+ P

(x + 3

4

)]which imply

P(x) =14

[P( x

4

)+ P

(x4+

14

)+ P

(x4+

12

)+ P

(x4+

34

)].

By induction on n we get

P(x) =12n ∑

0≤k≤2n−1P(

x2n +

k2n

)for any n ∈ N. Letting n→ ∞ yields

P(x) =∫ 1

0P(t)dt = ln Q(t)

∣∣∣∣10= ln Q(1)− ln Q(0) = 0

where we used the fact that

lim Q(0) = limx→0

ln Q(x) = limx→0

ln Q(x + 1) = ln Q(1),

which shows that Q(x) ≡ 1 in (0, ∞) because Q(1/2) = 1.

There is another characterization of Γ(s) due to Bohr, Mollerup, and Artin.

Theorem 15.3.7. (Bohr-Mollerup-Atin) If a function f : (0, ∞)→ R satisfying(a) f (1 + x) = x f (x),(b) ln f is convex,(c) f (1) = 1,

then f (x) = Γ(x) for any x ∈ (0, ∞).

Euler defined the Gamma function by

(15.3.13) Γ(s) :=1s ∏

n∈N

(1 + 1n )

s

1 + s/n= lim

n→∞

nsn!∏0≤k≤n(s + k)

, s ∈ R \ Z≤0,

in a letter (1729) to Goldbach. Let

(15.3.14) Γn(s) =∫ n

0

(1− t

n

)nts−1 dt.

As n → ∞ we get Γn(s) → Γ(s) defined in (15.3.13) while on the other hand,Γn(s)→ Γ(s) defined in (15.3.2).


Theorem 15.3.8. (1) (Weierstrass) For any s > 0, one has

(15.3.15)1

Γ(s)= seγs ∏

n∈N

[(1 +

sn

)e−s/n

]where γ is the Euler’s constant given by (15.3.4).

(2) For any s > 0, one has

(15.3.16) ln Γ(s) =(

s− 12

)ln s− s + ln

√2π + µ(s)

where

(15.3.17) µ(s) :=∫ ∞

0

(1

et − 1− 1

t+

12

)e−st dt

t.

Recall the Bernoulli numbers Bn given by

∑n≥0

Bn

n!tn =

tet − 1

(see also (13.3.21)). Hence

1et − 1

=1t− 1

2+

112

t− 1460

t3 + · · · .

In general

(15.3.18) µ(s) =∫ ∞

0

(∑n≥2

Bn

n!tn

)e−st dt

t2 .

In particular,

µ(1) = 1− ln√

2π, µ

(12

)=

12(1− ln 2), Γ(x) ≈

√2πxx− 1

2 e−x

as x → ∞. There are variant expressions for µ(s):(1) (Gudermann)

µ(x) = ∑n≥0

∫ 1

0

12 − u

u + x + ndu.

(2) If Λ(t) := 12 − t + brc with t ∈ R, then

µ(x) =∫ ∞

0

Λ(t)t + x

dt.

(3) One has

µ(x) = 2∫ ∞

0

tan−1(t/x)e2πt − 1

dt,

µ(x) =xπ

∫ ∞

0ln(

11− w−2πt

)dt

t2 + x2 ,

µ(x) = ∑n∈N

1nπ

∫ ∞

0

sin(2πt)t + x

dt.


Theorem 15.3.9. (Stirling) We have

n! ∼√

2πnn+ 12 e−n n→ ∞,(15.3.19)

Γ(x) =√

2πxx− 12 e−xeθx/12x for some θx ∈ (0, 1),(15.3.20)

0 < µ(x) <1

12x,(15.3.21)

µ(x) = ∑1≤k≤m

(−1)k−1 Bk(2k− 1)2k

1x2k−1

+ (−1)mθmBm+1

(2m + 1)(2m + 2)1

x2m+1 ,(15.3.22)

for some θm ∈ (0, 1), where m ∈ N.

Consequently

(15.3.23) m! ∼√

2πnn+ 12 e−ne

112n−

1360n3

as n→ ∞.

15.3.3. Gamma function and Riemann ζ-function. The Riemann ζ-functionis given by

(15.3.24) ζ(z) := ∑n∈N

1nz , Re(z) > 1.

Actually, we can meromorphically extend it to the entire complex plane with theonly singularity at z = 1. The reason is as follows: for Re(z) > 1, we have

ζ(z) = ∑n∈N

z∫ ∞

n

dttz+1 = z

∫ ∞

0

(∑n≤t

1

)dt

tn+1

= z∫ ∞

1

btctn+1 dt =

zz− 1

− z∫ ∞

1

〈t〉tn+1 dt.

Thus we can holomorphically extend the Riemann zeta function to Re(z) > 0 butz 6= 1. In order to extend to the entire complex plane we need Riemann’s equa-tion.

Since

(15.3.25) Γ(z)n−z =∫ ∞

0tz−1e−ntdt, Re(z) > 0,

it follows that

(15.3.26) Γ(z)ζ(z) =∫ ∞

0

tz−1

et − 1dt, Re(z) > 0.

If we set

(15.3.27) ξ(z) := π−z/2Γ(z/2)ζ(z),

then we have the following equation due to Riemann

(15.3.28) ξ(z) = ξ(1− z), z 6= 0, 1.

In particular

(15.3.29) ξ(z) =[2zπz−1 sin

(πz2

)Γ(1− z)

]ζ(1− z).


This implies that ζ(z) can be holomorphically extended to Re(z) ≤ 0. Actuallyone can show that

(15.3.30) ξ(z) =∫ ∞

0ϑ1(t)tz/2−1dt, Re(z) > 1,

where ϑ1(t) := (ϑ(t)− 1)/2 and

(15.3.31) ϑ(t) := ∑n∈Z

e−πn2t

is the Jacobi theta function (By Fourier analysis in the next chapter, we shall showthat ϑ(1/r) =

√tϑ(t) for t > 0). Letting z → 2n + 1 for some n ∈ N in (15.3.29),

we find ζ(−2n) = 0. Hence−2N are (trivial) zeros of ζ(z). Note that ζ(0) = −1/2and ζ ′(0) = − 1

2 ln 2π.

Conjecture 15.3.10. (Riemann, 1859) Any nontrivial zeros of ζ(z) all lie on the lineRe(z) = 1

2 .

When z = 2n we have the following explicit formula

(15.3.32) ζ(2n) = (−1)n−122n−1 B2n

(2n)!π2n ∈ Q

by residue technique.

Conjecture 15.3.11. Is ζ(2n + 1) irrational?

The above conjecture is at least true for ζ(3).

15.3.4. Gamma function and Hausdorff dimension. Given ε > 0 and ρ ≥ 0.For any set D ⊂ Rn define

(15.3.33) Hp,ε(D) :=

wp ((Si)i∈I) : D =

⋃i∈I

Si, diam(Si) < ε for any i ∈ I

.

Here (Si)i∈I is an open cover of D, diam(Si) := supx,y∈Si|x− y| is the diameter of

Si, and wp((Si)i∈I) stands for the quantity

(15.3.34) wp ((Si)i∈I) := ∑i∈I

Cp

(12

diam(Si)

)p, Cp := Vol(Bp) =

πp/2

Γ(2 + p/2)

(Bp ⊂ Rp is the unit open sphere in Rp so that it is the usual unit sphere when p isan integer). If p ∈ N we have

Vol(Bp) =∫|x|≤1, x∈Rp

dx =πp/2

Γ(1 + p

2) .

As ε→ 0+,Hp,ε(D) is increasing. Hence we can define

(15.3.35) Hp(D) := supε>0Hp,ε(D)

called the p-dimensional Hausdorff measure.

We discuss measures in general. Let X be an arbitrary set. A set A of subsetsof X is called a σ-algebra if it satisfies the following conditions:

(1) ∅, X ∈ A.(2) If A, B ∈ A, then A \ B ∈ A.


(3) If (Ai)i∈I is a finite or countable collection of elements of A, then theirunion ∪i∈I Ai is also an element of A.

For a countable collection (Ai)i∈N of A, we have

X \⋂

i∈N

Ai =⋃

i∈N

(X \ Ai)

and then a σ-algebra contains any intersection of a countable collection of its ele-ments.

Definition 15.3.12. A measure on a σ-algebra A is a function µ : A → R+ ∪+∞ = [0,+∞] such that

(1) µ(∅) = 0, and(2) (σ-additivity) if (Ai)i∈N is a finite or countable collection of elements of

A and the sets Ai are disjoint, then

µ

(⋃i∈N

Ai

)= ∑

i∈Nµ(Ai).

Note that the expression ∑i∈N µ(Ai) is either a finite sum or a series; its valueis well-defined and independent of the order of terms since the terms are nonneg-ative.

Proposition 15.3.13. Let µ be a measure on a σ-algebra A.(a) Let (Ai)i∈N be a sequence of measurable sets such that Ai ⊂ Ai+1 for all i. Then

the sequence (µ(Ai))i∈N is nondecreasing and

(15.3.36) limi→+∞

µ(Ai) = µ(∪i∈N Ai).

(b) Let (Ai)i∈N be a sequence of measurable sets such that Ai ⊃ Ai+1 for all i, andassume that µ(A1) < +∞. Then the sequence (µ(Ai))i∈N is non-increasingand

(15.3.37) limi→+∞

µ(Ai) = µ(∩i∈N Ai).

PROOF. (a) Since Ai+1 = Ai ä(Ai+1 \ Ai), it follows that µ(Ai+1) = µ(Ai) +µ(Ai+1 \ Ai) ≥ µ(Ai) and then the sequence (µ(Ai))i∈N is nondecreasing. LetA := ∪i∈N Ai. Write Bi := Ai \ Ai−1 where A0 = ∅. Then

Ai = ä1≤k≤i

Bk, A =⋃

i∈N

Ai = äk∈N

Bk, µ(A) = ∑k∈N

µ(Bk) = limk→+∞

∑1≤i≤k

µ(Bi).

Hence we proved (15.3.36).(b) Since Ai ⊃ Ai+1, we have µ(Ai) ≥ µ(Ai+1) and hence the sequence

(µ(Ai))i∈N is non-increasing. Let Bi := Ai \ Ai+1. Then

µ(Ai) = µ(Ai+1) + µ(Bi), µ(A1)− µ(Ai+1) = ∑1≤k≤i

µ(Bk) = µ(A1 \ Ai+1).

Write Ci := A1 \ Ai. Then (Ci)i∈N is nondecreasing so that by part (a) we obtain

µ

(A1 \

⋂i∈N

Ai

)= µ

(⋃i∈N

Ci

)= lim

i→+∞µ(Ci) = lim

i→+∞µ(A1 \ Ai).


Therefore

limi→+∞

[µ(A1)− µ(Ai)] = µ(A1)− µ

(⋂i∈N

Ai

).

Because µ(A1) < +∞, the above identity shows (15.3.37).

Note that the condition that µ(A1) < +∞ in Proposition ?? (b) is essential. Infact, consider Ai = (i,+∞) and µ is the usual length measure; then µ(Ai) = +∞for every i and ∩i∈N Ai = ∅.

If S is an arbitrary collection of subsets of a set X, there exists a unique min-imal σ-algebra containing S; it is called the σ-algebra generated by S. Indeed,consider the set

(15.3.38) σ(S) :=⋂A : A is a σ-algebra and A ⊃ S.

Since X is itself a σ-algebra, it follows that such a set A always exists. For anyσ-algebra A that contains S, we must have ∅, X ∈ A and hence ∅, X ∈ σ(S).If (Ai)i∈I is a finite or countable collection of elements of σ(S), then Ai ∈ A foreach i ∈ I and every σ-algebra A containing S. Then ∪i∈I Ai ∈ A and hence∪i∈I Ai ∈ σ(S). The minimum follows immediately from the definition (??).

(1) If X is a topological space, then the σ-algebra generated by its topology(i.e., generated by the set of all open sets in the topology) is called theBorel σ-algebra of X. Elements of the Borel σ-algebra are called Borelsets.

(2) A measure defined on the Borel σ-algebra is called a Borel measure overX.

Theorem 15.3.14. (Lebesgue) There exists a unique Borel measure Lm over Rm whichis invariant under parallel translations and such that Lm([0, 1]m) = 1.

The measure Ln is called Lebesgue measure. The uniqueness implies thatevery translation-invariant Borel measure on Rm with a finite value for a cube is aconstant multiple of Lm.

(1) Lebesgue measure is invariant under isometries of Rm.(2) If L : Rm → Rm is a linear map, then

(15.3.39) Lm(L(A)) = |det(L)|Lm(A)

for any measurable set A ⊂ Rm. In particular, a homothety with coeffi-cient C multiplies the Lebesgue measure by Cm.

Let (X, d) be a metric space and p be a nonnegative real number. We cansimilarly define the p-dimensional Hausdorff measure Hp(X) of X as (15.3.35).Since Hp,ε(X) has a (possibly infinite) limit as ε → 0+, Hp(X) is well-defined forany metric space (X, d). It may be either a nonnegative real number or +∞.

Theorem 15.3.15. Let (X, dX) and (Y, dY) be metric spaces, and let A and B be subsetsof X. Then

(1) If A ⊂ B thenHp(A) ≤ Hp(B).(2) Hp(∪i∈I Ai) ≤ ∑i∈I µd(Ai) for any finite or countable collection of sets (Ai)i∈I ⊂

X.(3) If dist(A, B) > 0, thenHp(A ∪ B) = Hp(A) +Hp(B).


(4) If f : X → Y is a Lipschitz map with a Lipschitz constant C, thenHp( f (X)) ≤CpHp(X).

(5) If f : X → Y is a C-homothety, i.e., dY( f (x1), f (x2)) = CdX(x1, x2) for allx1, x2 ∈ X, thenHp( f (X)) = CpHp(X).

By Caratheodory’s criterion, any nonnegative function on the Borel σ-algebraof X possessing the properties (1)–(3) from Theorem 15.3.15 is a measure.

Theorem 15.3.16. For any metric space (X, d) and any p ≥ 0, Hp is a measure on theBorel σ-algebra of X.

Example 15.3.17. 0-dimensional Hausdorff measure of a set is its cardinality. In otherwords,H0(X) is a number of points in X if X is finite andH0(X) = +∞ if X is infinite.First consider a finite set X = x1, · · · , xn. Then H0,ε(X) = |X| for each ε > 0 andhence H0(X) = |X|. For an infinite set X, we have H0(X) ≥ H0(A) = |A| for anyfinite subset A ⊂ X. Letting |A| → +∞, we getH0(X) = +∞.

Let I = [0, 1] denote the interval [0, 1] of R. Then Im = [0, 1]m is the unit cubein Rm. ThenHm(Im) = 1. Moreover, we have

Theorem 15.3.18. For each m ∈ N, we have Lm = Hm on Rm.

Theorem 15.3.19. (Vitali’s covering theorem) Let X be a bounded set in Rm and letB be a collection of closed balls in Rm such that for every x ∈ X and ε > 0 there is aball B ∈ B such that x ∈ B and diam(B) < ε. Then B contains a finite or countablesub-collection (Bi)i∈I of disjoint closed balls which covers X up to a set of zero measure,i.e., such that Bi ∩ Bj = ∅ if i 6= j andHm(X \ ∪i∈IBi) = 0.

PROOF. Without loss of generality, we may assume that every ball B ∈ Bcontains at least one point of X and exclude the balls with radius greater than1. Then all these closed balls are contained in the 2-neighborhood of X which isbounded and hence has finite volume.

We construct a sequence (Bi)i∈N of closed balls by induction. If B1, · · · , Bnare already constructed, we choose the next ball Bn+1 as follows. Let Bn denotethe set of balls from the collection that do not intersect any of B1, · · · , Bn. If Bn isempty, then B1 ∪ · · · ∪ Bn covers the entire set X. If Bn is not empty, choose Bn+1to be any element of Bn with

(15.3.40) diam(Bn+1) >12

supB∈Bn

diam(B).

The balls Bi are disjoint by the construction. We will now show that they cover Xup to a set of zero m-dimensional Hausdorff measure. Since the balls are disjointand are contained in a set of finite volume, we have

∑i∈NHm(Bi) < +∞.

Fix an ε > 0. There is an index n such that ∑i≥n+1Hm(Bi) < ε. Let x ∈ X \ ∪i∈NBiand let B be any ball from the collection that contains x and does not intersect theballs B1, · · · , Bn. Note that B must intersect ∪i∈NBi because otherwise B ∈ Bn forany n. Since from (15.3.40),

Hm(B) = Lm(B) = C(m)

(diam(B)

2

)m≤ C(m) (diam(Bn)

m = 2mHm(Bn).


Letting n → +∞ yields Hm(B) = 0 a contradiction. Let k > n be the minimalindex such that B ∩ Bk 6= ∅. Then B ∈ Bk−1 and hence

diam(Bk) >12

diam(B)

by (15.3.40). Let x, r and xk, rk denote the center and radius of B and Bk respec-tively. Then r ≤ 2rk and

|x− xk| ≤ |x− x0|+ |x0 − xk| ≤ r + (rk + r) ≤ 2r + rk ≤ 5rk.

Hence x belongs to the ball with the same center as Bk and radius 5 times larger.We denote this ball by 5Bk. We have proved that every point x ∈ X \ ∪i∈NBibelongs to a ball 5Bk for some k > n. Thus X \ ∪i∈NBi ⊂ ∪i≥n+15Bi and hence

Hm (X \ ∪i∈NBi) ≤ ∑i≥n+1

Hm(5Bi) = 5m ∑i≥n+1

Hm(Bi) < 5mε.

Since ε was arbitrary, it follows thatHm(X \ ∪i∈NBi) = 0.

There is a “critical dimension” below which the Hausdorff measure is infiniteand above which the Hausdorff measure is zero. This dimension is called theHausdorff dimension.

Theorem 15.3.20. For a metric space (X, d) there exists a p0 ∈ [0,+∞] such thatHp(X) = 0 for all p > p0 andHp(X) = +∞ for all p < p0.

PROOF. Let p0 := infp ≥ 0 : Hp(X) < +∞. Clearly that Hp(X) = +∞for all p < p0. If p > p0, there is a p′ ∈ [p0, p) such that Hp′(X) = M < +∞.Therefore, for any ε > 0 there exists a covering (Si)i∈I of X such that diam(Si) < ε

for all i ∈ I and ∑i∈I C(p′)(diam(Si))p′ < 2M. Then

∑i∈I

C(p) (diam(Si))p = ∑

i∈IC(p) (diam(Si))

p−p′ (diam(Si))p′

≤ εp−p′ C(p)C(p′) ∑

i∈IC(p′) (diam(Si))

p′

≤ 2C(p)C(p′)

εp−p′M.

ThusHp,ε(X) ≤ 2εp−p′C(p)M/C(p′). Letting ε→ 0+ yieldsHp(X) = 0.

The value p0 in the proof of Theorem 15.3.20 is called the Hausdorff dimen-sion of X and denoted by dimH(X). Note that when p = dimH(X), the measureHp(X) has three possibilities, i.e., zero, positive number (may not be an integer)or infinity.

Example 15.3.21. The Cantor set C is created by repeatedly deleting the open middle thirdof [0, 1]. One starts by deleting the open middle (1/3, 2/3) from the interval [0, 1], leavingtwo line segments [0, 1/3] ∪ [2/3, 1]. Next, the open middle third of each these remain-ing segments is deleted, leaving four line segments [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪

15.4. Γ, Ψ, Φ FUNCTIONS AND MELLIN TRANSFORM 119

[8/9, 1]. Finally, the Cantor set C is given by

C =⋂

m∈N

⋂0≤k≤3m−1−1

([0,

3k + 13m

]∪[

3k + 23m , 1

])

= [0, 1] \⋃

m∈N

⋃0≤k≤3m−1−1

(3k + 1

3m ,3k + 2

3m

).

Let Cn denote the remaining in the n-th step. ThenHp(Cn) = C(p)2n/3np and

Hp(C) = limn→+∞

Hp(Cn) = limn→+∞

C(p)2n(

13n

)p=

+∞, p < log3 2,C(p), p = log3 2,

0, p > log3 2.

Thus dimH(C) = log3 2 = ln 2/ ln 3, while the usual dimension of the Cantor set isequal to 0.

Proposition 15.3.22. Let (X, dX) and (Y, dY) be metric spaces. Then(1) If Y ⊂ X, then dimH(Y) ≤ dimH(X).(2) If X is covered by a finite or countable collection (Xi)i∈I of its subsets, then

dimH(X) = supi∈I dimH(Xi).(3) If f : X → Y is a Lipschitz map, then dimH( f (X)) ≤ dimH(X). In particular,

bi-Lipschitz equivalent metric spaces have the same Hausdorff dimensions.(4) dimH(Rm) = dimH([0, 1]m) = m.

Let (X, dX) and (Y, dY) are metric spaces and f : X → Y a map such that

(15.3.41) dY( f (x1), f (x2)) ≤ CdX(x1, x2)α

for all x1, x2 ∈ X, where C, α are some positive constants. Then

Hp( f (X)) ≤ CpHpα(X)

and thereforedimH( f (X)) ≤ 1

αdimH(X).

15.4. Γ, Ψ, Φ functions and Mellin transform

Let G be a Abel group and F (G) be the set of all complex-valued functions onG. For any f1, f2 ∈ F (G) and λ ∈ C define

( f1 + f2)(g) := f1(g) + f2(g), (c f )(g) := c · f (g), g ∈ G.

Then F (G) is a vector space over C.

Exercise 15.4.1. Prove that F (G) is a vector space over C. When G is finite,dimC F (G) = |G|.

We say χ ∈ F (G) is a character of G if it satisfies(i) |χ(g)| = 1 for any g ∈ G, and

(ii) χ(g1g2) = χ(g1)χ(g2) for any g1, g2 ∈ G.

The set of all characters of G is denoted by G. Observe that χ(e) = 1 for any χ ∈ Gwhere e is the identity element of G.

Exercise 15.4.2. Let T := z ∈ C : |z| = 1. Then T is a multiplicative group andG = HomC(G, T). That is G is exact the set of all group homomorphisms from Ginto T.


For any χ, ϕ ∈ G define χϕ ∈ G by

(χϕ)(g) := χ(g)ϕ(g), g ∈ G.

Then G is a Abel group. The identity element in G is ι given by

ι(g) := 1.

The inverse element of χ ∈ G is given by

χ−1(g) := χ(g) =1

χ(g), g ∈ G.

Suppose first that G is a finite Abel group. Then dimC F (G) = |G|. For anyχ ∈ G with χ 6= ι, we have χ(g0) 6= 1 for some g0 ∈ G with g0 6= e. Compute

∑g∈G

χ(g) = ∑g∈G

χ(g0g) = ∑g∈G

χ(g0)χ(g) = χ(g0) ∑g∈G

χ(g).

Hence∑

g∈Gχ(g) = 0.

Consequently, for any χ, ϕ ∈ G with χ 6= ϕ, we have

∑g∈G

(χϕ−1)(g) = ∑g∈G

χ(g)ϕ(g) = 0.

On the other hand, we have

∑g∈G

(χχ−1)(g) = ∑h∈G

ι(g) = ∑g∈G

1 = |G|.

Thus1|G| ∑

g∈G(χχ−1)(g) = 1.

The above computations motivate us to define an inner product on F (G) by

(15.4.1) 〈 f1, f2〉 :=1|G| ∑

g∈Gf1(g) f2(g), f1, f2 ∈ F (G).

Therefore

(15.4.2) 〈χ, ϕ〉 = δχϕ, χ, ϕ ∈ G.

Theorem 15.4.3. (a) If G = Zn = [0], [1], · · · , [n− 1] and ξ = e2π√−1/n, then

G = χ0, χ1, · · · , χn−1,where χ1([j]) = ξ j for 0 ≤ j ≤ n− 1 and χk = χk

1 for 0 ≤ k ≤ n− 1.(b) If G1 and G2 are finite Abel groups, then G1 × G2 = G1 × G2.(c) For any finite Abel group G we have |G| = |G|.

Exercise 15.4.4. Prove Theorem 15.4.3.

Theorem 15.4.5. Let G be a finite Abel group.(a) G is a family of orthonormal basis for F (G).(b) For any f ∈ F (G) we have

(15.4.3) f = ∑χ∈G

c(χ)χ, c(χ) := 〈 f , χ〉.


PROOF. By (a) we havef = ∑

χ∈G

cχ

for some c ∈ C. Then c = 〈 f , χ〉 = ∑ϕ∈G〈c(ϕ)ϕ, χ〉 = ∑ϕ∈G c(ϕ)δϕχ = c(χ).

We next consider the group G = T = z ∈ C : |z| = 1. The group G is also atopological space, that is, the operations (z, w) 7→ zw and z 7→ z−1 are continuous.Hence G is a topological group (in fact it is a Lie group). In this we consider allcontinuous characters of T, i.e., the set T. Observe that the original characters arealso continuous with respect to the discrete topology equipped on finite groups.

For any a ∈ R define

(15.4.4) µa(x) := ax, x ∈ R.

Observe that µa is a continuous homomorphism1.(A) Choose a continuous homomorphism ψ : R → R and write a := ψ(1) ∈ R.

Then ψ = µa. In fact, for any m ∈ Z, we have ψ(m) = ma. If ψ(1/n) =: b,then

a = ψ(1) = ψ

(n · 1

n

)= nψ

(1n

)= nb;

that is, b = 1n a. Consequently, ψ(x) = ax for any x ∈ Q. Since Q is dense

in R and ψ si continuous, it follows that ψ(x) = ax for any x ∈ R.(B) Define the exponential map by

(15.4.5) exp : R −→ T, x 7−→ e2π√−1t.

Clear that exp is a continuous homomorphism, a local isomorphism, andker(exp) = Z. In the terminology of topology, exp : R → T is a coveringmap. For any continuous map ϕ : R → T there exists a unique continuousmap ϕ : R → R satisfying ϕ = exp ϕ. This is topology result. The basicidea is to construct ϕ near the point x = 0 ∈ R by the local isomorphismof exp.

(C) R ∼= R. Here R is the set of all continuous characters of R. For any a ∈ R wedefine ϕa ∈ R by ϕa(x) := e2π

√−1ax for any x ∈ R. Conversely, if ϕ ∈ R

then ϕ : R → T is continuous and homomorphic. By (B), there exists aunique continuous homomorphism ϕ : R → R such that ϕ = exp ϕ.Now define a := ϕ(1) ∈ R.

Given any continuous homomorphism χ : T→ T define

(15.4.6) ϕχ := χ exp : R −→ T.

Then ϕχ is also a continuous homomorphism and ϕχ(0) = ϕχ(1) = χ(1) = 1. Bythe above discussion we have a continuous homomorphism ϕχ = µa : R → Rsatisfying

(15.4.7) exp ϕχ = ϕχ = χ exp .

1A map ψ : R→ R is called a homomorphism, if it is a group homomorphism on R.


Rϕχ−−−−→ Ryexp

yexp

Tχ−−−−→ T

In particular, for χ = 1, we arrive at

exp(a) = χ (exp(1)) = χ(1) = 1 =⇒ a = n ∈ Z

and a = n is called the degree of χ and written as deg(χ).. Hence, (15.4.7) becomes

(15.4.8) χ(z) = zn, n := deg(χ), z ∈ T.

Conversely, for any integer n ∈ Z, define

(15.4.9) χn := zn, z ∈ T.

Observe that χn ∈ T and χnχm = χn+m for any m, n ∈ Z.

Theorem 15.4.6. One has T = (χn)n∈Z ∼= Z.

For any f ∈ F (T) (equivalently, f is a complex-valued function defined on Rwith period 2π) and χ ∈ T define

(15.4.10) c(χ) := 〈 f , χ〉 = 1|T|

∫T

f (x)χ(x)dx.

Then informally we have

(15.4.11) f = ∑χ∈T

c(χ)χ = ∑n∈Z

c(χn)χn.

Write

(15.4.12) cn := c(χn) =1

2π

∫ 2π

0f (x)e−2π

√−1nxdx

Then (15.4.11) equals

(15.4.13) f (x) = ∑n∈Z

cne2π√−1nx.

This is the Fourier analysis on the circle that will be discussed in the next chapter(see Chapter 16).

For R, we shall replace ∑χ∈T in (15.4.11) by∫

R and obtain

(15.4.14) f (x) =∫ ∞

−∞c(ϕa)ϕa da, c(ϕa) =

∫ ∞

−∞f (t)ϕa(t)dt.

Since ϕa(t) = e2π√−1at, we have

(15.4.15) f (a) =∫ ∞

−∞f (x)e−2π

√−1axdx, f (x) =

∫ ∞

−∞f (a)e2π

√−1axda.

This is the classical Fourier analysis on R1. Let L1(R) denote the set of all (Riemannor Lebesgue) integrable functions on R.

(i) If f ∈ L1(R) ∩ C1(R), then the first integral in (15.4.15) is well-defined. If∫ ∞−∞ means lim∆→∞

∫ ∆−∆, then the second identity in (15.4.15) holds.


(ii) (Transform invariance) If G = R or T we have∫G

f (x)dx =∫

Gf (gx)dx, g ∈ G.

In particular, for G = R, we then get∫ ∞

−∞f (x + a)dx =

∫ ∞

−∞f (x)dx;

if G = T, then∫ 2π

0f (e2π

√−1(θ+α))dθ =

∫ α+2π

αf (e2π

√−1θ)dθ.

(iii) If f ∈ L1(R) then the function f given by (15.4.15) is boundedand continuous. For each h ∈ R define

Lh : L1(R) −→ L1(R), f 7−→ Lh( f )(x) := f (x + h).

ThenLh( f ) = ϕh f

where ϕh(x) = e2π√−1hx.

15.4.1. Harmonic analysis on R+. The set R+ of all positive real numbersforms a topological group under the multiplication. The map

(15.4.16) ε : R −→ R+, x 7−→ ex,

is a topological isomorphism (that is, ε is not only a group isomorphism, but alsoa topological homeomorphism). The inverse map λ of ε is given by

(15.4.17) λ : R+ −→ R, y 7−→ ln y.

Moreover, we have two induced maps

(15.4.18) ε : R+ −→ R, ϕ 7−→ ϕ ε,

and

(15.4.19) λ : R −→ R+, ψ 7−→ ψ λ.

Since R = 〈ϕa〉a∈R, it follows that

(15.4.20) ϕ+a (x) := λ(ϕa)(x) = ϕa(ln x) = e2π

√−1a ln x = x2π

√−1a, x ∈ R+.

Theorem 15.4.7. R+ = 〈ϕ+a 〉a∈R. As a topological group, R+ is isomorphic to R.

On the group R+ there is a natural invariant measure dx/x in the sense that∫ rb

ra

dxx

= lnrbra

= lnba=∫ b

a

dxx

for any a, b > 0. Define

(15.4.21) f (a) = c(ϕ+a ) =

∫R+

f (x)ϕ+a (x)

dxx

= 〈 f , ϕ+a 〉L2(R+ ,dx/x);

then we have

(15.4.22) f (x) =∫

R+c(ϕ+

a )ϕ+a =

∫R

c(ϕ+a )ϕ+

a da =∫ ∞

−∞f (a)x2π

√−1ada.


The formulas (15.4.21) and (15.4.22) are the Fourier analysis on R+. Introduce

(15.4.23) τ := −2πa, f (τ) := f(− τ

2π

)= f (a).

Then (15.4.22) and (15.4.22) become

(15.4.24) f (τ) =∫ ∞

0f (x)x

√−1τ dx

x, f (x) =

12π

∫ ∞

−∞f (τ)x−

√−1τdτ.

Theorem 15.4.8. (a) If f ∈ L1(R+, dx/x)∩C1(0, ∞), then the first integral in (15.4.24)is well-defined. Moreover if

∫ ∞−∞ means lim∆→∞

∫ ∆−∆, then the second identity in (15.4.24)

holds.(b) If F ∈ L1(−∞, ∞) ∩ C1(−∞, ∞), then the integral

f (x) :=1

2π

∫ ∞

−∞F(τ)x−

√−1τdτ

is well-defined. Moreover

F(τ) = lim∆→∞

∫ ∆

1∆

f (x)x√−1τ dx

x.

Given f ∈ C1(R+)and σ > 0, define

(15.4.25) fσ(x) := xσ f (x).

Note that fσ ∈ C1(R+). If furthermore fσ ∈ L1(R+, dx/x), then we can define

(15.4.26) fσ(τ) =∫ ∞

0f (x)xσ+

√−1τ dx

xby Theorem 15.4.8 (a).

Theorem 15.4.9. (Mellin transform) If f ∈ C1(R+) and fσ ∈ L1(R+, dx/x) for anyσ ∈ (a, b) ⊂ R+, then

(15.4.27) F(z) :=∫ ∞

0f (x)xz dx

xis holomorphic for z with a < Re(z) < b, and

(15.4.28) f (x) =1

2π√−1

∫ σ+√−1∞

σ−√−1∞

F(z)x−zdz

which is independent of the choice of σ ∈ (a, b).

PROOF. It is clear that F(z) := fσ(τ) is holomorphic for z := σ +√−1τ with

a < σ < b. According to Theorem15.4.8 (b) we have

xσ f (x) = fσ(x) =1

2π

∫ ∞

−∞fσ(τ)x−

√−1τdτ

Thus

f (x) =1

2π

∫ ∞

−∞fσ(τ)x−σ−

√−1τdτ =

12π√−1

∫ σ+√−1∞

σ−√−1∞

F(z)x−zdz.

By Cauchy’s theorem the last integral is independent of σ.

We call ( f (x), F(z)) a Mellin pair and usually write F(z) as M( f )(z). Iff (x) = e−x, then fσ ∈ L1(R+, dx/x) for any σ ∈ R+. Then F(z) = Γ(z).


Theorem 15.4.10. (a) (e−x, Γ(z)) is a Mellin pair.(b) We have

(15.4.29) Γ(z) =∫ ∞

0e−xxz dx

x, e−x =

12π√−1

∫ σ+√−1∞

σ−√−1∞

Γ(z)x−zdz

for Re(z) > 0.

Exercise 15.4.11. e−x − 1 + x1! −

x2

2! + · · ·+ (−1)n xn−1

(n−1)! and Γ(z) (−n < Re(z) <

−(n− 1)) is a Mellin pair.

Exercise 15.4.12. Let ( f (x), F(z)) is a Mellin pair. Prove that for any a, b > 0 andc, ` ∈ C the following pairs are Mellin pairs:

(a) f (xa) and 1a F( z

a ),(b) 1

x` f ( 1x ) and F(`− z),

(c) f (bx) and F(z)bz ,

(d) xc f (x) and F(z + c),(e) f (bxa) and F( z

a )/abz/a.

Exercise 15.4.13. Let ( f (x), F(z)) be a Mellin pair. Prove for each n ∈ Z thatf (x)(ln x)n and F(n)(z) is a Mellin pair.

If ( f (x), F(z) and (g(x), G(z)) are Mellin pairs, then

F(z)G(z) =∫ ∞

0f (t)tz dt

t·∫ ∞

0g(y)yz dy

y=

∫ ∞

0

[∫ ∞

0f (t)g(y)(ty)z dt

t

]dyy

=∫ ∞

0

[∫ ∞

0f (t)g

( xt

) dtt

]xz dx

x=

∫ ∞

0( f ∗ g)(x)xz dx

x.

Here f ∗ g stands for the convolution of f (x) and g(x) in L1(R+, d/x):

(15.4.30) ( f ∗ g)(x) :=∫ ∞

0f (t)g

( xt

) dtt

.

Theorem 15.4.14. For any f , g ∈ L1(R+, dx/x), we haveM( f ∗ g) =M( f )M(g).

Exercise 15.4.15. (a) If f , g ∈ L1(R+, dx/x), then f ∗ g ∈ L1(R+, dx/x).(b) If f , g, h ∈ L1(R+, dx/x), then ( f ∗ g) ∗ h = f ∗ (g ∗ h).(c) If f , g ∈ L1(R+, dx/x), then f ∗ g = g ∗ f .(d) If f1, f2, g ∈ L1(R+, dx/x) and c1, c2 ∈ R, then (c1 f1 + c2 f2) ∗ g = c1( f1 ∗

g) + c2( f2 ∗ g).

The above exercise shows that (L1(R+, dx/x), ∗) is a commutative algebraover R. On the other hand, the set M of all meromorphic functions over C isalso a commutative algebra over R.

Theorem 15.4.16. (a)M : L1(R+, dx/x)→M is a algebraic homomorphism.(b) If f1, · · · , fn ∈ L1(R+, dx/x), then

(15.4.31) ∏1≤i≤n

M( fi) =M(∗1≤i≤n fi

),


where (∗1≤i≤n fi

)(x) =

∫ ∞

0· · ·

∫ ∞

0︸︷︷︸n−1

f1(x1) f2(x2) · · · fn−1(xn−1)

fn

(x

x1x2 · · · xn−1

)dxn−1

xn−1· · · dx2

x2

dx1

x1.

In the below of Theorem 15.3.3, we have proved that

Γ(z)Γ(1− z) =∫ ∞

0

11 + y

yz−1dy.

Hence Γ(z)Γ(1− z) and 11+x is a Mellin pair. The following exercise gives another

proof of this fact.

Exercise 15.4.17. Prove that (a) 1x e−

1x and Γ(1− z) is a Mellin pair.

(b) e−x ∗ ( 1x e−

1x ) = 1

1+x .

(c) e−nx1/nand Γ(nz)/nnz−1 is a Mellin pair.

We give alternative proof of Theorem 15.3.5. Since (Γ(z), e−x) and (Γ(z +12 ),√

xe−x) are Mellin pairs, it follows that Γ(z)Γ(z + 12 ) and

f (x) :=∫ ∞

0

√te−te−x/t dt

t

is a Mellin pair. If u = x/t then

f (x) =∫ ∞

0

√x

ue−

xu−u du√

u= −√

x f ′(x) =⇒ f ′(x)f (x)

= − 1√x

.

Since the initial condition is f (0) = Γ(1/2) =√

π, we have

f (x) =√

πe−2√

x.

hence Γ(z)Γ(z + 12 ) and

√πe−2

√x is a Mellin pair. However

√πe−2

√x and

√πΓ(2z)22z−1

is also a Mellin pair, we obtain Legendre formula.

Theorem 15.4.18. (Gauss) For any n ∈ N we have

(15.4.32) ∏0≤k≤n−1

Γ(

z +kn

)=

(2π)n−1

2√

n· Γ(nz)

nnz−1 .

PROOF. For each k, we see that Γ(z + kn ) and xk/ne−x is a Mellin pair. Accord-

ing to (15.4.31), we conclude that ∏0≤k≤n Γ(z + kn ) and

f (x) =∫ ∞

0· · ·

∫ ∞

0︸︷︷︸n−1

e− x

x1x2 ···xn−1 x1nn−1e−xn−1 · · · x

n−2n

2 e−x2 xn−1

n1 e−x1

dx1dx2 · · · dxn−1

x1x2 · · · xn−1

=∫ ∞

0· · ·

∫ ∞

0︸︷︷︸n−1

e−(x1···+xn−1+

xx1 ···xn−1

) dx1 · · · dxn−1

x1n1 · · · x

n−1n

n−1


is a Mellin pair. Using (15.3.10) we arrive at

(2π)n−1

2√

n= ∏

1≤k≤n−1Γ(

kn

)=

∫ ∞

0e−x1 x

n−1n

1dx1

x1

∫ ∞

0e−x2 x

n−2n

2dx2

x2· · ·

∫ ∞

0e−xn x

1nn−1

dxn−1

xn−1

=∫ ∞

0· · ·

∫ ∞

0︸︷︷︸n−1

e−(x1+···+xn−1)dx1 · · · dxn−1

x1n1 · · · x

n−1n

n−1

.

Consider the function

(15.4.33) f (x)− (2π)n−1√

ne−n n√x.

The following exercise shows

f (x) =(2π)n−1√

ne−n n√x.

On the other hand, using Exercise 15.4.17 (c), we see that (e−n n√x, Γ(nz)/nnz−1) isa Mellin pair.

Exercise 15.4.19. Show that the function (15.4.33) is identically zero. (Hint: Con-sider its derivative).

Theorem 15.3.2 and Theorem 15.3.3 show that the value of Γ(s) is determinedby its value in the interval (0, 1/2]. We now show that the value of Γ(s) is deter-mined by its value in the interval (0, 1/3]. By Theorem 15.3.3 we have

Γ(

1 + x2

)Γ(

1− x2

)=

π

sin(πx2 + π

2 ).

By Theorem 15.3.5, we get

Γ( x

2

)Γ(

1 + x2

)=√

π21−xΓ(x).

Hence

(15.4.34) Γ(x) =√

π

21−x cos πx2·

Γ( x2 )

Γ( 1−x2 )

.

When x ∈ ( 13 , 1

2 ], we have x2 , 1−x

2 ∈ (0, 13 ]. Then for any x ∈ ( 1

3 , 12 ], Γ(x) is given by

(15.4.34).

Exercise 15.4.20. Show that the value of Γ(x) is determined by its value in theinterval (0, 1/4].

Γ(

13− 2x

)=

√π

213+2x cos(π

6 − πx)·

Γ( 16 − x)

Γ( 13 + x)

,

Γ(

23+ x)

Γ(

76+ x)

=

√π

213+2x

Γ(

43+ 2x

),

Γ(3x) =33x− 1

2 2−53−2x

√π cos(π

6 − πx)Γ( 2

3 + x)Γ(x)Γ( 16 − x)

Γ( 13 − 2x)

.


Hence

Γ(x) =

2−

5+2x3 3x− 1

2√

π

sin π+πx3 sin π−πx

3· Γ( x

3 )Γ(1−2x

6 )

Γ( 1−x3 )Γ( 1−2x

3 ), x ∈

(14 , 3

8

),

4−x√πsin(πx) ·

Γ( 1−2x2 )

Γ(1−2x) , x ∈[

38 , 1

2

).

Exercise 15.4.21. Use Theorem 15.4.18 to show that

sin x = 2n−1 ∏0≤k≤n−1

sinx + kπ

n.

(Hint: choose z = x/n.))

Theorem 15.4.22. If f , g ∈ L1(R+, dx/x), thenM( f g) =M( f ) ∗M(g).

PROOF. Write F =M( f ) and G =M(g). Since

f (x) =1

2π√−1

∫ σ+√−1∞

σ−√−1∞

F(w)x−wdw,

we have

M( f g)(z) =∫ ∞

0f (x)g(x)xz dx

x

=1

2π√−1

∫ ∞

0

[∫ σ+√−1∞

σ−√−1∞

F(w)x−wdw

]g(x)xz dx

x

=1

2π√−1

∫ σ+√−1∞

σ−√−1∞

F(w)

[∫ ∞

0g(x)xz−w dx

x

]dw

=1

2π√−1

∫ σ+√−1∞

σ−√−1∞

F(w)G(z− w)dw.

The last one is the definition of (F ∗ G)(z).

Letting y = nx in Exercise 15.4.17 (c), we see that (e−nx, Γ(z)/nz) is a Mellinpair for each n ∈ N. Recall the Riemann zeta function given by

ζ(z) := ∑n∈N

1nz .

Theorem 15.4.23. (1) If Re(z) > 1, then ( 1ex−1 , Γ(z)ζ(z)) is a Mellin pair.

(2) If Re(z) ∈ (0, 1), then ( 1ex−1 −

1x , Γ(z)ζ(z)) is a Mellin pair.

(3) If σ > 1, then

1ex − 1

=1

2π√−1

∫ σ+√−1∞

σ−√−1∞

Γ(z)ζ(z)x−zdz.

PROOF. It follows from the fact that 1u−1 = ∑n∈N u−n.

Exercise 15.4.24. Complete the proof of Theorem 15.4.23.

If a function f has the expansion

f (x) = ∑n∈N

ane−nx,


such that for any σ ∈ (a, ∞) we have fσ(x) = xσ f (x) ∈ L1(R+, dx/x), then

M( fσ)(τ) = Γ(σ +√−1τ) ∑

n∈N

an

nσ+√−1τ

, σ > a.

Theorem 15.4.25. If for any σ ∈ (a, ∞) ⊂ R+ we have xσ ∑n∈N ane−nx ∈ L1(R+, dxx ),

then ∑n∈N ane−nx and Γ(z)∑n∈Nannz , Re(z) > a, is a Mellin pair.

The series

(15.4.35) ∑n∈N

an

nz

is called a Dirichlet series associated to the sequence (an)n∈N.By integration by parts, we have

Γ(z)ζ(z) =∫ ∞

0

xz

ex − 1dxx

=∫ ∞

0

xex − 1

xz−2dx

= − 1z− 1

∫ ∞

0

ddx

(x

ex − 1

)xz dx

x;

in general, for each k = 0, 1, 2, · · · , one has

Γ(z)ζ(z) =(−1)k+1

(z− 1)z(z + 1) · · · (z + k− 1)

∫ ∞

0xz+k−1 dk+1

dxk+1

(x

ex − 1

)dx.

Equivalently, for each k = 0, 1, 2, · · · ,

(15.4.36) ζ(z) =(−1)k+1

(z− 1)Γ(z + k)

∫ ∞

0xz+k dk+1

dxk+1

(x

ex − 1

)dxx

.

When Re(z) > −k, the improper integral∫ ∞

0xz+k dk+1

dxk+1

(x

ex − 1

)dxx

is convergent, so that the right-hand side of (15.4.36) is a holomorphic function forany z with Re(z) > −k but z 6= 1 (because Γ(z + k) is holomorphic and identicallyzero). Consequently, we can holomorphically extend the Riemann zeta function tothe entire complex plane except z = 1.

Theorem 15.4.26. (a) ζ(z) is holomorphic on C \ z = 1.(b) limz→1(z− 1)ζ(z) = 1.(c) If Re(z) > −k, then ζ(z) is given by (15.4.36).(d) (z− 1)ζ(z) is a entire function.(e) ζ(0) = 1

2 .

PROOF. Compute

(z− 1)ζ(z) = − 1Γ(z)

∫ ∞

0

ddx

(x

ex − 1

)xz dx

x.

Letting z→ 1 yields

limz→1

(z− 1)ζ(z) = − 1Γ(1)

∫ ∞

0

ddx

(x

ex − 1

)dx = − x

ex − 1

∣∣∣∣∞0= 1.

Hence z = 1 is the simple pole of ζ(z).


We finally compute the value of ζ(0). Letting k = 1 and z = 0 in (15.4.36)imply

ζ(0) = −∫ ∞

0

d2

dx2

(x

ex − 1

)dx = − d

dx

(x

ex − 1

) ∣∣∣∣∞0

=(x− 1)ex + 1(ex − 1)2

∣∣∣∣∞0

= − limx→0

(x− 1)ex + 1(ex − 1)2

= − limx→∞

ex = (x− 1)ex

2(ex − 1)ex = − limx→0

x2(ex − 1)

= −12

.

Hence ζ(0) = −1/2.

Exercise 15.4.27. For a given positive number a > 0, define the Hurwitz zetafunction by

ζ(z, a) := ∑n≥0

1(n + a)z .

Show that ζ(z, a) is convergent for any z with Re(z) > 1, and is uniformly conver-gent for any z with Re(z) > 1+ δ (δ > 0). Moreover ( e−ax

1−e−x , Γ(z)ζ(z, a)) is a Mellinpair.

Define two functions I, J by

(15.4.37) I :=∫ ∞

0e−t cos(ut)dt, J :=

∫ ∞

0e−t sin(ut)dt.

Note thatI +√−1J =

∫ ∞

0e−t+

√−1utdt.

Compute

I = −∫ ∞

0cos(ut)de−t = −

[0− 1 + u

∫ ∞

0e−t sin(ut)dt

]= 1− uJ,

I =∫ ∞

0

e−t

ud sin(ut) =

∫ ∞

0

sin(ut)u

e−tdt =1u

J.

Hence

(15.4.38) I =1

1 + u2 , J =u

1 + u2 .

Write

(15.4.39) c(z) :=M(cos x), s(z) :=M(sin x)

the Mellin transform of cos x and sin x, respectively.

Exercise 15.4.28. (Euler) Show that2π

∫ ∞

0

u−z c(z)1 + u2 du =

∫ ∞

0e−ttz−1dt = Γ(z),∫ ∞

0

x−z

1 + x2 dx =π

2sec

πz2

,

c(z) = Γ(z) cosπz2

, Re(z) > 0,

s(z) = Γ(z) sinπz2

, Re(z) > −1.


Together with Exercise 15.4.12 (c), we have

Corollary 15.4.29. (Euler) (a) For each fixed t > 0, (cos(tx), Γ(z) cos πz2 · t−z) is a

Mellin pair. Hence

(15.4.40)∫ ∞

0cos(tx)xz dx

x=

Γ(z)tz cos

πz2

, Re(z) > 0,

and

(15.4.41) cos(tx) =1

2π√−1

∫ σ+√−1∞

σ−√−1∞

Γ(z) cosπz2(tx)−zdz.

(b) For each fixed t > 0, (sin(tx), Γ(z) sin πz2 · t−z) is a Mellin pair. Hence

(15.4.42)∫ ∞

0sin(tx)xz dx

x=

Γ(z)tz sin

πz2

, Re(z) > −1,

and

(15.4.43) sin(tx) =1

2π√−1

∫ σ+√−1∞

σ−√−1∞

Γ(z) sinπz2(tx)−zdz.

In particular, the above exercise gives us the identity that

(15.4.44)π

2Γ(z) sin πz2

= Γ(1− z) cosπz2

=∫ ∞

0sin x · x1−z dx

x=∫ ∞

0

sin xxz dx.

Letting z = 1 in (15.4.44) yields

(15.4.45)∫ ∞

0

sin xx

dx =π

2.

Theorem 15.4.30. (1) Suppose that ( f (x), F(z)) is a Mellin pair. Then TFAE

(i) f (x) satisfies

(15.4.46) f (x) =

√2π

∫ ∞

0f (t) cos(tx)dt.

(ii) F(z) satisfies

(15.4.47)F(z)

2z2 Γ( z

2 )=

F(1− z)

21−z

2 Γ( 1−z2 )

.

(2) Suppose that ( f (x), F(z)) is a Mellin pair. Then TFAE

(i) f (x) satisfies

(15.4.48) f (x) =

√2π

∫ ∞

0f (t) sin(tx)dt.

(ii) F(z) satisfies

(15.4.49)F(z)

2z2 Γ( 1

2 + z2 )

=F(1− z)

21−z

2 Γ( 12 + 1−z

2 ).


PROOF. We give a proof of (i). (i) ⇒ (ii) : Using (15.4.46) and (15.4.40) wehave

F(z) =∫ ∞

0f (x)xz dx

x=

√2π

∫ ∞

0

[∫ ∞

0f (t) cos(tx)dt

]xz dx

x

=

√2π

∫ ∞

0f (t)

[∫ ∞

0cos(tx)xz dx

x

]dt

=

√2π

Γ(z) cosπz2

∫ ∞

0f (t)t1−z dt

t=

√2π

Γ(z) cosπz2

F(1− z).

According to the identity (see Problems 15.6)

2z2 Γ( z

2 )

21−z

2 Γ( 1−z2 )

=

√2π

Γ(z) cosπz2

,

we obtain (15.4.47).(ii)⇒ (i) : The above identity together with (15.4.47) gives us

F(z) =

√2π

∫ ∞

0

[∫ ∞

0f (t) cos(tx)dt

]xz dx

x.

The uniqueness of Mellin transforms implies (15.4.46).

Let Kc(z) and Ks(z) denote the Mellin transform of√

2/π cos x and√

2/π sin xrespectively. Then

(15.4.50) Kc(z) =

√2π

Γ(z) cosπz2

, Ks(z) =

√2π

Γ(z) sinπz2

by Exercise 15.4.28.

Exercise 15.4.31. Let ( f (x), F(z)) and (g(x), G(z)) are Mellin pairs.(1) If f (x) and g(x) satisfy

f (x) =

√2π

∫ ∞

0g(t) cos(tx)dt, g(x) =

√2π

∫ ∞

0f (t) cos(tx)dt,

thenF(z) = G(1− z)Kc(z), G(z) = F(1− z)Kc(z).

(2) If f (x) and g(x) satisfy

f (x) =

√2π

∫ ∞

0g(t) sin(tx)dt, g(x) =

√2π

∫ ∞

0f (t) sin(tx)dt,

thenF(z) = G(1− z)Ks(z), G(z) = F(1− z)Ks(z).

To apply Theorem 15.4.30 we should find these functions satisfying (15.4.46)or (15.4.48). For example

(i) x−12 , e−

12 x2

, sech(√

π2 x), · · · , satisfy (15.4.46).

(ii) x−12 , xe−

12 x2

, 1e√

2πx−1− 1√

2πx, · · · , satisfy (15.4.48).

We will discard the function e−12 since it does not belong to L1(R+, dx/x).


(1) f (x) = 2e−12 x2

. Since (e−x, Γ(z)) is a Mellin pair, it follows from Exercise15.4.12 (e) that ( f (x), F(z)) is a Mellin pair, where F(z) = 2

z2 Γ( z

2 ). Hence(15.4.47) holds.

(2) f (x) =√

2xe−12 x2

. Since (2e−12 x2

, 2z2 Γ( z

2 )) is a Mellin pair, it followsfrom Exercise 15.4.12 (d) that ( f (x), F(z)) is a Mellin pair, where F(z) =2

z2 Γ( z+1

2 ). Hence (15.4.49) holds.

(3) f (x) = sech(√

π2 x). Recall that

12

sech(x) =1

ex + e−x =e−x

1 + e−2x = ∑n≥0

(−1)ne−(2n+1)x

so that∫ ∞

0

12

sech(x)dxx

= ∑n≥0

(−1)n∫ ∞

0e−(2n+1)xxz dx

x= Γ(z) ∑

n≥0

(−1)n

(2n + 1)z .

Define χ1 : N→ Z3 by

(15.4.51) χ1(2k) = 0, χ1(4k + 1) = 1, χ1(4k + 3) = −1.

Its associated Dirichlet series is denoted by

(15.4.52) L(z, χ1) := ∑n∈N

χ1(n)nz .

Hence we can conclude that

(15.4.53)(

12

sech(x), Γ(z)L(z, χ1)

)is a Mellin pair.

Consequently, ( f (x), 2( 2π )

z2 Γ(z)L(z, χ1)) is a Mellin pair.

(4) f (x) = cosh( 12√

πx)/cosh(√

πx) satisfies (15.4.46). Then ( f (x), F(z)) isa Mellin pair, where F(z) = L(z, χ2) = ∑n∈N χ2(n)/nz and

(15.4.54) χ2(n) :=

1, n ≡ 1, 3 (mod 8),−1, n ≡ 5, 7 (mod 8),0, otherwise.

(5) f (x) = sinh( 12√

πx)/sinh(√

πx) satisfies (15.4.48). Then ( f (x), F(z)) is aMellin pair, where F(z) = L(z, χ3) = ∑n∈N χ3(n)/nz and

(15.4.55) χ3(n) :=


(6) f (x) = 1

1+2cosh(√

2π3 x)

satisfies (15.4.46). Then ( f (x), F(z)) is a Mellin pair,

where F(z) = L(z, χ4) = ∑n∈N χ4(n)/nz and

(15.4.56) χ4(n) :=

1, n ≡ 1 (mod 3),−1, n ≡ −1 (mod 3),0, otherwise.


(7) f (x) =sinh(√

π6 x)

cosh(√

2π3 x)−1

satisfies (15.4.48). Then ( f (x), F(z)) is a Mellin pair,

where F(z) = L(z, χ5) = ∑n∈N χ5(n)/nz and

(15.4.57) χ5(n) :=


(4) f (x) = 1e√

2πx−1− 1√

2πx. Then ( f (x), F(z)) is a Mellin pair, where F(z) =

(2π)−z2 Γ(z)ζ(z) with Re(z) ∈ (0, 1). If we define

(15.4.58) λ(z) :=F(z)

2z2 Γ( 1

2 + z2 )

=1

2√

ππ−

z2 Γ( z

2

)ζ(z),

then

(15.4.59) λ(z) = λ(1− z), 0 < Re(z) < 1

by (15.4.49). Consequently, for any z ∈ C with 0 < Re(z) < 1, one has

(15.4.60) ζ(1− z) =[

2z+1Γ(z)π sin πz

2

]ζ(z), ζ(z) =

[2zπz−1 sin

πz2

Γ(1− z)]

ζ(1− z).

Moreover, compute

π−z2 Γ( z

2

)ζ(z) = π−

z2 Γ( z

2

) [2zπz−1 sin

πz2

Γ(1− z)]

ζ(1− z)

= πz2−12z sin

πz2

Γ( z

2

)Γ(1− z)ζ(1− z)

= πz2−12z π

Γ( z2 )Γ(1−

z2 )

Γ( z

2

)Γ(1− z)ζ(1− z)

= πz2 2z Γ(1− z)

Γ(1− z2 )

ζ(1− z) = πz2 2z Γ( 1−z

2 )√π2z ζ(1− z).

Thus we get

(15.4.61) π−z2 Γ( z

2

)ζ(z) = π−

1−z2 Γ

(1− z

2

)ζ(1− z), 0 < Re(z) < 1.

Theorem 15.4.32. (Riemann) Define

(15.4.62) ξ(z) :=

12 z(z− 1)π−

z2 Γ( z

2 )ζ(z), Re(z) > 0,12 z(z− 1)π−

1−z2 Γ( 1−z

2 )ζ(1− z), Re(z) < 1.

Then ξ(z) is an entire function and satisfies

(15.4.63) ξ(z) = ξ(1− z)

for any z ∈ C.

PROOF. By Theorem 15.3.2 (3) and Theorem 15.4.26 (b), we can conclude thatthe function 1

2 z(z− 1)π−z2 Γ( z

2 )ζ(z) is holomorphic for any z with Re(z) > 0. Sim-

ilarly, the function 12 z(z − 1)π−

1−z2 Γ( 1−z

2 )ζ(1− z) is holomorphic for any z withRe(z) < 1. Over the interval 0 < Re(z) < 1, these two functions are the same by(15.4.61).


Since (e−x, Γ(z)) is a Mellin pair, wee see that (e−πx2, 1

2 π−z2 Γ( z

2 ) is a Mellinpair.

(i) For each n ∈ N, (2e−πn2x2, π−

z2 Γ( z

2 )1nz ) is a Mellin pair. That is,

π−z2 Γ( z

2

) 1nz = 2

∫ ∞

0e−πn2x2

xz dxx

, Re(z) > 0,(15.4.64)

2e−πn2x2=

12π√−1

∫ σ+√−1∞

σ−√−1∞

π−z2 Γ( z

2

) x−z

nz dz, σ > 0.(15.4.65)

(ii) By (i), we see that (∑n∈N 2e−πn2x2, π−

z2 Γ( z

2 )ζ(z)) is a Mellin pair. Define

(15.4.66) θ(x) := ∑n∈Z

e−πn2x, Λ(z) := π−z2 Γ( z

2

)ζ(z).

Theorem 15.4.33. (θ(x2 − 1), Λ(z)) is a Mellin pair. That is,

Λ(z) =∫ ∞

0

[θ(x2)− 1

]xz dx

x, Re(z) > 1(15.4.67)

θ(x2)− 1 =1

2π√−1

∫ σ+√−1∞

σ−√−1∞

π−z2 Γ( z

2

)ζ(z)x−zdz, σ > 1.(15.4.68)

Hence, ( 12 [θ(x)− 1], Λ(2z)) is a Mellin pair for Re(z) > 1

2 .

From (15.4.68) we see that for any σ > 1

θ(x2)− 1 =1

2π√−1

∫ 1−σ+√−1∞

1−σ−√−1∞

π−z2 Γ( z

2

)ζ(z)x−zdz +

1x− 1

=1

2π√−1

∫ σ+√−1∞

σ−√−1∞

π−1−z

2 Γ(

1− z2

)ζ(1− z)x−1+zdz +

1x− 1

=1

2π√−1

∫ σ+√−1∞

σ−√−1∞

π−z2 Γ( z

2

)ζ(z)

(1x

)−zdz +

1x− 1

=1x

[θ

(1x2

)− 1]− 1

x− 1

by Theorem 15.4.32. Here the first identity follows from Cauchy’s integral formulaand Theorem 15.4.26.

Theorem 15.4.34. For any x > 0 we have

(15.4.69) θ(x) =1√x

θ

(1x

).

Thus θ(x) is a modular form of weight 1/2 (see Section 15.5 for more details).

Exercise 15.4.35. Verify that

12π√−1

(∫ σ+√−1∞

σ−√−1∞

−∫ 1−σ+

√−1∞

1−σ−√−1∞

)π−

z2 Γ( z

2

)ζ(z)x−zdz =

1x− 1.


15.4.2. Ψ and Φ functions. For any z ∈ C \ Z≤0 define

f (z) := − ∑n≥0

(1

z + n− 1

n + 1

).

Since for n 1, we have

1z + n

− 1n + 1

=1− z

n2 +z2 − 1

n3 +1− z3

n4 +z4 − 1

n5 + · · · ∼ 1− zn2

and hence the series ∑n≥0(1

z+n −1

n+1 ) is convergent. Note that

f (z + 1) =1z+ f (z).

If g′(z) = f (z) and h(z) = eg(z), then

h(z + 1) = Czh(z).

Hence h(z) looks like Γ(z) or f (z0 looks like Γ/(z)/Γ(z).

Define Ψ function by

(15.4.70) Ψ(z) := −γ− ∑n≥0

(1

z + n− 1

n + 1

)= ∑

n≥0

[ln(

1 +1n

)− 1

z + n

].

Theorem 15.4.36. Ψ(z) satisfies

Ψ(z + 1)−Ψ(z) =1z

,(15.4.71)

Ψ(1− z)−Ψ(z) = π cot(πz),(15.4.72)

Ψ(1 + z)−Ψ(1− z) =1z− π cot(πz),(15.4.73)

ψ

(12+ z)−Ψ

(12− z)

= π tan(πz).(15.4.74)

PROOF. The first identity is obvioud. Recall the identity

(15.4.75) π cot(πz) =1z+ ∑

n∈N

(1

z + n+

1z− n

).

Then (15.4.72) and (15.4.73) follow. By replacing z by 12 − z in (15.4.72) we obtain

(15.4.74).

Theorem 15.4.37. Ψ(k)(z) satisfy

Ψ′(z) = ∑n≥0

1(z + n)2 ,(15.4.76)

Ψ′(z) + Ψ′(1− z) = π2csc2(πz),(15.4.77)

Ψ(k)(z) = ∑n≥0

(−1)k+1k!(z + n)k+1 = (−1)k+1k!ζ(k + 1, z),(15.4.78)

Ψ(k)(1) = (−1)k+1k!ζ(k + 1).(15.4.79)

Here ζ(k + 1, z) is the Hurwitz zeta function given in Exercise 15.4.27.


PROOF. Recall that (using (15.4.75))

π2

sin2(πz)= π2csc2(πz) = ∑

n∈Z

1(z + n)2 .

Then we get (15.4.76) and (15.4.77). The last two identities are obvious.

Recall that Ψ(z) = −γ + f (z) where

f (z) = − ∑n≥0

(1

z + n− 1

n + 1

).

Compute

f2N+1(z) := ∑0≤n≤2N+1

(1

n + 1− 1

n + z

)= ∑

0≤n≤N

(1

2n + 1− 1

2n + z

)+ ∑

0≤n≤N

[1

(2n + 1) + 1− 1

(2n + 1) + z

]

=12 ∑

0≤n≤N

(1

n + 12− 1

n + z2

)+

12 ∑

0≤n≤N

(1

n + 1− 1

n + z+12

)

=12

fN

( z2

)+

12

fN

(z + 1

2

)+

12 ∑

0≤n≤N

(1

n + 12− 1

n + 1

)

=12

[fN

( z2

)+ fN

(z + 1

2

)]+ ∑

1≤n≤2N

(−1)n−1

n.

Letting N → ∞ yields

Theorem 15.4.38. Ψ satisfies

Ψ(z) =12

[Ψ( z

2

)+ Ψ

(z + 1

2

)]+ ln 2,(15.4.80)

Ψ(z) =1n ∑

0≤k≤n−1Ψ(

z + kn

)+ ln n, n ∈ N.(15.4.81)


We define

(15.4.82) Φ(z) :=∫ z

1Ψ(w)dw

with Φ(1) = 0, where the integral path does not intersect the negative real axis.

Theorem 15.4.40. We have

(15.4.83) Φ(2) = 0, Φ(

12

)= ln

√π.


PROOF. Compute

Φ(2) =∫ 2

1

[−γ− ∑

n≥0

(1

z + n− 1

n + 1

)]dz

= −γ− ∑n≥0

∫ 2

1

(1

z + n− 1

n + 1

)dz = −γ− ∑

n≥0

(ln

n + 2n + 1

− 1n + 1

)= −γ + lim

N→∞

[1 +

12+ · · ·+ 1

N− ln(N + 1)

]= −γ + γ = 0.

Recall the Wallis formula that

(15.4.84)π

2= lim

N→∞

[(2N − 2)!!(2N − 1)!!

]2

2N.

Compute

Φ(

12

)=

∫ 12

1

[−γ− ∑

n≥0

(1

z + n− 1

n + 1

)]dz

=12

γ− ∑n≥0

(ln

12 + n1 + n

+12· 1

n + 1

)

=12

γ− 12

limN→∞

(1 +

12+ · · ·+ 1

N− ln N

)− lim

N→∞

(∑

0≤n≤N−1ln

2n + 12n + 2

+12

ln N

)

= limN→∞

ln[

(2N)!!(2N − 1)!!

· 1√N

].

Hence we get 2Φ(1/2) = ln π.

Exercise 15.4.41. This exercise gives a proof of (15.4.84). Let

I(n) :=∫ π

0sinn xdx.

Then

I(n) =n− 1

nI(n− 2),

I(n)I(n− 2)

=n− 1

n,

I(2n− 1)I(2n + 1)

=2n + 1

2n.

Using I(2k + 1) < I(2k) < I(2k− 1) we get

(2k)!!(2k + 1)!!

<(2k− 1)!!(2k)!!

π

2<

(2k− 2)!!(2k− 1)!!

=⇒ 1 <π2[

(2k)!!(2k−1)!!

]2 12k+1

<2k + 1

2k.

Thusπ

2= lim

k→∞

[(2k)!!

(2k− 1)!!

]2 12k + 1

.

Equivalently, one has (2n)!!/(2n− 1)!! ∼√

πn as n→ ∞.


Integrating (15.4.71) we get

Φ(z + 1)−Φ(z) = ln z + C1.

Letting z = 1 yields C1 = 0. Integrating (15.4.72) we get

−Φ(1− z)−Φ(z) = ln sin(πz) + C2.

Letting z = 12 yields C2 = − ln π. Integrating (15.4.80) we get

Φ(z) = Φ( z

2

)+ Φ

(z + 1

2

)+ z ln 2 + C3.

Letting z = 1 yields C3 = − ln 2− ln√

π.

Theorem 15.4.42. Φ(z) satisfies

Φ(z + 1) = Φ(z) + ln z,(15.4.85)Φ(z) + Φ(1− z) = ln π − ln sin(πz),(15.4.86)

Φ(z) = Φ( z

2

)+ Φ

(1 + z

2

)+ ln(2z−1)− ln

√π.(15.4.87)

if we write F(z) := eΦ(z), then Theorem 15.4.42 states

F(1 + z) = zF(z),

F(z)F(1− z) =π

sin(πz),

F(z)F(

z +12

)=

√π

22z−1 F(2z).

By Theorem 15.3.6 we have F(z) = Γ(z).

Theorem 15.4.43. We have

(15.4.88) Φ(z) = ln Γ(z), Ψ(z) =Γ′(z)Γ(z)

.

Moreover

(15.4.89) Φ(z) = −γz− ln z + ∑n∈N

[ zn− ln

(1 +

zn

)].

PROOF. Integrating (15.4.70) yields

Φ(z) =∫ z

1Ψ(w)dw =

∫ z

1

[−γ− 1

w− ∑

n∈N

(1

w + n− 1

n

)]dw

= −γz + γ− ln z− ∑n∈N

(ln

z + n1 + n

− zn+

1n

)= −γz + γ− ln z + ∑

n∈N

(zn− ln

z + nn

)+ ∑

n∈N

(ln

n + 1n− 1

n

)= −γz + γ− ln z + ∑

n∈N

(zn− ln

z + nn

).

Note the series in (15.4.89) is absolutely convergent, since zn − ln(1+ z

n ) ∼z2

2n2 .

Exercise 15.4.44. Verify that the series in (15.4.89) is absolutely convergent.


Using the expansion ln(1 + 1n ) = ∑n∈N

(−1)k−1

knk , we get

ln(1 + N)− ∑1≤n≤N

1n= ∑

1≤n≤Nln(

1 +1n

)− ∑

1≤n≤N

1n= ∑

k≥2∑

1≤n≤N

(−1)k−1

knk .

Letting N → ∞ yields (15.4.90) below, because ζ(2) > ζ(3) > ζ(4) > · · · .

Proposition 15.4.45. We have

γ = ∑k≥2

(−1)kζ(k)k

,(15.4.90)

Ψ(1 + z) = −γ + ∑n∈N

(−1)n−1ζ(n + 1)zn, |z| < 1,(15.4.91)

Φ(1 + z) = ln Γ(1 + z)

= −γz + ∑n≥2

(−1)nζ(n)n

zn, |z| < 1,(15.4.92)

Ψ(z) = Ψ(z0)

+ ∑k∈N

(−1)k+1ζ(k + 1, z0)(z− z0)k, z0 ∈ C \ Z≤0,(15.4.93)

γ = ∑n≥2

(−1)n ζ(n)− 1n

+ 1− ln 2,(15.4.94)

Φ(1 + z) = ln Γ(1 + z)

= − ln(1 + z) + (1− γ)z + ∑n≥2

(−1)n ζ(n)− 1n

zn,(15.4.95)

ln Γ(1 + z) =12

lnπz

sin(πz)− γz

− ∑n∈N

ζ(2n + 1)2n + 1

z2n+1,(15.4.96)

ln Γ(1 + z) =12

lnπz

sin(πz)+

12

ln1− z1 + z

+ (1− γ)z

− ∑n∈N

ζ(2n + 1)− 12n + 1

z2n+1,(15.4.97)

γ = 1 + ln 2− ln 3− ∑n∈N

ζ(2n + 1)− 1(2n + 1)4n ,(15.4.98)

Ψ(1 + z) =12z− π

2cot(πz)− ∑

n≥0ζ(2n + 1)z2n,(15.4.99)

Ψ(1 + z) =12z− π

2cot(πz)− 1

1− z2 − ∑n≥0

[ζ(2n + 1)− 1] z2n.(15.4.100)

Here ζ(1) := γ.


PROOF. Compute

Ψ(1 + z) = −γ− ∑n≥0

(1

z + 1 + n− 1

n + 1

)= −γ− ∑

n∈N

(1

z + n− 1

n

)

= −γ− ∑n∈N

[∑k≥0

(−1)k

n

( zn

)k− 1

n

]= −γ− ∑

n∈N∑

k∈N

(−1)k

nk+1 zk

= −γ + ∑k∈N

(−1)k+1ζ(k + 1)zk.

Hence γ can be viewed as “ζ(1)”. Taking the integration of (15.4.91) we obtain(15.4.92).

For any z0 ∈ C \ Z≤0, we have

Ψ(z0 + z)−Ψ(z0) = ∑n≥0

(1

z0 + n− 1

z0 + z + n

)

= ∑n≥0

[1

z0 + n− 1

z0 + n ∑k≥0

(−1)k(

zz0 + n

)k]

= ∑n≥0

∑k∈N

(−1)k+1

(z0 + n)k+1 zk = ∑k∈N

(−1)k+1ζ(k + 1, z0)zk

Since

(15.4.101) 1 < ζ(n) < 1 +1

2n−2 , n ≥ 2,

we have

0 < ζ(n)− 1 <1

2n−2

and get Euler’s result (15.4.94). The formula (15.4.95) follows from (15.4.92) and

ln(1 + z) = ∑n∈N(−1)n−1

n zn. Replacing z by −z in (15.4.95) yields

ln Γ(1− z) = − ln(1− z)− (1− γ)z + ∑n≥2

ζ(n)− 1n

zn.

Hence

ln Γ(1 + z)− ln Γ(1− z)− ln1− z1 + z

+ 2(1− γ)z− 2 ∑n∈N

ζ(2n + 1)− 12n + 1

z2n+1.

Using the identity

ln Γ(1 + z) + ln Γ(1− z) = lnπz

sin(πz),

we conclude the formula (15.4.97). While the formula (15.4.96) follows from theidentity

(15.4.102) ln sin(πz) = ln(πz)− ∑n∈N

ζ(2n)n

z2n.

Letting z = 12 in (15.4.97) yields (15.4.98).

Exercise 15.4.46. Verify (15.4.101) and (15.4.102).


Theorem 15.4.47. (Binet) (1) We have

(15.4.103) Γ(1 + z) = ∑n≥0

anzn, |z| < 1,

where

(15.4.104) a0 = 1, an+1 =1

n + 1 ∑0≤k≤n

(−1)k+1ζ(k + 1)an−k, n ≥ 0.

Actually,

(15.4.105) an = (−1)n ∑1≤k≤n

∑r1+···+rk=n

1k!

ζ(r1) · · · ζ(rk)

r1 · · · rk.

Here ζ(1) := γ.(2) We have

(15.4.106)1

Γ(1 + z)= ∑

n≥0bnzn, |z| < 1,

where

(15.4.107) b0 = 1, bn+1 =1

n + 1 ∑0≤k≤n

(−1)kζ(k + 1)bn−k, n ≥ 0.

Actually

(15.4.108) bn = (−1)n ∑1≤k≤n

∑r1+···+rk=n

(−1)k

kζ(r1) · · · ζ(rk)

r1 · · · rk.

PROOF. Since Γ′(1 + z) = ∑n∈N nanzn−1 for |z| < 1 and Ψ(1 + z)Γ(1 + z) =Γ′(1 + z), we have

∑0≤k≤n

(−1)k+1ζ(k + 1)an−k = (n + 1)an+1, n ∈ N

by (15.4.91). The formula (15.4.105) follows immediately.

For example,

a1 = −ζ(1) = −γ, a2 =12

ζ(2) +12

ζ(1)2 =π2

12+

γ2

2and

b1 = ζ(1) = γ, b2 = −12

ζ(2) +12

ζ(1)2 = −π2

12+

γ2

2.

Letting z = x +√−1y in (15.4.70) we have

Ψ(x +√−1y) = −γ + ∑

n≥0

[1

n + 1− 1

(x + n) +√−1y

]= −γ + ∑

n≥0

[1

n + 1− x + n

(x + n)2 + y2

]+√−1 ∑

n≥0

y(x + n)2 + y2 .

Hence the function Ψ(z) has only real zeros. When x, y > 0 we gat

(15.4.109) Ψ(x)−Ψ(y) = (x− y) ∑n≥0

1(x + n)(y + n)

.


Thus Ψ(x) is monotone increasing in the interval (0, ∞). Since Ψ(1) = −γ < 0and Ψ(2) = 1− γ > 0, it follows that Ψ(r) = 0 for a unique r ∈ (1, 2). On otherhand, for each n ∈ N, the function Ψ(x) is monotone increasing in the interval(−n − 1,−n) and Ψ(x) has a simple pole at −n. Hence Ψ(rn) = 0 for a uniquern ∈ (−n− 1,−n). We also have

(15.4.110) Ψ(n) = −γ + 1 +12+ · · ·+ 1

n− 1.

Theorem 15.4.48. We have the following properties:

(a) Ψ(z) has only real zeros.(b) Ψ(x) is monotone increasing in the interval (0, ∞) and for each n ∈ N the

function Ψ(x) is monotone increasing in the interval (−n− 1,−n).(c) Ψ(x) has only one positive real zero r ∈ (1, 2).(d) Ψ(x) has only one zero rn ∈ (−n− 1,−n).(e) limx→∞[Ψ(x)− ln x] = 0.

Observe that

r = 1.4616321451105, r0 = −0.504083008, r1 = −1.57349847, r2 = −2.61072087.

Write

(15.4.111) rn := −n− 1 + εn, εn ∈ (0, 1).

Letting z = rn = −n− 1 + εn in (15.4.72) yields

π cot(πεn) = Ψ(n + 2− εn) = ln n + δn

by Theorem 15.4.48 (e), for some δn → 0. Therefore

εn =1π

tan−1(

π

ln n + δn

)=

1π

[π

ln n + δn− 1

3

(π

ln n + δn

)3+ · · ·

]

=1

ln n + δn− π2

3(ln n + δn)3 + · · · .

Corollary 15.4.49. (Hermite) One has

(15.4.112) rn = −n− 1 +1

ln n+ δ′n,

where δ′n = O(1/(ln n)3).

The following result is a consequence of Theorem 15.3.2.

Corollary 15.4.50. Γ(x) satisfies

(a) Γ(x) is positive in the interval (0, ∞) and attains the minimum at r.(b) Γ(x) is positive in the interval (−2n,−2n + 1) and attains the minimum at

r2n−1.(c) Γ(x) is negative in the interval (−2n − 1,−2n) and attains the maximum at

r2n−1.


15.4.3. Bohr-Mollerup-Artin theorem. Recall that

Γ(x + 1) = xΓ(x), ln Γ(x) = Φ(x), Φ′(x) = Ψ(x), Ψ′(x) > 0 in (0, ∞).

We say a function F(x) on (a, b) is convex if

F(λx1 + (1− λ)x2) ≤ λF(x1) + (1− λ)F(x2)

for any λ ∈ (0, 1) and x1, x2 ∈ (a, b). If F ∈ C2(a, b), then F is convex if and only ifF′′ > 0. In summary, we have

(i) Γ(x + 1) = xΓ(x).(ii) ln Γ(x) is convex in (0, ∞).

(iii) Γ(1) = 1.

Theorem 15.4.51. If a function f : (0, ∞)→ (0, ∞) satisfies(i) f (x + 1) = x f (x),

(ii) ln f (x) is convex in (0, ∞),(iii) f (1) = 1,

thenf (x) = lim

n→∞

nxn!x(x + 1) · · · (x + n)

.

PROOF. Suppose n ≥ 2 and x ∈ (0, 1]. Then f (n) = (n − 1)!. Let F(x) :=ln f (x). Since n + x = x(n + 1) + (1− x)n, we get

F(n + x) ≤ xF(n + 1) + (1− x)F(n)

so thatF(n + x)− F(n)

x≤ F(n + 1)− F(n).

Since n = x1+x (n− 1) + (1− x

1+x )(n + x), it follows that

F(n) ≤ x1 + x

F(n− 1) +1

1 + xF(n + x)

so that

F(n)− F(n− 1) ≤ F(n + x)− F(n)x

.

Using F(n + x)− F(n) = ln n, we conclude that

ln(n− 1) ≤ ln f (x + n)− ln(n− 1)!x

≤ ln n

or equivalently

ln [(n− 1)x(n− 1)!] ≤ ln f (x + n) ≤ ln [nx(n− 1)!] .

Because of f (x + n) = (x + n− 1)(x + n− 2) · · · (x + 1)x f (x), we arrive at

(n− 1)x(n− 1)!x(x + 1) · · · (x + n− 1)

≤ f (x) ≤ nx(n− 1)!x(x + 1) · · · (x + n− 1)

.

Letting

Gn(x) :=nxn!

x(x + 1) · · · (x + n)we have

Gn(x) ≤ f (x) ≤ x + nn

Gn(x)

for any n ≥ 2 and x ∈ (0, ∞). Hence f (x) = limn→∞ Gn(x).


Corollary 15.4.52. (Euler) One has

(15.4.113) Γ(z) = limn→∞

nzn!z(z + 1) · · · (z + n)

.

The formula (15.4.113) can be obtain by direct computation. From∫ 1

0tx−1(1− t)ndt =

n!x(x + 1) · · · (x + n)

we have

Gn(x) = nx∫ 1

0tx−1(1− t)ndt =

∫ n

0yx−1

(1− y

n

)ndy.

Hence Γ(x) = limn→∞ Gn(x).We observe that

Gn(z)Gn−1(z)

= 1 +cn(z)

n2

where limn→∞ cn(z) =z(z+1)

2 . Hence

(15.4.114) Γ(z) =1z ∏

n∈N

(1 +

cn(z)n2

).

Theorem 15.4.53. (Bohr-Mollerup-Artin) If a function f : (0, ∞)→ (0, ∞) satisfies(i) f (x + 1) = x f (x),

(ii) ln f (x) is convex in (0, ∞),(iii) f (1) = 1,

then f (x) = Γ(x) for any x ∈ (0, ∞).

We now give several applications of Theorem 15.4.53.

(A) Theorem 15.4.18: For each n ∈ N define

f (x) := nxΓ( x

n

)Γ(

x + 1n

)· · · Γ

(x + n− 1

n

).

Then f (x + 1) = x f (x) and ln f (x) is convex. According to Theorem15.4.53 we have f (x) = cnΓ(z), where

cn = f (1) = nΓ(

1n

)Γ(

2n

)· · · Γ

(n− 1

n

)=√

n(2π)n−1

2

by (15.3.10).(B) Theorem 15.3.3: Let

g(x) :=π

Γ(1− x) sin(πx), x ∈ (0, ∞).

Then ln g(x) is convex and g(x + 1) = xg(x). According to Theorem15.4.53 we have g(x) = cΓ(x) where

c =π

Γ(x)Γ(1− x) sin(πx).

Letting x → 0 yields c = 1.


(C) B(x, a) = Γ(x)Γ(a)/Γ(x + a) : Compute

B(x, a) = limn→∞

n!(x + a)(x + a + 1) · · · (x = a + n)x(x + 1) · · · (x + n)a(a + 1) · · · (a + n)

= limn→∞

(1 +

xa

)(1 +

xa + 1

)· · ·(

1 +x

a + n

)1x

(1− x

x + 1

)(1− x

x + 2

)· · ·(

1− xx + n

).

Corollary 15.4.54. One has

(15.4.115) B(a, x) =1x ∏

n∈N

[(1 +

xa + n− 1

)(1− x

x + n

)].

We now consider two polynomials

P(x) = ∏1≤i≤d

(x− αi), Q(x) = ∏1≤i≤d

(x− βi),

where α1, · · · , αd, β1, · · · , βd ∈ C and β1, · · · , βd /∈ Z≤0. Hence

(15.4.116) ∏n∈N

P(n)Q(n)

is convergent⇐⇒ ∑1≤i≤d

αi = ∑1≤i≤d

βi.

In this case

(15.4.117) ∏n∈N

P(n)Q(n)

=Γ(1− β1) · · · Γ(1− βd)

Γ(1− α1) · · · Γ(1− αd).

Exercise 15.4.55. Verify (15.4.116). Take P(x) = x2 and Q(x) = x2 − 14 to reprove

(15.4.84).

15.4.4. Euler and Weierstrass product formulas. Since

nz =

(1 +

11

)z· · ·(

1 +1

n− 1

)z (1 +

1n

)z· 1(1 + 1

n )z

,

(z + 1) · · · (z + n)n!

=(

1 +z1

) (1 +

z2

)· · ·(

1 +zn

),

it follows from (15.4.113) that

(15.4.118) Γ(z) =1z ∏

n∈N

(1 + 1n )

z

(1 + zn )

,

called the Euler product formula. Because of the expansion(1 +

zn

)e−

zn =

(1 +

zn

)∑k≥0

(−1)k

k!zk

nk = 1− ∑k≥2

(−1)k(k− 1)k!

zk

nk

and

ln[zGN(z)] = z ∑1≤n≤N

ln(

1 +1n

)− ∑

1≤n≤Nln(

1 +zn

)− z ln

(1 +

1N

)= ∑

1≤n≤N

[ zn− ln

(1 +

zn

)]− z ∑

1≤n≤N

[1n− ln

(1 +

1n

)]− z ln

(1 +

1N

),


we have

Γ(z) =1z ∏

n∈N

ez[ln(1+ 1n )−

1n ]

(1 + zn )e− z

n=

ez ∑n∈N [ln(1+ 1n )−

1n ]

z ∏n∈N[(1 + zn )e− z

n ]

=e−γz

z· 1

∏n∈N[(1 + zn )e− z

n ].

Theorem 15.4.56. (Weierstrass) One has

(15.4.119)1

Γ(z)= zeγz ∏

n∈N

[(1 +

zn

)e−

zn

].

The above shows that 1/Γ(z) is an entire function. Letting z = x +√−1y in

(15.4.119) we arrive at

1|Γ(x +

√−1y)|

=√

x2 + y2eγx ∏n∈N

[√(1 +

xn

)2+( y

n

)2e−

xn

]

=√

x2 + y2eγx ∏n∈N

[(1 +

xn

)e−

xn

√1 +

y2

(x + n)2

].

On the other hand,

1|Γ(x +

√−1y)|

=

√x2 + y2

Γ(x + 1)

√∏

n∈N

[1 +

y2

(x + n)2

]by

1Γ(x + 1)

=1

xΓ(x)= eγx ∏

n∈N

[(1 +

xn

)e−

xn

].

If x ∈ (0, 1), then we get

11 + y2 ∏

n∈N

(1 +

y2

n2

)< ∏

n∈N

[1 +

y2

(x + n)2

]< ∏

n∈N

(1 +

y2

n2

)=

eπy − e−πy

2π.

Theorem 15.4.57. For any x ∈ (0, 1) and y 6= 0, we have

(15.4.120) |Γ(x +√−1y)| = λ · Γ(x + 1)√

x2 + y2

√2πy

eπy − e−πy

for some λ = λ(x) ∈ (1,√

1 + y2).

Let ξ3 := −1+√−3

2 . Then ξ33 = 1.

Corollary 15.4.58. (Liouville) One has

(15.4.121)1

Γ(z)Γ(ζ3z)Γ(ζ23z)

= z3 ∏n∈N

(1 +

z3

n3

).

PROOF. Compute

1Γ(z)Γ(ζ3z)Γ(ζ2

3z)= z3 ∏

n∈N

[(1 +

zn

)(1 +

ζ3zn

)(1 +

ζ23zn

)]= z3 ∏

n∈N

(1 +

z3

n3

)since 1 + ζ3 + ζ2

3 = 0.


Exercise 15.4.59. If ζm := e2π√−1/m, then

∏0≤k≤m−1

1Γ(1− ζk

mz)= ∏

n∈N

(1− zm

nm

).

For any w ∈ C we have

∏0≤k≤m−1

(w− ζkmz) = wm − zm.

Therefore

∏0≤k≤m−1

1Γ(w− ζk

mz)

= ∏0≤k≤m−1

(w− ζk

mz)eγ(w−ζkmz) ∏

n∈N

[(1 +

w− ζkmz

n

)e−

w−ζkmz

n

]= (wm − zm)emγw ∏

n∈N

[(1 +

wn

)m−( z

n

)m]e−

mn w

= wm(

1− zm

wm

)emγw ∏

n∈N

[(1 +

wn

)me−

mn w]

∏n∈N

[1− zm

(w + n)m

]=

1[Γ(w)]m ∏

n≥0

[1− zm

(w + n)m

].

Theorem 15.4.60. (Mellin) One has

(15.4.122) ∏1≤k≤m−1

Γ(w)

Γ(w− ζkmz)

= ∏n≥0

[1−

(w

z + n

)m].

Since 1, ζm, · · · , ζm−1m are exact roots of the algebraic equation Xm − 1 = 0, we

can ask what is the infinite product of

∏1≤i≤m

1Γ(w− riz)

where r1, · · · , rm are roots of a polynomial?

Exercise 15.4.61. (Mellin) Prove that

wΨ(z) = −γw− wz− ∑

n∈N

(w

z + n− w

n

),

ewΨ(z) = e−γwe−wz ∏

n∈Ne−

wz+n +

wn ,

ewΨ(z) Γ(z)Γ(z + w)

= ∏n∈N

[(1 +

wz + n

)e−

wz+n

],

eΨ(z) = ∏n∈N

[(1 +

1z + n

)e−

1z+n

].


Exercise 15.4.62. (Liouville) Let

f (z) := z3 ∏n∈N

(1 +

z3

n3

), g(z) := ∏

n∈N

[1 +

8z3

(2n + 1)3

],

g1(z) := ∏n∈N

[1 +

27z3

(3n + 1)3

], g2(z) := ∏

n∈N

[1 +

27z3

(3n + 2)3

].

Thenf (2z) = 8 f (z)g(z), f (3z) = 27 f (z)g1(z)g2(z).

15.4.5. Asymptotics of Γ(z). If we replace 1k by∫ 1

0uk−1du

in (15.4.110), then we obtain

Theorem 15.4.63. (Legendre) When Re(z) > 0, we have

Ψ(z) = −γ +∫ 1

0

1− uz−1

1− udu(15.4.123)

= −γ +∫ ∞

0

[1

1 + t− 1

(1 + t)z

]dtt

(15.4.124)

= −γ +∫ ∞

0

(e−x − e−zx

1− e−x

)dx.(15.4.125)

PROOF. The second identity follows from the first one by the change of vari-ables u = 1

1+t , while the third one by u = e−x.

Exercise 15.4.64. Show that

Ψ(

32

)= −γ + 2− ln 2,

Ψ(

43

)= −γ + 3− 3

2ln 3− π

6

√3,

Ψ(

34

)= −γ +

π

2− 3 ln 2,

Ψ(

14

)= −γ− π

2− 3 ln 2,

Ψ(

23

)= −γ +

π

2

√13− 3

2ln 3,

Ψ(

13

)= −γ− π

2

√13− 3

2ln 3.

Corollary 15.4.65. (1) If Re(z), Re(w) > 0, then

Ψ(z)−Ψ(w) =∫ 1

0

uw − uz

1− uduu

(15.4.126)

=∫ ∞

0

[1

(1 + t)w −1

(1 + t)z

]dtt

(15.4.127)

=∫ ∞

0

e−wx − w−zx

1− e−x dx.(15.4.128)


(2) If −1 < Re(z) < 1, then

1z− π cot(πz) =

∫ 1

0

u−z − uz

1− udu

=∫ ∞

0

[(1 + t)z+1 − 1

(‘ + t)z+1

]dtt

(15.4.129)

=∫ ∞

0

e(1+z)x − e(1−z)x

1− e−x dx.

(3) One has

(15.4.130) lnsin(πz)

πz=∫ 1

0

uz − u−z − 21− u

duln u

.

Exercise 15.4.66. Prove that

lnπ

2=

∫ 1

0

v− 1v + 1

dvln v

=∫ 1

0

(u14 − u−

14 )2

u− 1du

ln u,

lnπ

2√

2=

∫ 1

0

(y2 − 1)y2

(1 + y)(1 + y2)

dyln y

.

Using (15.3.5) we obtain

γ = −∫ ∞

0e−t ln tdt = lim

n→∞

∫ ∞

1n

(−e−t ln t

)dt

= limn→∞

(e−t ln t

∣∣∣∣∞1n

−∫ ∞

1n

e−t

tdt

)= lim

n→∞

(e−

1n ln n−

∫ ∞

1n

e−t

tdt

On the other hand, since e−1/n ln n ∼ ln n ∼ ln(n + 1), it follows that

γ = limn→∞

[∫ ∞

0

1t(t + 1)

dt−∫ ∞

1n

e−t

tdt]

.

Theorem 15.4.67. (Euler) We have

(15.4.131) γ =∫ ∞

0

(1

1 + t− e−t

)dtt

.


γ =∫ ∞

0

(1

ex − 1− 1

xex

)dx(15.4.132)

=∫ 1

0

(1

1− v− 1

ln(1/v)

)dv(15.4.133)

=∫ 1

0

(1u− e1− 1

u

u2

)du.(15.4.134)

Corollary 15.4.69. (Dirichlet-Gauss) When Re(z) > 0 we have

Ψ(z) =∫ ∞

0

[e−t − 1

(1 + t)z

]dtt

(15.4.135)

=∫ ∞

0

(e−x

x− e−zx

1− e−x

)dx.(15.4.136)


Exercise 15.4.70. Give an alternative proof of (15.4.135) by considering the doubleintegral ∫∫

[0,∞)×[0,∞)e−uuz−1 (e−y − e−uy) dydu

y.


Ψ(x) =∫ 1

0

(e1− 1

y − yz)

dyy(1− y)

=∫ 1

0

[1

ln(1/v)− vz−1

1− v

]dv.

Integrating (15.4.135) and (15.4.136) yield

Corollary 15.4.72. (Plana) When Re(z) > 0 we have

ln Γ(z) =∫ ∞

0

[e−t(z− 1)− (1 + t)−1 − (1 + t)−z

ln(1 + t)

]dtt

,(15.4.137)

ln Γ(z) =∫ ∞

0

[e−x(z− 1) +

e−zx − e−x

1− e−x

]dxx

.(15.4.138)

Recall that1

ex − 1− 1

x+

12= ∑

n∈N

(−1)n−1B′n(2n)!

x2n−1 = − 1x+

12

cothx2

, 0 < |x| < 1,

or

(15.4.139)1

22πt − 1− 1

2πt+

12= ∑

n∈N

(−1)n−1ζ(2n)π

t2n−1, 0 < |t| < 12π

,

where B′n := B2n = 2(2n)!ζ(2n)/(2π)2n the 2n-th Bernoulli number given by(15.3.32) (A proof can be found in 16.2.5). For example,

B′1 =16

, B′2 =1

30, B′3 =

142

, B′4 =1

30, B′5 =

566

, B′6 =691

2730, B′7 =

76

, B′8 =3617510

.

Recall also that

(15.4.140) x coth(x) = 1 + 2x2 ∑n∈N

1x2 + (πn)2 .

Motivated by (15.4.139), we can rewrite (15.4.136) as

Ψ(z + 1) =∫ ∞

0

(e−x

x− e−zx

ex − 1

)dx

=∫ ∞

0

12

e−zxdx +∫ ∞

0

e−x − e−zx

2dx−

∫ ∞

0

(1

ex − 1− 1

x+

12

)e−zxdx.

Theorem 15.4.73. When Re(z) > 0 we have

Ψ(z + 1) =12z

+ ln z−∫ ∞

0

(1

ex − 1− 1

x+

12

)e−zxdx,(15.4.141)

Ψ(z) = − 12z

+ ln z−∫ ∞

0

(1

ex − 1− 1

x+

12

)e−zxdx.(15.4.142)

PROOF. It follows from Ψ(z + 1)−Ψ(z) = 1z ,

According to Theorem 15.4.48 we define

(15.4.143) ν(z) := ln z−Ψ(z).


Corollary 15.4.74. When Re(z) > 0 we have

ν(z) =12z

+∫ ∞

0

(1

ex − 1− 1

x+

12

)e−zxdx,(15.4.144)

ν(z) = ∑n≥0

[1

z + n− ln

(1 +

1z + n

)],(15.4.145)

ln Γ(z + 1) =12

ln z + z(ln z− 1) + 1 + µ(z)− µ(1).(15.4.146)

Here

(15.4.147) µ(z) :=∫ ∞

0

(1

ex − 1− 1

x+

12

)e−zx dx

x, µ′(z) =

12z− ν(z).

Note that µ(z) can be written as

µ(z) =∫ ∞

0

(1

1− e−x −1x− 1

2

)e−zx dx

x.

PROOF. Using (15.4.70) and limN→∞ ln z+N+1N+1 = 0, we obtain

ν(z) = ln z− ∑n≥0

[ln(

1 +1n

)− 1

z + n

]= ln z− lim

N→∞∑

1≤n≤N

(ln

n + 1n− 1

z + n

)= lim

N→∞∑

0≤n≤N

[1

z + n− ln

(1 +

1z + n

)]+ lim

N→∞∑

0≤n≤N

(ln

z + n + 1z + n

− lnn + 1

n

)+ ln z

= ∑n≥0

[1

z + n− ln

(1 +

1z + n

)]+ lim

N→∞ln

z + N + 1N + 1

= ∑n≥0

[1

z + n− ln

(1 +

1z + n

)].

Integrating (15.4.142) and using ln Γ(z + 1) = ln Γ(z + 1) − ln Γ(2) =∫ z

1 Ψ(w +1)dw we obtain (15.4.146).

From (15.4.146) we have

Γ(z + 1) = zz+ 12 e1−zeµ(z)−µ(1).

Note that limx→∞ µ(x) = 0.

Consider the double integral

I :=∫ c

a

∫ ∞

0

e−vy − e−by

ydydv.


By Froullani integrals2, we have

lnbv=∫ ∞

0

e−vy − e−by

ydy.

By Fubini’s theorem, we arrive at(15.4.148)∫ ∞

0

[e−ax − e−cx

x− (c− a)e−bx

]dxx

= (c− a)(1 + ln b)− (c ln c− a ln a).

Observe that

µ(1) =∫ ∞

0

(1

ex − 1− 1

x+

12

)e−x dx

x=∫ ∞

0

(1

1− e−x −1x− 1

2

)e−x dx

x.

Since

µ(1) =∫ ∞

0

(1

1− e−2x −1

2x− 1

2

)e−2x dx

xand

11− e−2x =

11− e−x −

e−x

1− e−2x ,

it follows that

0 =∫ ∞

0

(1

1− e−2x −2− e−x

2x− 1− e−x

2

)e−x dx

x.

Lemma 15.4.75. We have

(15.4.149) µ

(12

)=

12− 1

2ln 2, µ(1) = 1− ln

√2π.

PROOF. Using the above identity and the expression

µ

(12

)=

∫ ∞

0

(1

1− e−x −1x− 1

2

)e−

12 x dx

x

=∫ ∞

0

(1

1− e−2x −1

2x− 1

2

)e−x dx

x

=12

∫ ∞

0

(1− e−x

x− e−x

)e−x dx

xwe have

µ

(12

)=

12[(2− 1)(1 + ln 2)− (2 ln 2− 1 ln 1)] =

1− ln 22

by taking a = 1 and b = c = 2 in (15.4.148). Letting z = 1/2 in (15.4.146) weconclude that

ln Γ(

32

)=

12

ln12+

12

(ln

12− 1)+ 1 + µ

(12

)− µ(1)

which implies µ(1) = 12 − ln

√π + µ(1/2) = 1− ln

√2π.

Exercise 15.4.76. Another method to compute C := 1− µ(1). Show that

2If f ∈ C[0, ∞) and f (∞) := limx→∞ f (x) exists, then∫ ∞

0[ f (ax)− f (bx)]

dxx

= [ f (0)− f (∞)] lnba

for any a, b > 0.


(a) ∑1≤k≤n ln k = C + (n + 12 ) ln n− n + µ(n).

(b) ∑1≤k≤2n ln k = C + (2n + 1) ln(2n)− 2n + µ(2n).(c) ∑1≤k≤n ln(2k) = n ln 2 + C + (n + 1

2 ) ln n− n + µ(n).(d) ∑1≤k≤n(2k− 1) = n ln n + (n + 1

2 ) ln 2− n + µ(2n)− µ(n).(e) As n→ ∞ we have

(2 ln 2− ln 1) + ∑2≤k≤n−1

[2 ln(2k)− 2 ln(2k− 1)] + [ln(2n)− 2 ln(2n− 1)]→ lnπ

2.

(f) C = ln√

2π.

Corollary 15.4.77. (1) When Re(z) > 0 we have

ln Γ(z + 1) =

(z +

12

)ln z− z + ln

√2π + µ(z),(15.4.150)

ln Γ(z) =

(z− 1

2

)ln z− z + ln

√2π + µ(z).(15.4.151)

(2) When x 1 we have

ln Γ(x + 1) ∼(

x +12

)ln x− x + ln

√2π,(15.4.152)

Γ(x) ∼√

2πxx− 12 e−x.(15.4.153)

(3) (Stirling) When n 1 we have

(15.4.154) n! ∼√

2πnn+ 12 e−n =

√2πn

(ne

)n.

PROOF. The first two identities follow from (15.4.146), (15.4.149), and Γ(z +1) = zΓ(z). The last three identities follows from µ(x)→ 0 as x → ∞.

Proposition 15.4.78. (1) ν(z) satisfies

(15.4.155) ν(z + 1) = ν(z) + ln(

1 +1z

)− 1

z.

(2) (Binet) µ(z) satisfies

(15.4.156) µ(z + 1) = µ(z) +[

1−(

z +12

)ln(

1 +1z

)].

(3) (Gudermann) We have

(15.4.157) µ(z) = ∑n≥0

[(z + n +

12

)ln(

1 +1

z + n

)− 1]

.

(4) We have

(15.4.158) ν(z) = ∑k≥2

(−1)k

kζ(k, z).

(5) We have

(15.4.159) µ(z) = ∑k≥2

(−1)k(k− 1)2k(k + 1)

ζ(k, z).


(6) (Binet) We have

(15.4.160) µ(z) = ∑k≥2

k− 12k(k + 1)

ζ(k, z + 1).

(7) We have

(15.4.161) ν(z) =1z− ∑

k≥2

1k

ζ(k, z + 1).

PROOF. (1) follows from (15.4.145).(2) By (15.4.147) we have

µ′(z+ 1)−µ′(z) =1

2(z + 1)− ν(z+ 1)− 1

2z+ ν(z) =

12(z + 1)

+12z− ln

(1 +

1z

).

Integrating over [1, z] yields

µ(z + 1)− µ(2)− µ(z) + µ(1) =ln(z + 1)− ln 2

2+

12

ln z + z(ln z− 1) + 1

− (z + 1)[ln(z + 1)− 1] + 2(ln 2− 1).

However the formula (15.4.146) tells µ(2)− µ(1) = − 32 ln 2 + 1 and hence

µ(z + 1)− µ(z) = 1−(

z +12

)ln(z + 1) +

(z +

12

)ln z.

(3) For each n ≥ 0 one has

µ(z) = µ(z + n + 1)− ∑0≤k≤n

[µ(z + k + 1)− µ(z + k)]

= µ(z + n + 1)− ∑0≤k≤n

[1−

(z + k +

12

)ln(

1 +1

z + k

)]= ∑

0≤k≤n

[(z + k +

12

)ln(

1 +1

z + k

)− 1]+ µ(z + n + 1).

Letting n→ ∞ we obtain (15.4.157).(4) Each term in (15.4.145) is equal to

1z + n

− ln(

1 +1

z + n

)=

1z + n

− ∑k∈N

(−1)k−1

k1

(z + n)k = ∑k≥2

(−1)k

k1

(z + n)k .

Therefore we get (15.4.158).(5) Similarly, each term in (15.4.157) is equal to(

z + n +12

)ln(

1 +1

z + n

)− 1 = ∑

k≥2

(−1)k(k− 1)2k(k + 1)

1(z + n)k .

Therefore we get (15.4.159).(6) Since (

11− e−x −

1x− 1

2

)1x=

ex(x− 2) + (x + 2)2x2(ex − 1)

,

it follows that

µ(x) =12

∫ ∞

0

ex(x− 2) + (x + 2)x2 · e−zx

ex − 1dx =

12

∫ ∞

0∑k≥1

k(k + 2)!

xke−zx

ex − 1dx.


Using xke−zx/(ex − 1) = ∑m≥0 xkw−(z+m+1)x we obtain (15.4.160). The same ar-gument can be applied to (15.4.161).

Exercise 15.4.79. (Lerch) Prove that

|Γ(x +√−1y)| =

√2π(x2 + y2)

12 (x− 1

2 )e−x−y tan−1 yx (1 + ε)

where ε→ 0 when x or y tends to infinity.

For any k ∈ N the inequality k2 − 5k + 6 = (k− 2)(k− 3) ≥ 0 implies

k− 1k(k + 1)

≤ 16

.

(Here we can take > when l > 3.) Hence, for any x > 0, we arrive at

(15.4.162) 0 < µ(x) <1

12 ∑k≥2

ζ(k, x + 1).

On the other hand,

∑k≥2

ζ(k, x + 1) = ∑k≥2

∑n≥1

1(x + n)k = ∑

n≥1∑k≥2

1(x + n)k

= ∑n≥1

1(x + n)2

11− 1

x+n= ∑

n≥1

1(x + n)(x + n− 1)

= ∑n≥1

(1

x + n− 1− 1

x + n

)=

1x

.

Consequently, we obtain

(15.4.163) 0 < µ(x) <1

12x.

Theorem 15.4.80. (Stirling) For any x > 0 we have

(15.4.164) Γ(x) =√

2πxx− 12 e−xe

θx12x

for some θx ∈ (0, 1). In particular

(15.4.165) n! =√

2πnn+ 12 e−ne

θn12n .

Recall from (15.4.151) that

Γ(x) =√

2πxx− 12 e−xeµ(x).

From (15.4.140) we see that(1

ex − 1− 1

x+

12

)1x= 2 ∑

n∈N

1x2 + (2πn)2 .

However1

x2 + (2πn)2 =1

(2πn)2 ·1

1 + x2

(2πn)2

=1

(2πn)2

[∑

0≤k≤m−1(−1)k x2k

(2πn)2k + (−1)m rn

(2πn)2m x2m

],


where rn := 4n2π2/(4n2π2 + x2) ∈ (0, 10. Consequently for each m ∈ N we have(1

ex − 1− 1

x+

12

)1x

= 2 ∑1≤k≤m

∑n∈N

(−1)k−1x2k−2

(2πn)2k

+ 2(−1)mθmx2m ∑n∈N

1(2πn)2m+2

where θm ∈ (0, 1). From (15.3.32) and (15.4.147) we get

µ(z) = 2 ∑1≤k≤m

∑n∈N

(−1)k−1

(2πn)2k

∫ ∞

0x2k−2e−zxdx

+ 2(−1)mθm ∑n∈N

1(2πn)2m+2

∫ ∞

0x2me−zxdx

= 2 ∑1≤k≤m

∑n∈N

(−1)k−1

(2πn)2k(2k− 2)!

z2k−1 + 2(−1)mθm ∑n∈N

1(2πn)2m+2

(2m)!z2m+1

= 2 ∑1≤k≤m

(−1)k−1(2k− 2)!(2π)2kz2k−1 ζ(2k) + 2(−1)mθm

(2m)!z2m+1

1(2π)2m+2 ζ(2m + 2)

= ∑1≤k≤m

(−1)k−1B2k

2k(2k− 1)z2k−1 + (−1)mθmB2m+2

(2m + 2)(2m + 1)z2m+1 .

Corollary 15.4.81. Given Re(z) > 0. For each m ∈ N there exists θm ∈ (0, 1) such that

(15.4.166) µ(z) = ∑1≤k≤m

(−1)k−1B2k

2k(2k− 1)z2k−1 + (−1)mθmB2m+2

(2m + 2)(2m + 1)z2m+1 .

An immediate consequence of (15.4.166) is

(15.4.167) n! ≈√

2πnn+ 12 e−ne

112n−

1360n3 =

√2πn

(ne

)ne

112n−

1360n3 .

Exercise 15.4.82. (Cauchy) Show for each n ∈ N that

2(2n)2n+ 1

2

(2π)2n− 12

e−2n < B2n <1

12(2n)2n+ 1

12

(2π)2n− 52

e−2n+ 124n .

Taking the integral on both sides of the identity12 − uu + z

=z + 1

2u + z

− 1

yields ∫ 1

0

12 − uu + z

du =

(z +

12

)ln(

1 +1z

)− 1.

In general, for each n ≥ 0, one has∫ n+1

n

12 − u + n

u + zdu =

(z + n +

12

)ln(

1 +1

z + n

)− 1.

Hence

µ(z) = ∑n≥0

∫ n+1

n

12 − u + n

u + zdu = ∑

n≥0

∫ 1

0

12 − t

u + z + ndu.


Define

(15.4.168) Λ(t) :=12− t + btc, t ∈ R.

Theorem 15.4.83. When Re(z) > 0 we have

(15.4.169) µ(z) =∫ ∞

0

Λ(t)t + z

dt.

Exercise 15.4.84. (Binet-Liouville-Bourgoret) Show that

µ(x) = 2∫ ∞

0

tan−1(t/x)e2πt − 1

dt

=xπ

∫ ∞

0ln(

11− e−2πt

)dt

t2 + x2

= ∑n∈N

1nπ

∫ ∞

0

sin(2nπt)t + x

dt.

15.5. Modular forms

15.6. Problems

Problem 15.6.1. Prove

γ = −∫ 1

0ln(− ln x)dx

=

[1 +

12+

13+ · · ·+ 1

n− ln

(n +

12

)]− 2 ∑

k≥n+1

[13

1(2k)3 +

15

1(2k)5 + · · ·

],

where γ is the Euler constant.

Problem 15.6.2. (a) (Euler) Show that for any n ∈ N we have

Γ(

n +12

)=∫ 1

0

[ln(

1x

)] 2n−12

dx =(2n− 1)!!

2n

√π.

(b) Compute Γ(−n + 12 ).

(c) If z ∈ C \ Z≤0, then Γ(z) = Γ(z).(d) Prove that for any real number y 6= 0 we have

|Γ(√−1y)| =

√2π

y(eπy − e−πy).

(e) Prove that for any real number y we have∣∣∣∣Γ(12+√−1y

)∣∣∣∣ =√

2π

eπy + eπy ≤√

π.

(f) Prove that for any real number y we have

lim|y|→∞

Γ(

12+√−1y

)= 0.

15.6. PROBLEMS 159

(g) (Euler-Laplace) Prove that∫ ∞

−∞x2ne−x2

dx =(2n)!4nn!

√π.

(h) Prove that

Γ(1− ε1)Γ(1− ε5) = Γ(1− ε2)Γ(1− ε4)

where εk := ekπ√−1/3.

Problem 15.6.3. Prove for 0 < Re(z) < 1 that∫ ∞

0

x−z

1 + x2 dx =π

2sec

πz2

,∫ ∞

0

xz−1

1 + xdx = π cot(πz).

Problem 15.6.4. (a) Prove for a, b > 0 that

B(a, b) =∫ 1

0

xa−1 + xb−1

(1 + x)a+b dx.

(b) Prove for a, b > 0 that∫ ∞

0e−at (1− e−nt) b

n−1 dt =1n

B(

an

,bn

).

(c) Prove for any n ∈ N that

Γ(2n + 1)Γ( 12 )

Γ(n + 1)Γ(n + 12 )

= 22n.

(d) Prove that

2z2 Γ( z

2 )

21−z

2 Γ( 1−z2 )

=

√2π

Γ(z) cosπz2

,2

z2 Γ( 1

2 + z2 )

21−z

2 Γ(1− z2 )

=

√2π

Γ(z) sinπz2

.

(e) Prove that [Γ(

23

)]2= 21/3

(π

3

)1/2Γ(

16

).

(d) Prove that

Γ(

12− z)=

22z−1√π

cos(πz)Γ(z)

Γ(2z), Γ(z) =

√π

4z sin(πz)Γ( 1

2 − z)Γ(1− 2z)

.

(e) (Knar) Prove for any z ∈ C \ Z≤0 and n ∈ N that

Γ(z) =2z(2−21−n)

(2√

π)n Γ( z

2n

)∏

1≤k≤nΓ(

12+

z2k

).

(f) Prove that for any x > 0 that

Γ(x) =2n√Γ(2nx)2nx−1+2−n ∏

1≤k≤n

2k

√B(

2k−1x,12

).

(g) (Dirichlet formula) Prove for any a, b, c, k > 0 that

Γ(a)∫ ∞

0

e−kxxb−1

(x + c)a dx = Γ(b)∫ ∞

0

e−cyya−1

(y + k)b dy.


In particular,

Γ(a + b)∫ ∞

0

ya−1

(1 + y)a+b dy = Γ(a)Γ(b).

(h) Define F on (0, ∞) by

F(x) := exp

[∑

n∈N

sin(2nπx)2n

]Γ(x).

Show that F ∈ C(0, ∞), F(x + 1) = xF(x), F(x) is not identically zero, and F(x)F(x +12 ) =

√π21−2xF(2x). However F(x) 6= Γ(x).

(i) Verify that the function

f (x) :=

[2Γ( z+3

4 )

Γ( z+14 )

]2

satisfies f (z + 1) f (z− 1) = z2.(j) Prove for 0 < Re(z) < 1 that∫ ∞

0e−xxz dx

x·∫ ∞

0e−xx1−z dx

x=∫ 1

0

xz

1 + xdxx·∫ 1

0

x1−z

1 + xdxx

.

(k) Prove that ∫ 1

0

x2dx√1− x4

·∫ 1

0

dx√1− x4

=∫ 1

0

xdx√1− x4

.

Problem 15.6.5. (a) Let

Ω :=

(x1, · · · , xn) ∈ Rn : x1, · · · , xn ≥ 0, ∑

1≤i≤n

(xi

ai

)mi

≤ 1

where a1, · · · , an > 0 and m1, · · · , mn ∈ N. Prove that∫

Ω∏

1≤i≤n(xi)qi−1dx1 · · · dxn = ∏

1≤i≤n

aqii

mi·

∏1≤i≤n Γ( qimi)

Γ(1 + ∑1≤i≤mqimi)

.

(b) Prove that the volume of (x, y, z) ∈ R3 : (x/a)2 + (y/b)2 + (z/c)2 ≤ 1 witha, b, c > 0 equals 4π

3 abc.(c) Prove that the volume of (x, y, z) ∈ R3 : (x/a)4 + (y/b)4 + (z/c)4 ≤ 1 with

a, b, c > 0 equals abc√

26 [B( 1

4 , 12 )]

2.

Problem 15.6.6. If ( f (x), F(z)) is a Mellin pair, then

12π

∫ ∞

−∞

∣∣∣∣F(12+√−1τ

)∣∣∣∣2 dτ =∫ ∞

0f 2(x)dx.

Problem 15.6.7. (Euler) Given p, q > 0, define

α := tan−1 qp∈(−π

2,

π

2

).

Show that(e−px cos(q− x),

Γ(z)(p2 + q2)

z2

cos(αz)

),

(e−px sin(q− x),

Γ(z)(p2 + q2)

z2

sin(αz)

)are Mellin pairs.

15.6. PROBLEMS 161

Problem 15.6.8. Prove that

ζ(z) =(2π)z

2Γ(z) cos πz2

ζ(1− z).

Problem 15.6.9. Define

θ(z) := ∑n∈Z

eπ√−1n2z.

Show that the series is abstractly convergent in H := z ∈ C : Im(z) > 0 and

θ(z + 2) = θ(z), θ

(−1

z

)=

√−√−1zθ(z), z ∈ H.

Problem 15.6.10. Show that

8πhc3

∫ ∞

0

ν3dν

chνkT − 1

=8π4k4T4

15c3h3 ,

where c, ν, T, k, h > 0.


Ψ(

12

)= −γ− ln 2, Ψ

(32

)= 2− γ− 2 ln 2,

Γ′(

12

)= −

√π(γ + 2 ln 2), Γ′

(32

)=√

π(

1− γ

2− ln 2

),∫ ∞

0e−x2

ln xdx =14

Γ′(

12

),∫ ∞

0e−xn

ln xdx =1n2 Γ′

(1n

),

Γ′′(1) = γ2 +π2

6, Γ′′

(12

)=√

π

[(γ + 2 ln 2)2 +

π2

2

],∫ ∞

0e−x2

(ln x)2dx =14

Γ′′(

12

),∫ ∞

0e−xn

(ln x)2dx =1n2 Γ′′

(1n

).

Problem 15.6.12. For each fixed x > 0 prove that the quadratic form Γ(x)t2 + 2Γ′(x)t+Γ′′(x) is positive definite. As a consequence, ln Γ(x) is convex in (0, ∞).


Γ(z) =1z ∏

n∈N

(n + 1)z

nz−1(z + n).


∏n∈N

n2 + 9n + 18n2 + 9n + 20

=57

, ∏n∈N

2n2 + 5n + 22n2 + 5n + 3

=34

.


1Γ(z)Γ(

√−1z)Γ(−z)Γ(−

√−1z)

= −z4 ∏n∈N

(1− z4

n4

),

1Γ(√−1z)Γ(−

√−1z)

= z2 ∏n∈N

(1 +

z2

n2

)=

z sinh(πz)π

.


Problem 15.6.16. If η1, · · · , ηm are exact roots of the equation Xm + 1 = 0 and ζ :=e2π√−1/m, then

∏n≥2

nm − 1nm + 1

=2Γ(1− ηm)

m ∏1≤k≤m−1

Γ(1− ηk)

Γ(1− ζk).

Problem 15.6.17. (Gram) Show that

∏n≥2

n3 − 1n3 + 1

=23

.

Problem 15.6.18. Show that

lnπ

2=

∫ 1

0

v− 1v + 1

dvln v

=∫ 1

0

(u14 − u−

14 )2

u− 1du

ln u,

lnπ

2√

2=

∫ 1

0

(y2 − 1)y2

(1 + y)(1 + y2)

dyln y

.

Problem 15.6.19. (Euler) Show that

π cot(πz) =∫ 1

0

uz − u1−z

1− uduu

=∫ 1

:0

uz−1 − u−z

1− udu.

Problem 15.6.20. For any c > 0 show that

2 ln 2 =∫ 1

0

uc + uc+ 12 − 2u2c

1− uduu

,

3 ln 3 =∫ 1

0

uc + uc+ 13 + uc+ 2

3 − 3u3c

1− uduu

,

ln n =∫ 1

0

(uc

1− u− nunc

1− un

)duu

.

Problem 15.6.21. Show that ζ ′(0) = − ln√

2π.

Problem 15.6.22. Show that∫ 1

0

12 − u

(u + z)2 du =1z− ln

(1 +

1z

)− 1

2z(z + 1),

ν(z) =12z

+ ∑n≥0

∫ 1

0

12 − u

(u + z + n)2 du,

ν(z) =12z

+∫ ∞

0

Λ(t)(t + z)2 dt.

Problem 15.6.23. (Kummer) Show for any x ∈ (0, 1) that

ln Γ(x) =(

12− z)(γ + ln 2) + (1− x) ln π − 1

2ln sin(πx) + ∑

n∈N

ln nnπ

sin(2nπx).

Problem 15.6.24. (Stieltjes) Show that∣∣∣µ (ρe√−1θ)∣∣∣ ≤ 1

12ρ cos2 θ2

,

∣∣∣∣∣ν (ρe√−1θ)− 1

2ρe√−1θ

∣∣∣∣∣ ≤ 112ρ2 cos3 θ

2

.

CHAPTER 16

Fourier analysis

16.1. Basic properties

Suppose that f , g are integrable over [a, b]. Define

f ⊥ g⇐⇒∫ b

af (x)g(x)dx = 0.

LetR[a, b] denote the set of all (Riemann) integrable functions over [a, b] and con-sider an inner product

〈 f , g〉L2[a,b] :=∫ b

af (x)g(x)dx, f , g ∈ R[a, b],

onR[a, b]. Observe that for any m, n ∈ N,

〈sin mx, sin nx〉L2[−π,π] = 〈cos mx, cos nx〉L2[−π,π] = πδm,n,

〈sin mx, cos nx〉L [−π,π] = 0.

Thus, (cos nx, sin nx)n∈N, together with 1, forms a “basis” for the vector spaceR[−π, π]. Define Fourier series by

(16.1.1)a0

2+ ∑

n∈N(an cos nx + bn sin nx) , x ∈ [−π, π].

Here a0, an, bn are constants. Recall e√−1x = cos x +

√−1 sin x. Hence (16.1.1) can

be written as

(16.1.2) ∑n∈Z

Ane√−1nx

where A0 = 12 a0, An

12 an −

√−12 bn, and A−n = 1

2 an +√−12 bn.

If the Fourier series (16.1.1) converges uniformly to a function f (x) in [−π, π],then f ∈ C[−π, π] and

an =1π

∫ π

−πf (x) cos(nx)dx,(16.1.3)

bn =1π

∫ π

−πf (x) sin(nx)dx.(16.1.4)

Consequently

(16.1.5) An =1

2π

∫ π

−πf (x)e−

√−1nxdx, n ∈ Z.

163

164 16. FOURIER ANALYSIS

In general, for any function f ∈ R[−π, π] we define the Fourier series of f as(16.1.1) with an, bn are given by (16.1.3) and (16.1.4), and formally write it as

(16.1.6) f (x) ∼ a0

2+ ∑

n∈N(an cos nx + bn sin nx) = ∑

n∈ZAne

√−1nx, x ∈ [−π, π].

Example 16.1.1. Compute the Fourier series of

f (x) = x, x ∈ [−π, π].

In this case an = 0 for any n ≥ 0 and bn = (−1)n−1 2n , it follows that

f (x) ∼ ∑n∈N

(−1)n−1 2n

sin(nx).

Suppose f ∈ R(0, π) and consider the odd extension fodd of f by defining

fodd(x) := − f (−x), x ∈ (−π, 0).

Then “an” for fodd are zero while “bn” are

1π

∫ π

−πfodd(x) sin(nx)dx =

2π

∫ π

0f (x) sin(nx)dx.

Hence we get Fourier sin series of f

(16.1.7) f (x) ∼ ∑n∈N

[2π

∫ π

0f (x) sin(nx)dx

]sin(nx).

Similarly, we can consider the even extension feven of f by defining

feven(x) := f (−x), x ∈ (−π, 0).

In this case we obtain Fourier cos series of f

(16.1.8) f (x) ∼ 1π

∫ π

0f (x)dx + ∑

n∈N

[2π

∫ π

0f (x) cos(nx)dx

]cos(nx).

Example 16.1.2. Find the Fourier sin series of

f (x) = x2, 0 < x < π.

The answer is

x2 ∼ 2π ∑n∈N

(−1)n

nsin(nx)− 8

π ∑n∈N

sin[(2n− 1)x](2n− 1)3 .

Example 16.1.3. Find the Fourier cos series of

f (x) = x2, 0 < x < π.

The answer is

x2 ∼ π2

3+ 4 ∑

n∈N

(−1)n

n2 cos(nx).

Exercise 16.1.4. Complete the computations in Example 16.1.2 and Example 16.1.3.


If the symbol “∼” in Example 16.1.3 is “=”, then, taking x = π, we arrive at

π2 =π2

3+ 4 ∑

n∈N

1n2 =⇒ ∑

n∈N

1n2 =

π2

6.

Thus we give an alternative proof of Euler’s formula provided that we can takethe equality.

Let f (x) be integrable over [−T, T]. Letting x = Tπ t, we get a function ϕ(t) :=

f ( Tπ t) with t ∈ [−π, π]. Since

ϕ(t) ∼ a0

2+ ∑

n∈N

[an cos(nt) + bn sin(nt)

]where

an :=1π

∫ π

−πϕ(t) cos(nt)dt, bn :=

1π

∫ π

−πϕ(t) sin(nt)dt,

we get

(16.1.9) f (x) ∼ a0

2+ ∑

n∈N

[an cos

(nπ

Tx)+ bn sin

(nπ

Tx)]

with

(16.1.10) an =1T

∫ T

−Tf (x) cos

(nπ

Tx)

dx, bn =1T

∫ T

−Tf (x) sin

(nπ

Tx)

dx.

Example 16.1.5. Compute the Fourier series of

f (x) = x cos x, −π

2≤ x ≤ π

2.

In this example, T = π/2, an = 0, and

bn =2T

∫ T

0x cos x · sin(2nx)dx =

(−1)n−1

π

16n(4n2 − 1)2 .

Hence

x cos x ∼ ∑n∈N

(−1)n−116nπ(4n2 − 1)2 sin(2nx).

16.1.1. Partial sum. If f is an integrable function given on an interval [a, b] oflength L := b− a, then the n-th Fourier coefficient of f is defined by

(16.1.11) f (n) :=1L

∫ b

af (x)e−2π

√−1nx/Ldx, n ∈ Z.

The Fourier series of f is given formally by

(16.1.12) f (x) ∼ ∑n∈N

f (n)e2π√−1nx/L.

For example, if f is an integrable function on the interval [−π, π] then the n-thFourier coefficient of f is

f (n) = An =1

2π

∫ π

−πf (x)e−

√−1nxdx, n ∈ Z,

and the Fourier series of f is

f (x) ∼ ∑n∈Z

Ane√−1nx.


If f is an integrable function on [0, 2π], then

(16.1.13) f (n) =1

2π

∫ 2π

0f (x)e−

√−1nxdx and f (x) ∼ ∑

n∈Zf (n)e

√−1nx.

If f is a function defined on the circle S1, then we can think of f as a 2π-periodic function on R. We may restrict the function f to any interval of length2π, for instance [0, 2π] or [−π, π], and compute its Fourier coefficient. Observethat f is periodic shows that the resulting integrals are independent of the choseninterval. Thus the Fourier coefficients of f on the circle are well-defined.

If g is an integrable function on [0, 1], then

(16.1.14) g(n) =∫ 1

0f (x)e−2π

√−1nxdx and g(x) ∼ ∑

n∈Zg(n)e2π

√−1nx.

If f is given on [0, 2π], then g(x) := f (2πx) is defined on [0, 1] and a changeof variables shows that the n-th Fourier coefficient of f equals the n-th Fouriercoefficient of g.

The trigonometric series is an expression of the form

∑n∈Z

cne2π√−1nx/L

where cn ∈ C. If a trigonometric series involves only finitely many nonzero terms,that is, cn = 0 for all large |n|, it is called a trigonometric polynomial; its degreeis the largest value of |n| for which cn 6= 0.

The N-th partial sum of the Fourier series of f defined on [a, b] with L = b− a,for N ∈ N, is a trigonometric polynomial given by

(16.1.15) SN( f )(x) := ∑−N≤n≤N

f (n)e2π√−1nx/L.

Note that by definition the above sum is symmetric since n ranges from −N to N.A basic and natural question arises as follows:

Problem: In what sense does SN( f ) converge to f as N → ∞?We first might ask whether SN( f ) converge to f pointwise. That is, do we havelimN→∞ SN( f )(x) = x for every x?. We see easily that we cannot expect this reusltto be true for every x, since we can change an integrable function at one point with-out changing its Fourier coefficients. As a result, we might ask the same questionassuming that f is continuous and periodic. For a long time it was believed thatunder these additional assumptions the answer would be “yes”. It was a surprisewhen Du Bois-Reymond showed that there exists a continuous function whoseFourier series diverges at a point. When f is continuously differentiable, we willsee that then the Fourier series of f converges to f uniformly.

If f is integrable, we will show that SN( f ) converges to f in the mean squaresense. That is,

12π

∫ π

−π|SN( f )(x)− f (x)|2 dx → 0

as N → ∞. In 1913, Lusin proposed a famous conjecture that if f in integrable thenthe Fourier series of f converges to f except possibly on a set of “measure zero”.This conjecture was settled by Carleson in 1966.


Example 16.1.6. (Dirichlet kernel) The N-th Dirichlet kernel is defined for x ∈[−π, π] by

(16.1.16) DN(x) := ∑−N≤n≤N

e√−1nx.

Writing ω := e√−1x we see that

DN(x) = ∑0≤n≤N

ωn + ∑−N≤n≤−1

ωn =1−ωN+1

1−ω+

ω−N − 11−ω

=ω−N −ωN+1

1−ω.

Hence

(16.1.17) DN(x) =sin[(N + 1

2 )x]sin(x/2)

.

Example 16.1.7. (Poisson kernel) The Poisson kernel is defined for x ∈ [−π.π] and0 ≤ r < 1 by

(16.1.18) Pr(x) := ∑n∈Z

r|n|e√−1nx.

Observe that the above series is convergent absolutely and uniformly. Letting ω = re√−1x

we have

Pr(x) = ∑n≥0

ωn + ∑n≥1

ωn =1

1−ω+

ω

1−ω=

1− |ω|2|1−ω|2 .

Thus

(16.1.19) Pr(x) =1− r2

1− 2r cos x + r2 .

Example 16.1.8. (Fejer kernel) The N-th Fejer kernel is defined for x ∈ [−π, π] by

(16.1.20) FN(x) :=1N ∑

0≤n≤N−1Dn(x).

Writing ω = e√−1x we get

FN =1N ∑

0≤n≤N−1

ω−n −ωn+1

1−ω=

1N

11−ω

[1−ω−N

1−ω− ω(1−ωN)

1−ω

].

Thus

(16.1.21) FN(x) =1N

sin2(Nx/2)sin2(x/2)

, x ∈ [π, π].

16.1.2. Uniqueness of Fourier series. Suppose that f and g are integrablefunctions on the circle S1 with the same Fourier coefficients f (n) = g(n) for n ∈ Z.Then the Fourier coefficients of f − g are zero.

Theorem 16.1.9. Suppose f is an integrable function on the circle S1 with f (n) = 0 forn ∈ Z. If f is continuous at x0, then f (x0) = 0.


PROOF. Suppose the theorem holds for any real-valued function f . In general,write f (x) = u(x) +

√−1v(x) where u and v are real-valued. If we define f (x) :=

f (x), then

u(x) =f (x) + f (x)

2, v(x) =

f (x)− f (x)2√−1

,

and since ˆf (x) = f (−n), we conclude that the Fourier coefficients of u and v allvanish, hence f = 0 at its points of continuity.

We may assume that f is a real-valued function defined on [−π, π], x0 = 0,and f (0) > 0. Since f is continuous at 0, we can choose δ ∈ (0, π/2] so thatf (x) > f (0)/2 whenever |x| < δ. Let

p(x) := ε + cos x

where ε > 0 is chosen so small that |p(x)| < 1− ε2 for any δ ≤ |x| ≤ π. Choose

η ∈ (0, δ) so that p(x) > 1 + ε2 for |x| < η. Set

pk(x) := [p(x)]k, B := max|x|≤π

| f (x)|.

Since f (0) = 0 for any n ∈ Z, it follows that

0 =∫ π

−πf (x)pk(x)dx

for any k ∈ N. On the other hand,∫ π

−πf (x)pk(x)dx =

∫|x|<η

f (x)pk(x)dx +∫

η≤|x|<δf (x)pk(x)dx

+∫|x|≥δ

f (x)pk(x)dx

≥ 2ηf (0)

2

(1 +

ε

2

)k+ 0−

∣∣∣∣∫|x|≥δf (x)pk(x)dx

∣∣∣∣≥ η f (0)

(1 +

ε

2

)k− 2πB

(1− ε

2

)k.

Letting k→ ∞ yields a contradiction. Hence f (0) = 0.

Corollary 16.1.10. If f ∈ C(S1) and f (n) = 0 for all n ∈ N, then f ≡ 0.

Corollary 16.1.11. If f ∈ C(S1) and the Fourier series of f is absolutely convergent,then the Fourier series converges uniformly to f , that is,

limN→∞

SN( f )(x) = f (x)

uniformly in x.

PROOF. Since ∑n∈Z | f (n)| is absolutely convergent, it follows that the function

g(x) := ∑n∈Z

f (x)e√−1nx.

is continuous on S1. Moreover the Fourier coefficients of g are exactly f (n). ByCorollary 16.1.10, we obtain f ≡ g on S1.


WE write f = O(g) as x → a (where a ∈ R ∪ ±∞) if for some positiveconstant C we have | f (x)| ≤ C|g(x)| as x tends to a. In particular, f (x) = O(1)means that f is bounded.

Corollary 16.1.12. If f ∈ Ck(S1) for some k ≥ 2, then f (n) = O(1/|n|k) as |n| → ∞.In particular, the Fourier series of f converges absolutely and uniformly to f .

PROOF. We may assume that k = 2 and use the integration by parts for n 6= 0.Compute

2π f (n) =∫ 2π

0f (x)e−

√−1nxdx

=

[f (x)−e−

√−1nx

√−1n

]2π

0

+1√−1n

∫ 2π

0f ′(x)e−

√−1nxdx

=1√−1n

∫ 2π

0f ′(x)e−

√−1nxdx

=1√−1n

[f ′(x)

−e−√−1nx

√−1n

]2π

0

+1

(√−1n)2

∫ 2π

0f ′′(x)e−

√−1nxdx

=−1n2

∫ 2π

0f ′′(x)e−

√−1nxdx.

If B := maxx∈S1 | f (x)|, then

| f (n)| ≤ C2πn2 .

Thus f (n) = O(1/n2).

16.1.3. Convolution. Given two 2π-periodic integrable functions f and g onR, we define their convolution f ∗ g on [−π, π] by

(16.1.22) f ∗ g(x) :=1

2π

∫ π

−πf (y)g(x− y)dy.

It is clear that f ∗ g = g∗ by definition (16.1.22). Observe that

( f ∗ DN)(x) =1

2π

∫ π

−πf (y)

(∑

−N≤n≤Ne√−1n(x−y)

)dy

= ∑−N≤n≤N

(1

2π

∫ π

−πf (y)e

√−1n(x−y)

)dy

= ∑−N≤n≤N

f (n)e√−1nx = SN( f )(x).

Thus

(16.1.23) f ∗ DN = SN( f ).

Proposition 16.1.13. Suppose that f , g, and h are 2π-periodic integrable functions. Then(i) f ∗ (g + h) = f ∗ g + f ∗ h.

(ii) (c f ) ∗ g = c( f ∗ g) = f ∗ (cg) for any c ∈ C.(iii) ( f ∗ g) ∗ h = f ∗ (g ∗ h).(iv) f ∗ g is continuous.


(v) f ∗ g(n) = f (n)g(n).

PROOF. We only prove (iv) and (v). Assume first that f , g are continuous. Forany n ∈ Z, one has

f ∗ g(n) =1

2π

∫ π

−π( f ∗ g)(x)e−

√−1nxdx

=1

2π

∫ π

−π

[∫ π

−πf (y)g(x− y)dy

]e−√−1nxdx

=1

2π

∫ π

−πf (y)e−

√−1ny

[1

2π

∫ π

−πg(x− y)e−

√−1n(x−y)dx

]dy

= f (n)g(n).

Since g is continuous and periodic, it follows that g is uniformly continuous inR. Hence for any ε > 0 there exists a positive number δ > 0 so that |g(x)− g(y)| <ε whenever |x− y| < δ. Using the identity

( f ∗ g)(x1)− ( f ∗ g)(x2) =1

2π

∫ π

−πf (y)[g(x1 − y)− g(x2 − y)]dy,

we arrive at

|( f ∗ g)(x1)− ( f ∗ g)(x2)| ≤1

2π

∫ π

−π| f (y)||g(x1 − y)− g(x2 − y)|dy

≤ ε

2π

∫ π

−π| f (y)|dy ≤ ε · max

[−π,π]| f |,

whenever |x1 − x2| < δ.In general, when f and g are merely integrable, we may the the following

lemma to prove (iv).

Lemma 16.1.14. Suppose f is integrable on the circle and bounded by B. Then thereexists a sequence ( fk)k∈N of continuous functions on the circle so that

sup[−π,π]

| fk| ≤ B for any k ∈ N and∫ π

−π| f (x)− fk(x)|dx → 0

as k→ ∞.

PROOF. Assume f is real-valued. Given ε > 0 we may choose a partition−π = x0 < x1 < · · · < xN = π of the interval [−π, π] so that the upper and lowersums of f differ by at most ε. Denote by f ∗ the step function defined by

f ∗(x) := supy∈[xj−1,xj ]

f (y), if x ∈ [xj−1, xj) for 1 ≤ j ≤ N.

By construction we have | f ∗| ≤ B and∫ π

−π| f ∗(x)− f (x)|dx =

∫ π

−π[ f ∗(x)− f (x)]dx < ε.

Now we can modify f ∗ to make it continuous and periodic. For small δ > 0 letf (x) := f ∗(x) when the distanxe of x from any of the division points x0, · · · , xNis ≥ δ. In the δ-neighborhood of xj for j = 1, · · · , N − 1, define f (x) to be thelinear function for which f (xj ± δ) = f ∗(xj ± δ). Near x0 = −π, f is linear withf (−π) = 0 and f (−π + δ) = f ∗(−π + δ). Similarly near xN = π the function


f is linear with f (π) = 0 and f (π − δ) = f ∗(π − δ). Since f (−π) = f (π), wemay extend f to a continuous and 2π-periodic function on R. Moreover | f | ≤ Band f differs from f ∗ only in the N intervals of length 2δ surrounding the divisionpoints. Thus ∫ π

−π| f ∗(x)− f (x)|dx ≤ 4BNδ.

Therefore∫ π

−π| f (x)− f (x)|dx ≤

∫ π

−π| f ∗(x)− f (x)|dx +

∫ π

−π| f ∗(x)− f (x)|dx < 2ε

for δ ∈ (0, ε/4BN). Denoting by fk the f so constructed, when 2ε = 1/k, we seethat the sequence ( fk)k∈N has the properties required by the lemma.

Applying Lemma 16.1.14 to f and g we get sequences ( fk)k∈N and (gk)k∈N ofcontinuous functions on S1 satisfying

sup[−π,π]

| fk| ≤ sup[−π,π]

| f |, sup[−π,π]

|gk| ≤ sup[−π,π]

|g|

and ∫ π

−π| f (x)− fk(x)|dx → 0,

∫ π

−π|g(x)− gk(x)|dx → 0

as k→ ∞. From

f ∗ g− fk ∗ gk = ( f − fk) ∗ gk + fk ∗ (g− gk),

and

|[( f − fk) ∗ g](x)| ≤ 12π

∫ π

−π| f (x− y)− fk(x− y)||g(y)|dy

≤ 12π

sup[−π,π]

|g|∫ π

−π| f (y)− fk(y)|dy → 0

as k → ∞, we conclude that ( f − fk) ∗ g → 0 uniformly in x. Similarly, fk ∗ (g−gk) → uniformly, and therefore fk ∗ gk tends uniformly to f ∗ g. Since each fk ∗ gkis continuous, it follows that f ∗ g is also continuous.

For each fixed n ∈ Z, we must have fk ∗ gk(n) → f ∗ g(n) as k → ∞ sincefk ∗ gk converges uniformly to f ∗ g. On the other hand, fk ∗ gk(n) = fk(n)gk(n). Toprove f ∗ g(n) = f (n)g(n), we need only to verify fk(n)→ f (n) and gk(n)→ g(n)as k→ ∞. Because

| f (n)− fk(n)| =1

2π

∣∣∣∣∫ π

−π[ f (x)− fk(x)]e−

√−1nxdx

∣∣∣∣ ≤ 12π

∫ π

−π| f (x)− fk(x)|dx,

we get fk(n)→ f (n) as n→ ∞.

Exercise 16.1.15. Complete the proof of Proposition 16.1.13.


16.1.4. Good kernels. A family of kernels (Kn(x))n∈N on the circle is said tobe a family of good kernels if it satisfies the following properties:

(a) For all n ∈ N,1

2π

∫ π

−πKn(x)dx = 1.

(b) There exists M > 0 such that for all n ∈ N∫ π

−π|Kn(x)|dx ≤ M

(c) For every δ > 0, ∫δ≤|x|≤π

|Kn(x)|dx → 0

as n→ ∞.

Example 16.1.16. Let DN(x) the N-th Dirichlet kernel. Then (DN)N∈N is not a familyof good kernels. For any N ∈ N

12π

∫ π

−πDn(x)dx = ∑

−N≤k≤N, k 6=0

12π

∫ π

−πe√−1kxdx + 1

= ∑−N≤k≤N, k 6=0

sin(kπ)

kπ+ 1 = 1.

Let

LN :=1

2π

∫ π

−π|DN(x)|dx =

12π

∫ π

−π

∣∣∣∣∣ sin[(N + 12 )x]

sin x2

∣∣∣∣∣ dx.

For any x ∈ [−π/2, π/2], we have 2π x ≤ sin x ≤ x. Hence

LN ≥ 2π

∫ π

0

| sin[(N + 12 )x]|

|x| dx =2π

∫ Nπ+ π2

0

| sin θ|θ

dθ

=2π ∑

0≤k≤N−1

∫ (k+1)π

kπ

| sin θ|θ

dθ +∫ Nπ+ π

2

Nπ

| sin θ|θ

dθ

≥ 2π ∑

0≤k≤N−1

1(k + 1)π

∫ (k+1)π

kπ| sin θ|dθ +

2π(2N + 1)

∫ Nπ+ π2

Nπ| sin θ|dθ

=4

π2 ∑1≤k≤N

1k+

4π2

12N + 1

.

Since the sequence (aN)N∈N with aN = ∑1≤k≤N1k − ln N is monotone decreasing, it

follows that ∑1≤k≤N ≥ ln N + γ, where γ is the Euler constant. Consequently,

(16.1.24) LN ≥4

π2 ln N +4

π2

(γ +

12N + 1

).

Thus (DN)N∈N is nota a family of good kernels.

Exercise 16.1.17. Try to prove LN = 4π2 ln N + O(1) as N → ∞.

Suppose we are given a series of complex numbers c0 + c1 + c2 + · · · = ∑k≥0 ck.We define the n-th partial sum sn by

sn := ∑0≤k≤n

ck


and say that the series converges to s if limn→∞ sn = s. The N-th Cesaro mean ofthe sequence (sk)k≥0 or the N-th Cesaro sum of the series ∑k≥0 ck is defined by

(16.1.25) σN :=1N ∑

0≤k≤N−1sk.

We say that the series ∑k≥0 ck is Cesaro summable to σ if σN converges to σ asN → ∞.

(i) The series ∑k≥0(−1)k is Cesaro dummable to 1/2.(ii) If a series is convergent to s, then it is also Casaro summable to the same

limit s.(iii) (Tauber’s theorem) If the series ∑n≥0 cn is Cesaro summable to σ and

cn = o(1/n), then ∑n≥0 cn converges to σ.

Exercise 16.1.18. Prove the above Tauber’s theorem.

If f ∈ R(S1), thenSN( f ) = f ∗ DN .

We form the N-th Cesaeo mean of the Fourier series

(16.1.26) σN( f ) :=S0( f ) + S1( f ) + · · ·+ SN−1( f )

N= f ∗ FN ,

where FN(x) is the N-th Fejer kernel given by

(16.1.27) FN =D0 + D1 + · · ·+ DN−1

N=

1N

sin2(Nx/2)sin2(x/2)

.

Exercise 16.1.19. (FN)N∈N is a family of good kernels.

A series of complex numbers ∑k≥0 ck is said to be Abel summable to s if forevery r ∈ [0, 1) the series

A(r) := ∑k≥0

ckrk

converges, and limr→1 A(r) = s. The quantities A(r) are called the Abel means ofthe series ∑k≥0 ck.

(i) The series ∑k≥0(−1)k(k + 1) is Abel summable to 1/4.(ii) If a series is Cesaro summable to σ then it is Abel summable to σ.

(iii) (Tauber’s theorem) If the series ∑n≥0 cn is Abel summable to σ and cn =o(1/n), then ∑n≥0 cn converges to σ.

Exercise 16.1.20. Prove the above Tauber’s theorem. (Hint: For A(r) = ∑n≥0 cnrn

and 0 < r < 1, we have∣∣∣∣∣ ∑0≤n≤N

cn − f (x)

∣∣∣∣∣ =

∣∣∣∣∣ ∑1≤n≤N

cn(1− rn)− ∑n≥N+1

cnrn

∣∣∣∣∣≤ (1− x) ∑

1≤n≤N|ncn|+

1N(1− x)

supn>N|ncn|.

Setting ∑0≤n≤N cn = sN and x = 1− 1/N we get∣∣∣∣sN − A(

1− 1N

)∣∣∣∣ ≤ 1N ∑

1≤n≤N|ncn|+ sup

n>N|ncn|.

If ncn → 0 and A(r)→ s, then sN → s.)


Tauber in 1897 also derived his “second theorem” which gives a necessary andsufficient condition for the step from Abel summability to convergence.

Theorem 16.1.21. (Tauber, 1897) An Abel summable series ∑n≥0 cn is convergent ifand only if

1n ∑

1≤k≤nkak = sn −

1n ∑

0≤k≤n−1sk → 0

as n→ ∞.

Exercise 16.1.22. This exercise shows the following implications about series:

(16.1.28) convergent =⇒ Cesaro summable =⇒ Abel summable,

and the fact that none of the arrows can be reversed. Consider the series ∑n≥0(−1)n

and ∑n≥0(−1)n−1n.

If f is integrable on S1 with f (x) ∼ ∑n∈Z Ane√−1nx, we define the Abel means

of f by

(16.1.29) Ar( f )(x) := ∑n∈Z

r|n|Ane√−1nx.

Since An is uniformly bounded, it follows that Ar( f ) converges absolutely anduniformly for each r ∈ [0, 1). Observe that

(16.1.30) Ar( f ) = f ∗ Pr

where Pr(x) is the Poisson kernel given by

(16.1.31) Pr(x) := ∑n∈Z

r|n|e√−1nx =

1− r2

1− 2r cos x + r2 , r ∈ [0, 1).

Exercise 16.1.23. Show that (Pr)r∈[0,1) is a family of good kernels. That is,

(a) For any r ∈ [0, 1),

12π

∫ π

−πPr(x)dx = 1.

(b) There exists M > 0 such that for all r ∈ [0, 1)∫ π

−π|Pr(x)|dx ≤ M

(c) For every δ > 0, ∫δ≤|x|≤π

|Pr(x)|dx → 0

as r → 1.

Theorem 16.1.24. Let (Kn)n∈N be a family of good kernels, and f an integrable functionon the circle. Then

limn→∞

( f ∗ Kn)(x) = f (x)

whenever f is continuous at x. If f is continuous everywhere, then the above limit isuniform.


PROOF. By definition of good kernels, we have

f ∗ Kn(x)− f (x) =1

2π

∫ π

−π[ f (y)Kn(x− y)− f (x)Kn(y)]dy

=1

2π

∫ π

−πKn(y)[ f (x− y)− f (x)]dy

for any x. The continuity of f at x implies that for any ε > 0 there exists a positivenumber δ > 0 such that | f (x− y)− f (x)| < ε whenever |y| < δ. Hence

| f ∗ Kn(x)− f (x)| ≤ 12π

∫ π

−π|Kn(y)|| f (x− y)− f (x)|dy

≤ 12π

∫δ≤|y|≤π

|Kn(y)|| f (x− y)− f (x)|dy

+1

2π

∫|y|<δ

|Kn(y)|| f (x− y)− f (x)|dy

≤ ε

2π

∫|y|<δ

|Kn(y)|dy +1

2π· 2 max

[−π,π]| f | ·

∫δ≤|y|≤π

|Kn(y)|dy

≤ ε

2πM +

1π

max[−π,π]

| f | ·∫

δ≤|y|≤π|Kn(y)|dy.

Letting n→ ∞ yields

lim supn→∞

| f ∗ Kn(x)− f (x)| ≤ εM2π

,

thus f ∗ Kn(x)→ f (x) whenever f is continuous at x.

Corollary 16.1.25. (1) If f is integrable on the circle, then the Fourier series of f is Cesarosummable to f at every point of continuity of f . Moreover, if f is continuous on the circle,then the Fourier series of f is uniformly Cesaro summable to f .

(2) If f is integrable on the circle, then the Fourier series of f is Abel summable to f atevery point of continuity of f . Moreover, if f is continuous on the circle, then the Fourierseries of f is uniformly Abel summable to f .

In Exercise 16.1.8 and Exercise 16.1.20 we assumed that ncn → 0. Hardy askeda question whether the above condition in Tauber’s theorems could be relaxed toboundedness of the sequence (ncn)n∈N. According to the hint in Exercise 16.1.20,we see that the boundedness of the sequence (ncn)n≥0 together with the Abelsummability of the series ∑n≥0 cn implies that the sequence (sN)N≥0 is bounded.

Theorem 16.1.26. (Hardy-Landau, 1910) If the series ∑n≥0 cn is Crsaro summable andeither |ncn| ≤ C (Hardy) or ncn ≥ −C (Landau), then ∑n≥0 cn converges.

PROOF. We give a proof due to Kloosterman (1940). We consider real cn andset

sn = c0 + · · ·+ cn =: c(−1)n , c(−2)

n := c(−1)0 + · · ·+ c(−1)

n .

For any integers h > 0 and n < k ≤ n + h we have

c(−2)n+h = c(−2)

n +[c(−1)

n+1 + · · ·+ c(−1)n+h

]= c(−2)

n + hc(−1)n + [hcn+1 + (h− 1)cn+2 + · · ·+ cn+h] ;


the final sum does not exceed h(h+1)2 maxn+1≤k≤n+h ck; etc. Hence we get the dis-

crete Taylor’s formula:

c(−2)n+h = c(−2)

n + hc(−1)n +

h(h + 1)2

c∗ξ ,

where c∗ξ is a number between minn<k≤n+h ck and maxn<k≤n+h ck.

We may assume that the series ∑n≥0 cn is Cesaro summable to 0 so that c(−2)n /n→

0 as n → ∞. The inequalities ncn ≥ −C with C > 0 and |c(−2)n | ≤ nε with small

ε > 0 and n ≥ n0 for some sufficiently large integer n0, imply that for h ≈ 2n√

ε/C

sn =c(−2)

n+h − c(−2)n

h− h(h + 1)

2c∗ξ ≤

2n + hh

ε + Ch + 1

2n< 3√

Cε.

For an estimate in the other direction we may take h ≈ −2n√

ε/C; for h < 0the discrete Taylor’s formula needs some adjustment. The conclusion is now thatsn → 0 as n→ ∞.

Hardy used Theorem 16.1.26 to deduce the following results for Fourier series:

(i) Let f be a continuous function of period 2π whose Fourier coefficientsf (n) = O(1/|n|) (compared with Corollary 16.1.12 where we can deducethat f (n) = O(1/|n|) for any f ∈ C1(S1)). Then the Fourier series of fconverges to f at every point.

(ii) Let f be a periodic function of bounded variation over a period. Thenthe Fourier coefficients f (n) are O(1/n), and hence the Fourier series off converges at every point x to the value 1

2 [ f (x−) + f (x+)].

Theorem 16.1.27. (1) (Littlewood, 1911) If the series ∑n≥0 cn is Abel summable and|ncn| ≤ C, then ∑n≥0 cn converges.

(2) (Hardy-Littlewood, 1914) If the series ∑n≥0 cn is Abel summable and ncn ≥−C, then ∑n≥0 cn converges.

16.2. Convergence

If f ∈ C1()S1), we have f (n) = O(1/|n|) by Corollary 16.1.12. In general, wehave the following

Theorem 16.2.1. (Riemann’s lemma) If ψ is integrable [a, b], then

(16.2.1) limp→∞

∫ b

aψ(x) sin(px)dx = lim

p→∞

∫ b

aψ(x) cos(px)dx = 0.

PROOF. Since ψ is integrable, it follows that ψ is bounded. For any ε > 0 wecan find a partition a = x0 < x1 < · · · < xn = b such that

∑1≤i≤n

ωi∆xi <ε

2, ∆xi := xi − xi−1, ωi := Mi −mi,

16.2. CONVERGENCE 177

where mi := min[xi−1,xi ]ψ and Mi := max[xi−1,xi ]

ψ. Compute∣∣∣∣∫ b

aψ(x) sin(px)dx

∣∣∣∣ =

∣∣∣∣∣ ∑1≤i≤n

∫ xi

xi−1

[ψ(x)−mi] sin(px) + mi sin(px)

dx

∣∣∣∣∣≤ ∑

1≤i≤n

∫ xi

xi−1

|ψ(x)−mi|| sin(px)|dx

+ ∑1≤i≤n

|mi|∣∣∣∣∫ xi

xi−1

sin(px)dx∣∣∣∣

≤ ∑1≤i≤n

∫ xi

xi−1

|ψ(x)−mi|dx +2p ∑

1≤i≤n|mi|

≤ ∑1≤i≤n

ωi∆xi +2p ∑

1≤i≤n|mi| ≤

ε

2+

ε

2= ε,

for p ≥ 4ε ∑1≤i≤n |mi|.

16.2.1. Dirichlet-Jordan and Dini-Lipachitz theorems. Recall that

Sn( f )(x) =a0

2+ ∑

1≤k≤n

[ak cos(kx) + bk sin(kx)

],

ak =1π

∫ π

−πf (x) cos(kx)dx,

bk =1π

∫ π

−πf (x) sin(kx)dx

for any integrable function f with period 2π. Then

Sn( f )(x) =1π

∫ π

−πf (t)

[12+ ∑

1≤k≤ncos n(t− x)

]dt

=1

2π

∫ π

−πf (t)

sin 2n+12 (t− x)

sin t−x2

dt(16.2.2)

=1

2π

∫ π

0[ f (x + t) + f (x− t)]

sin 2n+12 t

sin t2

dt.

Given a function σ(x) we then get

(16.2.3) Sn( f )(x)− σ(x) =1π

∫ π

0

[f (x + t) + f (x− t)

2− σ(x)

]sin 2n+1

2 tsin t

2dt.

Corollary 16.2.2. (1) If f is a integrable function with period 2π, then the convergenceof (Sn( f )(x))n∈N depends only on f |(x−δ,x+δ) for some sufficiently small number δ > 0.

(2) If ψ is a integrable function over [0, δ], then

limn→∞

∫ δ

0ψ(x)

sin 2n+12 t

2 sin t2

dt = limn→∞

∫ δ

0ψ(t)

sin 2n+12 t

tdt.


PROOF. (1) For any given δ > 0 we have sin(t/2) > Cδ for some Cδ > 0 andany δ ≤ t ≤ π. By Theorem 16.2.1 we see that

limn→∞

∫ π

δ[ f (x + t) + f (x− t)]

sin 2n+12 t

sin t2

dt = 0.

So according to (16.2.2), the convergence of Sn( f )(x) depends only on f |(x−δ,x+δ).(2) Since 2 sin t

2 ∼ t as t→ 0, we may consider the function

g(t) =

1

2 sin t2− 1

t , 0 < t ≤ δ,

0, t = 0,

which is a continuous function over [0, δ]. Now the result follows from Theorem16.2.1.

As a consequence of Corollary 16.2.2 we arrive at

limn→∞

∫ π

0

sin 2n+12 t

tdt = lim

n→∞

∫ π

0

sin 2n+12 t

2 sin t2

dt.

For any t ∈ [0, π], we have

12+ cos t + cos(2x) + · · ·+ cos(nt) =

sin 2n+12 t

2 sin t2

.

Integrating over [0, π] yields

∫ π

0

sin 2n+12 t

2 sin t2

dt =π

2, n ∈ N.

Therefore we give alternative of (15.2.8).

Given a domain D ⊂ Rn and a function f defined on D. We say f ∈ C0,α(D)(C0,α(D) is called Holder’s space with order α) if | f (x) − f (y)| ≤ L|x − y|α forsome α ∈ (0, 1] and any x, y ∈ D. Write Lip(D) := C0,1(D). Observe that

(16.2.4) C1(D) ( Lip(D) ( C(D).


A function f : [a, b] → R is said to be piecewise monotone if there exists apartition a = x0 < x1 < · · · < xN = b such that f |(xi−1,xi)

, 1 ≤ i ≤ N, is monotone.

Lemma 16.2.4. (Dirichlet’s lemma) If ψ is monotone on [0, δ], then

limp→∞

∫ δ

0

ψ(t)− ψ(0+)

tsin(pt)dt = 0.

PROOF. We may assume that ψ is monotone increasing. For any ε > 0 thereexists a number η ∈ (0, δ) such that 0 ≤ ψ(t)− ψ(0+) < ε whenever t ∈ (0, η].


Compute ∫ δ

0

ψ(t)− ψ(0+)

tsin(pt)dt

=∫ η

0

ψ(t)− ψ(0+)

tsin(pt)dt +

∫ δ

η

ψ(t)− ψ(0+)

tsin(pt)dt

= [ψ(η)− ψ(0+)]∫ η

ξ

sin(pt)t

dt +∫ δ

η

ψ(t)− ψ(0+)

tsin(pt)dt

≤ ε

∣∣∣∣∫ pη

pξ

sin uu

du∣∣∣∣+ ∫ δ

η

ψ(t)− ψ(0+)

tsin(pt)dt

for some ξ ∈ [0, η]. By (15.2.8), there exists p1 > 0 such that∣∣∣∣∫ pη

pξ

sin uu

du∣∣∣∣ < 1

for any p ≥ p1. Since the function [ψ(t)− ψ(0+)]/t is integrable, it follows fromTheorem 16.2.1 that ∣∣∣∣∫ δ

η

ψ(t)− ψ(0+)

tsin(pt)dt

∣∣∣∣ < ε

for any p ≥ p2, where p2 is a sufficiently large number. Consequently, one gets∣∣∣∣∫ δ

0

ψ(t)− ψ(0+)

tsin(pt)dt

∣∣∣∣ < 2ε

whenever p ≥ max(p1, p2).

Let f : [a, b] → R be a function and x ∈ [a, b] be a continuous point or a firstkind of discontinuous point of f . We say f ∈ C0,α(x) if | f (x± t)− f (x±)| ≤ Ltα

for any sufficiently small number t > 0, where L > 0 and α ∈ (0, 1].

Theorem 16.2.5. Let f be a integrable function with period 2π. If f satisfies either(1) (Dirichlet-Jordan) f is piecewise monotone on some interval (x− δ, x + δ), or(2) (Dini-Lipschitz) f ∈ C0,α(x) for some α ∈ (0, 1],

then

(16.2.5)a0

2+ ∑

n∈N

[an sin(nx) + bn sin(nx)

]=

f (x+) + f (x−)2

.

PROOF. (1) Lemma 16.2.4 implies that∫ δ

0

f (x + t)− f (x + 0)t

sin(pt)dt→ 0,∫ δ

0

f (x− t)− f (x− 0)t

sin(pt)dt→ 0.

Using (16.2.3), Corollary 16.2.2, and Theorem 16.2.1, we arrive at

Sn( f )(x)− f (x + 0) + f (x− 0)2

=1π

∫ π

0[ f (x + t) + f (x− t)− f (x + 0)− f (x− 0)]

sin 2n+12 t

2 sin t2

dt

∼ 1π

∫ π

0

[f (x + t)− f (x + 0)

t+

f (x− t)− f (x− 0)t

]sin

2n + 12

tdt

which tends to zero.


(2) In this case we have | f (x± t)− f (x± 0)|/t ≤ L/t1−α. Hence∣∣∣∣ f (x + t) + f (x− t)− f (x+)− f (x−)t

∣∣∣∣ ≤ 2Lt1−α

.

Thus the left-hand side is integrable and the conclusion follows again by Theorem16.2.1.

Example 16.2.6. The Fourier cos series of f (x) = x2 with −π ≤ x ≤ π is

feven(x) ∼ π2

3+ 4 ∑

n∈N

(−1)n

n2 cos(nx).

Using Theorem 16.2.5 yields

(16.2.6) ∑n∈N

1n2 =

π2

6.

Example 16.2.7. The Fourier sin series of f (x) = x with −π ≤ x ≤ π is

fodd(x) ∼ ∑n∈N

(−1)n−1 2n

sin(nx).

Using Theorem 16.2.5 yields

∑n∈N

(−1)n−1 2n

sin(nx) =

x, −π < x < π,0, x = π or − π.

Example 16.2.8. The Fourier series of f (x) = cos(ax) with −π ≤ x ≤ π and α /∈ Z is

(16.2.7)π

2cos(ax)sin(aπ)

=12a

+ ∑n∈N

(−1)n a cos(nx)a2 − n2

by Theorem 16.2.5

We present an elegant proof, due to N. I. Lobatshewski, of (15.2.8). Settingx = 0 in (16.2.7) we get

1sin(aπ)

=1

aπ+ 2 ∑

n∈N

(−1)naπ

(aπ)2 − (nπ)2

and putting aπ = t yields

(16.2.8)1

sin t=

1t+ ∑

n∈N(−1)n

(1

t− nπ+

1t + nπ

).

We write

I =∫ ∞

0

sin xx

dx = ∑k≥0

∫ (k+1)π/2

kπ/2

sin xx

dx.

Since ∫ (2m+1)π/2

2mπ/2

sin xx

dx = (−1)m∫ π/2

0

sin tmπ + t

dt,∫ 2mπ/2

(2m+1)π/2

sin xx

dx = (−1)m−1∫ π/2

0

sin tmπ − t

dt,


we conclude that

I =∫ π/2

0sin t

[1t+ ∑

m∈N(−1)m

(1

t−mπ+

1t + mπ

)]dt

=∫ π/2

0sin t · 1

sin tdt =

π

2.

16.2.2. Uniform convergence. Consider a nonnegative sequence (an)n∈N.(1) We say (an)n∈N ∈ MS (monotone sequences) if it is decreasing.(2) (Shah, 1962) We say (an)n∈N ∈ QMS (quasi-monotone sequences) if for

some α ≥ 0 the sequence (an/nα)n∈N ∈ MS.(3) We say that (an)n∈N ∈ AMS (almost monotone sequences) if ak ≤ Man

for any k ≥ n.(4) We say (an)n∈N ∈ RQMS (regularly-varying quasi-monotone sequences)

if the sequence (an/R(n))n∈N ∈ MS for some regularly-varying sequence(R(n))n∈N. Here a regularly-varying sequence (R(n))n∈N means that(R(n))n∈N is a positive, increasing sequence and limn→∞

R(bλnc)R(n) < ∞

for some λ > 1.(5) We say (an)n∈N ∈ BVS (bounded variation sequences) if limn→∞ an = 0

and∑

n∈N|∆an| < ∞,

where ∆an := an − an+1.(6) (Leindler, 2001) We say that (an)n∈N ∈ RBVS (rest bounded variation

sequences) if limn→∞ an = 0 and

∑k≥n|∆ak| ≤ Man

for any n ∈ N, where M is a nonnegative constant.(7) (Le-Zhou, 2005) We say that (an)n∈N ∈ GBVS (group bounded variation

sequences) if

∑n≤k≤2n

|∆ak| ≤ Man

for any n ∈ N, where M is a nonnegative constant.(8) (Yu-Zhou, 2007) We say that (an)n∈N ∈ NBVS (non-onesided bounded

variation sequences) if

∑n≤k≤2n

|∆ak| ≤ M(an + a2n)

for any n ∈ N, where M is a nonnegative constant.(9) (Zhou, 2010) We say that (an)n∈N ∈ MVBVS (mean value bounded vari-

ation sequences) if

∑n≤k≤2n

|∆ak| ≤Mn ∑bn/λc≤k≤bλnc

ak

for any n ∈ N and some λ ≥ 2, where M is a nonnegative constant.

Exercise 16.2.9. A sequence (an)n∈N ∈ QMS if and only if there is α > 0 such thatan+1 ≤ an(1 + α/n) for n 1.


Exercise 16.2.10. Let (R(n))n∈N is a positive and increasing sequence. Show thatlim supn→∞

R(bλnc)R(n) < ∞ for some λ > 1 is equivalent to lim supm→∞

R(2n)R(n) < ∞.

We now consider a sequence (cn)n∈N of complex numbers. Given an angleθ0 ∈ [0, 2π) define K(θ0) := z ∈ C : |arg(z)| ≤ θ0.

(1) We say that (cn)n∈N ∈ MS if ∆cn ∈ K(θ0) for some θ0 ∈ [0, π/2) andany n ∈ N. If (cn)n∈N is a real sequence and ∆cn ∈ K(θ0) for someθ0 ∈ [0, π/2), then cn ≥ cn+1 for any n ∈ N. Thus the definition coincideswith the usual one in the real case.

(2) We say that (cn)n∈N ∈ RQMS if ∆(cn/R(n)) ∈ K(θ0) for some regularly-varying sequence (R(n))n∈N and θ0 ∈ [0, π/2).

(3) We say that (cn)n∈N ∈ GBVS if cn ∈ K(θ0) for some θ0 ∈ [0, π/2) and

∑n≤k≤2n

|∆ck| ≤ M|cn|

for any n ∈ N.(4) We say that (cn)n∈N ∈ MVBVS if cn ∈ K(θ0) for some θ0 ∈ [0, π/2) and

∑n≤k≤2n

|∆ck| ≤Mn ∑bn/λc≤k≤bλnc

|cn|

for any n ∈ N and some λ ≥ 2.

Remark 16.2.11. It is clear that

(i) MS ( QMS ( RQMS’(ii) MS ( RBVS.

Moreover we have

(iii) RBVS * RQMS and QMS * RBVS. For the first inclusion we consider thesequence

a2n−1 = 2−n, a2n = 2−n−2, n ∈ N.

For the second inclusion we consider the sequence

bn =

1

m2vm+1, n = vm,

bvm ∏vm≤`≤n−1

(1 + 1

`

), vm + 1 ≤ n ≤ mvm,

bmvm , mvm < n < vm+1,

where vm := 22m.

(iv) RBVS ( GBVS and RQMS ( GBVS.(v) GBVS ( NBVS.

(vi) NBVS ( MVBVS.

Thus

MS ( RBVS ∪QMS ( RBVS ∪ RQMS ( GBVS ( NBVS ( MVBVS.

(vii) RQMS ∪ RBVS ( AMS.

Theorem 16.2.12. (Chaundy-Jolliffe, 1916) If (an)n∈N ∈ MS then the series ∑n∈N ansin nx is uniformly convergent if and only if limn→∞ nan = 0.


PROOF. Let Sn(x) := ∑1≤k≤n ak sin(kx) and write || f || := maxR | f |. Then

||S2n − Sn−1|| ≥ ∑n≤k≤2n

ak sin(

kπ

4n

)≥√

22 ∑

n≤k≤2nak ≥

n + 1√2

a2n.

If the series ∑n∈N an sin nx is uniformly convergent, then limn→∞ 2na2n = 0. Sim-ilarly we have limn→∞(2n + 1)a2n+1 = 0. Thus limn→∞ nan = 0.

Conversely, we assume that limn→∞ nan = 0. To prove the uniform conver-gence,we suffice to verify that the sequence (Sn(x))n∈N is Cauchy for each x ∈ R.By periodicity, we may assume that x ∈ (0, π). For any ε > 0, there is an integern0 such that nan < ε whenever n > n0. For any x ∈ (0, π) take N := b1/xc andchoose n > n0. Then

∑k≥n

ak sin(kx) = ∑n≤k≤N−1

ak sin(kx) + ∑k≥N

ak sin(kx) =: I1 + I2.

Here we may assume that N > n. Using | sin x| ≤ |x| we have∣∣∣∣∣ ∑n≤k≤N−1

ak sin(kx)

∣∣∣∣∣ ≤ ∑n≤k≤N−1

ak| sin(kx)| ≤ maxn≤k≤N−1

kak ≤ ε.

On the other hand, we have|I2| ≤ 4πNε.

In fact, by Abel’s transform, we arrive at

∑k≥N

ak sin(kx) = −aN DN − 1(x) + ∑k≥N

∆ak Dk(x),

where

Dn(x) := ∑1≤k≤n

sin(kx) =cos x

2 − cos 2n+12 x

2 sin x2

.

Note that |Dn(x)| ≤ π/x for any x ∈ (0, π). Hence∣∣∣∣∣ ∑k≥N

ak sin(kx)

∣∣∣∣∣ ≤ π

x

(aN + ∑

k≥N|∆ak|

)≤ 2πN

(aN + ∑

k≥N|∆ak|

).

Since limn→∞ an = limn→∞(nan cot 1n ) = 0, it follows that∣∣∣∣∣∑k≥K

ak sin(kx)

∣∣∣∣∣ ≤ 4πN ∑k≥N|∆ak| = 4πNaN < 4πε.

Hence |∑k≥n ak sin(kx)| ≤ (1+ 4π)ε and then limn→∞ ||∑k≥n ak sin(kx)|| = 0.

Theorem 16.2.13. (Shah, 1962; Nurcombe, 1992) If (an)n∈N ∈ QMS then the series∑n∈N an sin nx is uniformly convergent if and only if limn→∞ nan = 0.

It is clear that the above theorem follows from the following

Theorem 16.2.14. (Xie-Zhou, 1994) For a complex sequence (cn)∈N ∈ RQMS writef (x) := ∑n∈Z cne

√−1nx. If cn + c−n ∈ K(θ0) for some θ0 ∈ [0, π/2), then f is contin-

uous, and limn→∞ || f − Sn|| = 0 if and only if

limn→∞

ncn = 0, ∑n∈N|cn + c−n| < ∞.


Corollary 16.2.15. If (an)n∈N ∈ RQMS then the series ∑n∈N an sin nx is uniformlyconvergent if and only if limn→∞ nan = 0.

Theorem 16.2.16. (Leindler, 2001) If (an)n∈N ∈ RBVS then the series ∑n∈N an sin nxis uniformly convergent if and only if limn→∞ nan = 0.

Theorem 16.2.17. (Le-Zhou; 2005; Tikhonov, 2007) For a complex sequence (cn)∈N ∈GBVS write f (x) := ∑n∈Z cne

√−1nx. If cn + c−n ∈ K(θ0) for some θ0 ∈ [0, π/2), then

f is continuous, and limn→∞ || f − Sn|| = 0 if and only if

limn→∞

ncn = 0, ∑n∈N|cn + c−n| < ∞.

Theorem 16.2.18. (Yu-Zhou; 2007) For a complex sequence (cn)∈N ∈ NBVS writef (x) := ∑n∈Z cne



limn→∞

ncn = 0, ∑n∈N|cn + c−n| < ∞.

Theorem 16.2.19. (Zhou, 2010) For a complex sequence (cn)∈N ∈ MVBVS writef (x) := ∑n∈Z cne



limn→∞

ncn = 0, ∑n∈N|cn + c−n| < ∞.

16.2.3. Some counterexamples: I. We given an example of a continuous pe-riodic function whose Fourier series diverges at a point. Thus, Theorem 16.2.5 (2)fails if the Dini-Lipschitz condition is replaced by the weaker condition of conti-nuity.

Consider the sawtooth function f which is odd in x and which equals√−1(π−

x) when 0 < x < π. Then

(16.2.9) f (x) ∼ ∑n 6=0

e√−1nx

n= ∑

n≤−1

e√−1nx

n+ ∑

n≥1

e√−1nx

n.

We shall see that the series

∑n≥−1

e√−1nx

n

is no longer the Fourier series of a Riemann integrable function. Indeed, sup-pose it were the Fourier series of an integrable function f , where in particular f isbounded. Using the Abel means, we have

|Ar( f )(0)| = ∑n∈N

rn

n,

which tends to infinity as r tends to 1. This gives the desired contradiction since

|Ar( f )(0)| ≤ 12π

∫ π

−π| f (x)|Pr(x)dx ≤ sup

x∈[−π,π]

| f (x)|

where Pr(x) denotes the Poisson kernel.


For each N ∈ N we define the following two functions on [−π, π],

(16.2.10) fN(x) := ∑1≤|n|≤N

e√−1nx

n, fN(x) := ∑

−N≤n≤−1

e√−1nx

n.

Since

∑1≤n≤N

1n≥ ∑

1≤n≤N−1

∫ n+1

n

dxx

=∫ N

1

dxx

= ln N,

it follows that

| fN(0)| =∣∣∣∣∣ ∑−N≤n≤−1

1n

∣∣∣∣∣ = ∑1≤n≤N

1n≥ ln N.

Lemma 16.2.20. Suppose that the Abel means Ar = ∑n∈N cnrn of the series ∑n∈N cnare bounded as r tends to 1 (with r < 1). If cn = O(1/n), then the partial sums SN :=∑1≤n≤N cn are bounded.

PROOF. Let r := 1− N−1 and choose M so that n|cn| ≤ M and |Ar| ≤ M.Using

SN − Ar = ∑1≤n≤N

(1− rn)cn − ∑n≥N+1

cnrn,

we arrive at

|SN − Ar| ≤ ∑1≤n≤N

|cn|(1− rn) + ∑n≥N+1

|cn|rn

≤ M ∑1≤n≤N

(1− r) +MN ∑

n≥N+1rn

≤ MN(1− r) +MN

11− r

= 2M.

Then |SN | ≤ 3M.

Using (16.2.9) we see that the series ∑n 6=0 e√−1nx/n is the Fourier series of the

sawtooth function f . Here

cn =e√−1nx

n+

e−√−1nx

−n= O

(1|n|

).

Since Ar( f ) = f ∗ Pr and f is bounded, it follows that Ar is bounded as r → 1 andby Lemma 16.2.20 then SN( f ) = fN is uniformly bounded in N and x.

Define

(16.2.11) PN(x) := e2N√−1x fN(x), PN(x) := e2N

√−1x fN(x).

Then

SM(PN) =

PN , M ≥ 3N,PN , M = 2N,0, M < N.

Choose Nk := 32kand αk := 1/k2. Then

Nk+1 > 3Nk, αk ln Nk → ∞.

Finally, we define

(16.2.12) g(x) := ∑k∈N

αkPNk (x).


Due to the uniform boundedness of the PN , the series above converges uniformlyto a continuous periodic function. However

|S2Nm( f )(0)| ≥ cαm ln Nm + O(1)→ ∞

as m→ ∞. Thus we produces a continuous periodic function whose Fourier seriesdiverges at 0.

16.2.4. Some counterexamples: II. Recall the conjugate Dirichlet kernel DNdefined by

(16.2.13) DN(x) := ∑|n|≤N

sign(x)e√−1nx =

cos x2 − cos 2N+1

2 xsin x

2.

As in the computation in Exercise 16.1.16 we have

(16.2.14)∫ π

−π

∣∣∣DN(x)∣∣∣ dx ≤ c ln N

for some c > 0. If f is Riemann integrable, then

f ∗ DN = O(ln N).

As consequence, we have(i) For each α ∈ (0, 1) the series

∑n∈N

sin nxln n

converges for every x but is not the Fourier series of a Riemann integrablefunction.

(ii) For each α ∈ (0, 1) the series

∑n∈N

sin nxnα

converges for every x but is not the Fourier series of a Riemann integrablefunction.

If both cases were true, then we get

(16.2.15) ∑1≤n≤N

1ln n

= O(ln N), ∑1≤n≤N

1nα

= O(ln N) (0 < α < 1).


Exercise 16.2.22. Verify (16.2.15) can not be true.

16.2.5. Bernoulli numbers and Riemann zeta functions. In this subsectionwe try to prove the formula (15.3.32). Setting x = π in (16.2.7) yields

(16.2.16) x cot x = 1 + 2 ∑n∈N

x2

x2 − (nπ)2 , x /∈ πZ.

Consequently,

(16.2.17) x cot x = 1− 2 ∑m∈N

ζ(2m)

π2m x2m.

16.3. THE FOURIER TRANSFORM ON R 187

According to Exercise 13.3.17 and (13.3.21) (which holds also for y ∈ C) we get

(16.2.18) x cot x =√−1x +

2√−1x

e2√−1x − 1

= 1 + ∑m∈N

(−1)m 22mB2m

(2m)!x2m.

Together with (16.2.17) and (16.2.18) we get (15.3.32).Recall the Bernoulli polynomials Bn(x) (defined by (13.3.17)–(13.3.19)) by the

formula

Bn(x) = ∑0≤k≤n

(nk

)Bkxn−k.

For example,

B1(x) = x− 12

, B2(x) = x2 − x +16

.

The Fourier sin series of B1(x) for 0 < x < 1 is

B1(x) = − 1π ∑

k∈N

sin(2πkx)k

.

Integrating it yields

B2n(x) = (−1)n+1 2(2n)!(2π)2n ∑

k∈N

cos(2πkx)k2n ,(16.2.19)

B2n+1(x) = (−1)n+1 2(2n + 1)!(2π)2n+1 ∑

k∈N

sin(2πkx)k2n+1 .(16.2.20)

Finally, for 0 < x < 1, one has

(16.2.21) Bn(x) = − n!(2π√−1)n ∑

k 6=0

e2π√−1kx

kn .

Thus, the Bernoulli polynomials are, up to normalization, successive integrals ofthe sawtooth function.

Exercise 16.2.23. Verify (16.2.19)–(16.2.21).

16.3. The Fourier transform on R

Recall that for any f ∈ C(S1) we have

f (x) ∼ ∑n∈Z

Ane√−1nx,

whereAn =

12π

∫ π

−πf (x)e−

√−1nxdx.

Letting y = x/2π yields

An =∫ 1

0f (y)e−2π

√−1nydy, f (y) ∼ ∑

n∈ZAne2π

√−1ny.

If f is a function defined on R, we may define

f (ξ) :=∫ ∞

−∞f (x)e−2π

√−1xξ dx, f (x) ∼

∫ ∞

−∞f (ξ)e2π

√−1xξ dξ

where the last one is the Fourier inverse formula.


For any function f on R, define∫ ∞

−∞f (x)dx := lim

N→∞

∫ N

−Nf (x)dx = p.v.

∫ ∞

−∞f (x)dx.

This limit may not exist. For example, it is clear if f (x) = 1/(1 + |x|).

A function f defined on R is said to be of moderate decrease if f is continuousand there exists a constant A > 0 so that

| f (x)| ≤ A1 + x2 , x ∈ R.

We denote by M(R) the set of functions of moderate decrease on R. Under theusual addition of functions and multiplication by scalars,M(R) is a vector spaceover C.

If f ∈ M(R) we have∫ ∞

−∞| f (x)|dx = lim

N→∞

∫ N

−N| f (x)|dx ≤ 2 lim

N→∞

∫ N

0

A1 + x2 dx

= 2 limN→∞

[∫ 1

0

A1 + x2 dx +

∫ N

1

A1 + x2 dx

]≤ 2 lim

N→∞

(A +

∫ N

1

Ax2 dx

)= 4A.

Hence the integral∫

R f (x)dx exists.

Proposition 16.3.1. The integral of a function of moderate decrease defined above satisfiesthe following properties:

(i) (Linearity) If f , g ∈ M(R) and a, b ∈ C, then∫R[a f (x) + bg(x)]dx = a

∫R

f (x)dx + b∫

Rg(x)dx.

(ii) (Translation invariance) For every a ∈ R we have∫ ∞

−∞f (x− a)dx =

∫ ∞

−∞f (x)dx.

(iii) (Scaling under dilations) If δ > 0 then

δ∫ ∞

−∞f (δx)dx =

∫ ∞

−∞f (x)dx.

(iv) (Continuity) If f ∈ M(R) then∫ ∞

−∞| f (x− a)− f (x)|dx → 0

as a→ 0.

PROOF. (i)–(iii) are obvious. We now prove (iv). Choose |a| ≤ 1. For anyε > 0 there is an integer N ∈ N such that∫

|x|≥N| f (x)|dx <

ε

4,∫|x|≥N

| f (x− a)|dx ≤ ε

4.


Since f is continuous on R, it follows that f is uniformly continuous in [−N −1, N + 1]. There exists δ > 0 so that

sup|x|≤N

| f (x− a)− f (x)| ≤ ε

4N

whenever |a| < δ. Hence∫ ∞

−∞| f (x− a)− f (x)|dx ≤

∫ N

−N| f (x− a)− f (x)|dx

+∫|x|≥N

| f (x− a)|dx +∫|x≥N

| f (x)|dx

≤ ε

4N· 2N +

ε

4+

ε

4= ε

for |a| ≤ δ ≤ 1.

16.3.1. Fourier transform. If f ∈ M(R) we define its Fourier transform forξ ∈ R by

(16.3.1) f (ξ) :=∫ ∞

−∞f (x)e−2π

√−1xξ dx.

It is clear that the integral exists. Moreover

(i) f is bounded and continuous.(ii) f (ξ)→ 0 as |ξ| → ∞. In fact,

f (ξ) =12

∫ ∞

−∞

∣∣∣∣ f (x)− f(

x− 12ξ

)∣∣∣∣ e−2π√−1xξdx → 0

by Proposition 16.3.1, as |ξ| → ∞.However there is a function f of moderate decrease whose Fourier transform

is not of moderate decrease.

Exercise 16.3.2. Let f be a continuous function on R which vanishes for |x| ≥ 1,with f (0) = 0, and which is equal to 1/ ln(1/|x|) for all x in a neighborhood ofthe origin. Prove that f is not of moderate decrease.

The Schwartz space S(R) on R consists of all smooth functions f so that allits derivatives f (`) are rapidly decreasing in the sense that

supx∈R|x|k| f (`)(x)| < ∞

for every k, ` ≥ 0. Observe that S(R) (M(R).

Exercise 16.3.3. Prove that S(R) is a vector space over C. Moreover, if f ∈ S(R)then

f ′(x) ∈ S(R), x f (x) ∈ S(R).Thus the Schwarz space is closed under differentiation and multiplication by poly-nomials.

Example 16.3.4. (Gaussian functions) This Gaussian function

f (x) = e−ax2, a > 0

belongs to S(R).


Example 16.3.5. (Bump functions) Let a < b and define

f (x) =

0, x ≤ a,

e−1/(x−a)e−1/(b−x), a < x < b,0, x ≥ b,

belongs to S(R).

Exercise 16.3.6. Show that e−|x| /∈ S(R).

Since S(R) (M(R) we can still define the Fourier transform of f ∈ S(R) by

f (ξ) =∫ ∞

−∞f (x)e−2π

√−1xξ dx.

We use the notion f (x)→ f (ξ) to mean that f denotes the Fourier transform of f .

Proposition 16.3.7. If f ∈ S(R), then

(i) f (x + a)→ f (ξ)e2π√−1aξ whenever a ∈ R,

(ii) f (x)e−2π√−1xa → f (ξ + x) whenever a ∈ R.

(iii) f (δx)→ δ−1 f (δ−1ξ) whenever δ > 0.(iv) f ′(x)→ 2π

√−1ξ f (ξ).

(v) −2π√−1x f (x)→ d

dξ f (ξ).

PROOF. We only give a proof of (v). Let F(x) = −2π√−1x f (x). Then

F(ξ) =∫ ∞

−∞−2π√−1x f (x)e−2π

√−1xξ dx.

Compute

f (ξ + a)− f (ξ)a

− F(ξ) =∫ ∞

−∞f (x)e−2π

√−1xξ

(e−2π

√−1xa − 1a

+ 2π√−1x

)dx.

Since f (x), x f (x) ∈ S(R), there exists an integer N so that∫|x|≥N

| f (x)|dx,∫|x|≥N

|x f (x)|dx ≤ ε.

Moreover, for |x| ≤ N there exists a0 so that |a| < a0 implies∣∣∣∣∣ e−2π√−1xa − 1a

+ 2π√−1x

∣∣∣∣∣ ≤ ε

N.

Hence for |a| < a0 we have∣∣∣∣∣ f (ξ + a)− f (ξ)a

− F(ξ)

∣∣∣∣∣≤

∫ N

−N

∣∣∣∣∣ f (x)e−2π√−1xξ

(e−2π

√−1xa − 1a

+ 2π√−1x

)∣∣∣∣∣ dx + Cε

≤ C′ε.

Here C, C′ are constants.

Theorem 16.3.8. If f ∈ S(R) then f ∈ S(R).


PROOF. Since f ∈ S(R) we have Ck,`( f ) := supx∈R |x|k| f (`)(x)| < ∞. Inparticular, | f (ξ)| ≤ C < ∞. From(

1 + |x|2)| f (x)| ≤ sup

x∈R| f (x)|+ sup

x∈R|x|2| f (x)| = C0,0( f ) + C2,0( f ) < ∞

we can find a constant C > 0 so that

| f (x)| ≤ C1 + |x|2

and hence| f (ξ)| ≤

∫ ∞

−∞| f (x)|dx ≤ 2C

∫ ∞

0

dx1 + x2 < ∞.

In general, for each pair (k, `), the expression ξk f (`)(ξ) is bounded, since it is theFourier transform of 1

(2π√−1)k (

ddx )

k[(−2π√−1x)` f (x)].

Recall that ∫ ∞

−∞e−πx2

dx = 1.

Theorem 16.3.9. (1) If f (x) = e−πx2then f (ξ) = f (ξ).

(2) If δ > 0 and

Kδ(x) :=1√δ

e−πx2

δ

then Kδ(ξ) = e−πδξ2.

(3) (Kδ)δ>0 is a family of good kernels as δ→ 0. That is∫ ∞

−∞Kδ(x)dx = 1,

∫ ∞

−∞|Kδ(x)|dx ≤ M,

and for any η > 0,

limδ→0

∫|x|>η

|Kδ(x)|dx = 0.

PROOF. (1) Define

F(ξ) := f (ξ) =∫ ∞

−∞e−πx2

e−2π√−1xξ dx.

Then F(0) = 1 and

F′(ξ) =∫ ∞

−∞f (x)

(−2π√−1x

)e−2π

√−1xξ dx =

√−1

∫ ∞

−∞f ′(x)e−2π

√−1xξ dx

=√−1(2π

√−1ξ) f (ξ) = −2πξF(ξ).

LetG(ξ) := F(ξ)eπξ2

.We get G′(ξ) = 0 and G(0) = 1 which implies G(ξ) ≡ 1.

(2) Write f (x) = w−πx2as before. Then

Kδ(ξ) =∫ ∞

−∞Kδ(x)e−2π

√−1xξ dx =

∫ ∞

−∞

1√δ

f(

x√δ

)e−2π

√−1xξ dx

=∫ ∞

−∞f(

x√δ

)e−2π

√−1 x√

δ(√

δξ)d(

x√δ

)= f (

√δξ) = w−πδξ2

.

The last statement is obvious.


If f , g ∈ S(R) their convolution is defined by

(16.3.2) ( f ∗ g)(x) :=∫ ∞

−∞f (x− t)g(t)dt.

Theorem 16.3.10. If ( f ∗ Kδ)(x)→ f (x) uniformly in x as δ→ 0.

PROOF. For any ε > 0 there exists R > 0 such that | f (x)| < ε/4 whenever|x| ≥ R, since | f (x)| ≤ C/|x| for some constant C > 0. The continuity of fimplies that f is uniformly continuous on [−R, R]. Hence we can find η > 0 sothat | f (x)− f (y)| < ε/2 whenever |x− y| < η and x, y ∈ [−R, R]. If x ∈ [−R, R]and |y| > R with |x− y| < δ, then

| f (x)− f (y)| ≤ | f (x)− f (R)|+ | f (y)− f (R)| ≤ ε

2+

ε

4< ε.

If |x|, |y| > R with |x − y| < η, then | f (x) − f (y)| < ε/2. Consequently, f isuniformly continuous on R. Compute

|( f ∗ Kδ)(x)− f (x)| ≤(∫|t|≥η

+∫|t|<η

)Kδ(t)| f (x− t)− f (x)|dt

< ε + 2 maxR| f | ·

∫|t|≥η

Kδ(t)dt < 2ε

for δ 1, by Theorem 16.3.9

Given f , g ∈ S(R) and consider

F(x, y) := f (x)g(y)e−2π√−1xy.

Since | f (x)| ≤ C f /(1 + |x|2) and |g(y)| ≤ Cg/(1 + |y|2), it follows that

|F(x, y)| ≤C f Cg

(1 + |x|2)(1 + |y|2) .

Consequently the following two functions

F1(x) :=∫ ∞

−∞F(x, y)dy, F2(y) :=

∫ ∞

−∞F(x, y)dx

are continuous and of moderate decrease. BY Fubini’s theorem, Theorem 13.2.2,we have ∫ ∞

−∞F1(x)dx =

∫ ∞

−∞F2(y)dy.

Thus

(16.3.3)∫ ∞

−∞f (x)g(x)dx =

∫ ∞

−∞f (y)g(y)dy, f , g ∈ S(R).

Theorem 16.3.11. (Fourier inversion) If f ∈ S(R), then

(16.3.4) f (x) =∫ ∞

−∞f (ξ)e2π

√−1ξxdξ.

PROOF. Let Gδ(x) := e−πδx2for δ > 0. According to Proposition 16.3.7 and

Theorem 16.3.9 we have

Gδ(ξ) =1√δ

f(

1√δ

ξ

)=

1√δ

f(

1√δ

ξ

)=

1√δ

e−πξ2

δ = Kδ(x)


where f (x) = e−πx2. Using (16.3.3) yields∫ ∞

−∞f (x)Kδ(x)dx =

∫ ∞

−∞f (ξ)Gδ(ξ)dξ.

Letting δ→ 0 we conclude that

f (0) =∫ ∞

−∞f (ξ)dξ.

In general, let F(y) := f (y + x). By the above argument we get

f (x) = F(0) =∫ ∞

−∞F(ξ)dξ =

∫ ∞

−∞f (ξ)e2π

√−1xξdξ.

Thus we proved (16.3.4).

Define

(16.3.5) F : F (R) −→ S(R), f 7−→ F ( f )

and

(16.3.6) F ∗ : S(R) −→ S(R), g 7−→ F ∗(g),

where

F ( f )(ξ) :=∫ ∞

−∞f (x)e−2π

√−1xξdx = f (ξ),(16.3.7)

F ∗(g)(x) :=∫ ∞

−∞g(ξ)e2π

√−1xξ dξ = g(x).(16.3.8)


(16.3.9) F F = 1 = F ∗ F on S(R).

Therefore the Fourier transform F is bijective mapping on the Schwartz space.

Proposition 16.3.12. If f , g ∈ S(R) then

(16.3.10) f ∗ g ∈ S(R), f ∗ g = g ∗ f , ( f ∗ g)(ξ) = f (ξ)g(ξ).

PROOF. (1) Recall that

( f ∗ g)(x) =∫ ∞

−∞f (y)g(x− y)dy.

Hence|x|k|( f ∗ g)(x)| ≤

∫ ∞

−∞| f (y)|

(|x|k|g(x− y)|

)dy.

We claim first that

supx∈R|x|k|g(x− y)| = sup

z∈R|z + y|k|g(z)| ≤ Ak (1 + |y|)k

for some constant Ak depending only on g and k. Indeed, if |y| ≤ |z|, then

|z + y|k|g(z)| ≤ 2k|z|k|g(z)| ≤ 2kCk,0(g);

if |y| ≥ |z|, then

|z + y|k|g(z)| ≤ 2k|y|6k|g(z)| ≤ 2kC0,0(g)|y|k.

For both cases, we get the desired result. Hence

supx∈R|x|k|( f ∗ g)(x)| ≤ Ak

∫ ∞

−∞| f (y)| (1 + |y|)k dy < ∞


because of f ∈ S(R). Using the identities(d

dx

)`

( f ∗ g)(x) =(

f ∗ g(`))(x), ` ∈ N,

which can be proved by induction on `, we arrive at f ∗ g ∈ S(R).The second result is obvious.To prove the last one we define

F(x, y) := f (y)g(x− y)e−2π√−1xξ

because

( f ∗ g)(ξ) =∫ ∞

−∞

[∫ ∞

−∞f (y)g(x− y)dy

]e−2π

√−1xξdx.

We also consider two functions

F1(x) := ( f ∗ g)(x)e−2π√−1xξ , F2(y) := f (y)e−2π

√−1yξ g(ξ).

By Fubini’s theorem, Theorem 13.2.2, we arrive at

( f ∗ g)(ξ) =∫ ∞

−∞F1(x)dx =

∫∫R2

F(x, y)dxdy =∫ ∞

−∞F2(y)dy = f (ξ)g(ξ).

Thus f ∗ g = f g.

The Schwartz space S(R) can be equipped with a Hermitian inner product

(16.3.11) ( f , g) :=∫ ∞

−∞f (x)g(x)dx, || f || :=

√( f , f ).

Theorem 16.3.13. (Plancherel) If f ∈ S(R), then || f || = || f ||.

PROOF. Let g(x) := f (−x) and h(x) := ( f ∗ g)(x). Then

g(ξ) =∫ ∞

−∞g(x)e−2π

√−1xξ dx =

∫ ∞

−∞f (−x)e−2π

√−1xξdx

=∫ ∞

−∞f (−x)e2π

√−1xξ dx = f (ξ).

Moreover

h(ξ) = ( f ∗ g)(ξ) = f (ξ)g(ξ) = | f (ξ)|2,

h(0) = ( f ∗ g)(0) =∫ ∞

−∞f (y)g(−y)dy =

∫ ∞

−∞| f (y)|2dy.

Consequently, we get

|| f ||2 =∫ ∞

−∞| f (y)|2dy = h(0) =

∫ ∞

−∞h(ξ)dξ =

∫ ∞

−∞| f (ξ)|2dξ = || f ||2

since h ∈ S(R).


16.3.2. Heat kernel on R. Consider the heat equation given by

(16.3.12)∂

∂tu(x, t) =

∂2

∂x2 u(x, t), u(x, 0) = f (x) ∈ S(R).

Define the heat kernel by

(16.3.13) Ht(x) = H(x, t) := K4πt(x) =1√4πt

e−x2

t2 , x ∈ R, t > 0.

Then

(16.3.14) Ht(x) = e−4π2tξ2.

Applying the Fourier transform on (16.3.12) yields

∂tu(ξ, t) = −4π2ξ2u(ξ, t), u(ξ, t) = A(ξ)e−4π2ξ2t, u(ξ, 0) = f (ξ), A(ξ) = f (ξ).

Theorem 16.3.14. Given f ∈ S(R), let

u(x, t) = ( f ∗ Ht)(x), t > 0

where Ht is the heat kernel. Then(i) The function u is C2 when x ∈ R and t > 0, and u solves the heat equation.

(ii) u(x, t) → f (x) uniformly in x as t → 0. Hence if we set u(x, 0) = f (x), thenu is continuous on the closure of the upper half-plane R2

+ = (x, t) ∈ R2 : x ∈R, t > 0.

(iii)∫ ∞−∞ |u(x, t)− f (x)|2dx → 0 as t→ 0.

(iv) u(·, t) belongs to S(R) uniformly in t, in the sense that for any T > 0

(16.3.15) supx∈R, t∈(0,T)

|x|k∣∣∣∣∣ ∂`

∂x`u(x, t)

∣∣∣∣∣ < ∞

for each k, ` ≥ 0.

PROOF. (i) Because u = f ∗ Hr, taking the Fourier transform in the x-variablegives

u(ξ, t) = f (ξ)e−4π2ξt

and so by the Fourier inversion formula

u(x, t) =∫ ∞

−∞u(ξ, t)e2π

√−1ξxdξ =

∫ ∞

−∞f (ξ)e−4π2ξte2π

√−1ξxdξ.

By differentiating under the integral sign, we get (i). In fact, we can prove that u issmooth.

(ii) It follows from u = f ∗ Ht = f ∗ K4πt ⇒ f as t→ 0.(iii) By Theorem 16.3.13, we get∫ ∞

−∞|u(x, t)− f (x)|2dx =

∫ ∞

−∞|u(ξ, t)− f (ξ)|2dξ =

∫ ∞

−∞| f (ξ)|2

∣∣∣e−4π2tξ2 − 1∣∣∣2 dξ.

Since |w−4π2tξ2 − 1|2 ≤ 4 and f ∈ S(R) it follows that for any ε > 0 there existsN ∈ N so that ∫

|ξ|≥N| f (ξ)|2dξ <

ε

8.


Since f is bounded, we can find δ > 0 so that

sup|ξ|≤N

| f (ξ)|2∣∣∣e−4π2tξ2 − 1

∣∣∣2 <ε

4N

whenever |t| < δ. Therefore we arrive at∫ ∞

−∞|u(x, t)− f (x)|2dx ≤

∫|ξ|≤N

| f (ξ)|2∣∣∣e−4π2tξ2 − 1

∣∣∣2 dξ

+∫|ξ|≥N

| f (ξ)|2∣∣∣e−4π2tξ2 − 1

∣∣∣2 dξ

≤ ε

4N2N +

ε

84 = ε

for |t| < δ.(iv) This result follows from

|u(x, t)| ≤∫|y|≤ |x|2

| f (x− y)|Ht(y)dy +∫|y|≥ |x|2

| f (x− y)|Ht(y)dy

≤ CN

(1 + |x|)N +C√

te−

cx2t .

Since f is rapidly decreasing, we have | f (x − y)| ≤ CN/(1 + |x|)N when |y| ≤|x|/2. If |y| ≥ |x|/2 then Ht(y) ≤ Ct−1/2e−cx2/t, and we then obtain the aboveinequality. We see that u(x, t) is rapidly decreasing uniformly for 0 < t < T. Thesame argument can be applied for higher derivatives.

Theorem 16.3.15. Suppose that u(x, t) satisfies the following conditions:(i) u is continuous on the closure of the upper half-plane.

(ii) u satisfies the heat equation for t > 0.(iii) u satisfies the boundary condition u(x, 0) = 0.(iv) u(·, t) ∈ S(R) uniformly in t as in (16.3.15).

Then u ≡ 0.

PROOF. Consider the energy of u given by

E(t) :=∫ ∞

−∞|u(x, t)|2dx ≥ 0.

since ∂tu = ∂2xu we arrive at

E′(t) =∫ ∞

−∞

[∂tu(x, t)u(x, t) + u(x, t)∂tu(x, t)

]dx

=∫ ∞

−∞

[∂2

xu(x, t)u(x, t) + u(x, t)∂2xu(x, t)

]dx

= −∫ ∞

−∞

[∂xu(x, t)∂xu(x, t) + ∂xu(x, t)∂xu(x, t)

]dx

= −2∫ ∞

−∞|∂xu(x, t)|2dx ≤ 0,

because u and its x-derivatives decrease rapidly as |x| → ∞. Since E(0) = 0, itfollows that E(t) ≡ 0 and then u ≡ 0.


16.3.3. Poisson’s summation formula. Given f ∈ S(R) and define

(16.3.16) F1(x) := ∑n∈Z

f (x + n).

Since | f (x)| ≤ A/|x|2, F1 is absolutely and uniformly convergent on every com-pact open subset of R. Hence F ∈ C(R). Note that F1(x + 1) = F1(x). We call thefunction F1 is the periodization of f .

Another continuous and periodic function

(16.3.17) F2(x) := ∑n∈Z

f (n)e2π√−1nx

comes from the identity

f (x) =∫ ∞

−∞f (ξ)e2π

√−1ξxdξ.

Exercise 16.3.16. Verify that F1 and F2 are continuous and periodic.

Theorem 16.3.17. (Poisson’s summation formula) If f ∈ S(R), then

(16.3.18) ∑n∈Z

f (x + n) = ∑n∈Z

f (n)e2π√−1nx

for any x ∈ R. In particular,

(16.3.19) ∑n∈Z

f (n) = ∑n∈Z

f (n).

PROOF. Compute

F1(m) =∫ 1

0F1(x)e−2π

√−1mxdx =

∫ 1

0

(∑

n∈Zf (x + n)

)e−2π

√−1mxdx

= ∑n∈Z

∫ 1

0f (x + n)e−2π

√−1mxdx = ∑

n∈Z

∫ n+1

nf (y)e−2π

√−1mydy

=∫ ∞

−∞f (y)e−2π

√−1mydy = f (m).

Thus F1(m) = F2(m). By Corollary 16.1.10, we get F1 ≡ F2.

Exercise 16.3.18. The formula (16.3.18) also holds for both f and f ′ being of mod-erate decrease.

Example 16.3.19. Consider the function f (x) given by

f (x) =

1− |x|, |x| ≤ 1,0, |x > 1.

Then

f (ξ) =(

sin πξ

πξ

)2

and

(16.3.20) ∑n∈Z

1(n + α)2 =

( π

sin πα

)2

for α ∈ R \ Z.



(16.3.21) ∑n∈Z

1n + α

=π

tan πα

whenever α ∈ R \ Z. (Hint: What is the precise meaning of the series on theleft-hand side of (16.3.21)? Evaluate at α = 1/2.)

Exercise 16.3.21. We use the Poisson summation formula to give another proof of(15.3.32. Apply Theorem 16.3.17 to f (x) = t

π(x2+t2)and f (ξ) = e−2πt|ξ| where

t > 0 in order to get

1π ∑

n∈Z

tt2 + n2 = ∑

n∈Ze−2πt|n|.

Prove that1π ∑

n∈Z

tt2 + n2 =

1πt

+2π ∑

m∈N(−1)m+1ζ(2m)t2m−1

as well as

∑n∈Z

e−2πt|n| =2

1− e−2πt − 1.

Use the fact thatz

ez − 1= 1− z

2+ ∑

m∈N

B2m

(2m)!z2m,

where Bk are the Bernoulli numbers to deduce that

2ζ(2m) = (−1)m+1 (2π)m+1

(2m)!B2m.

16.3.4. Theta and zeta functions. We define the (Jacobi) theta function ϑ(s)for s > 0 by (also see (15.3.31))

(16.3.22) ϑ(s) := ∑n∈Z

e−πn2s.

Theorem 16.3.22. One has

(16.3.23) s−1/2ϑ(1/s) = ϑ(s)

whenever s > 0.

PROOF. Define f (x) := e−πsx2. Then f (ξ) = s−1/2e−πξ2/s. By Theorem

16.3.17) we haver

ϑ(s) = ∑n∈Z

f (n) = ∑n∈Z

f (n) = s−1/2θ(1/s)

which implies (16.3.23).



16.3.5. Heat kernel on S1. Define the generalization Θ(z|τ) by

(16.3.24) Θ(z|τ) := ∑n∈Z

e√−1πn2τe2π

√−1nz

whenever Im(τ) > 0 and z ∈ C. Note that

Θ(0|√−1s) = ϑ(s).

A solution to the equation

∂tu(x, t) = ∂2xu(x, t)

subject to u(x, 0) = f (x), where f is periodic of period 1, is given by

u(x, t) =(

f ∗ HS1

t

)(x)

where HS1

t (x) is the heat kernel on the circle, that is,

(16.3.25) HS1

t (x) = ∑n∈Z

e−4π2n2te2π√−1nx = Θ(x|4π

√−1t).

Recall the heat kernel on R given by

Ht(x) =1√4πt

e−x24t , Ht(ξ) = e−4π2ξ2t.


∑n∈Z

Ht(x + n) = ∑n∈Z

Ht(n)e2π√−1nx = ∑

n∈Ze−4π2n2te2π

√−1nx.

Theorem 16.3.24. The heat kernel on the circle is the periodization of the heat kernel onthe real line:

(16.3.26) HS1

t (x) = ∑n∈Z

Ht(x + n).

The kernel (HS1

t (x))t>0, |x| ≤ 1/2, is a family of good kernels.

PROOF. It is clear that∫|x|≤1/2 HS1

t (x)dx = 1 and HS1

t ≥ 0. We claim thatwhen |x| ≤ 1/2

HS1

t (x) = Ht(x) + Et(x)where the error term satisfies |Et(x)| ≤ c1e−c2/r with c1, c2 > 0 and 0 < t ≤ 1. Infact,

Et(x) := ∑|n|≥1

Ht(x + n) =1√4πt

∑|n|≥1

e−(x+n)2/4t ≤ Ct−1/2 ∑n≥1

e−cn2/t

since |x| ≤ 1/2. Note that n2/t ≥ n2 and n2/t ≥ 1/t whenever 0 < t ≤ 1, soe−cn2/t ≤ e−cn2/2e−c/2t. Hence

|Et(x)| ≤ Ct−1/2e−c/2t ∑n≥1

e−cn2/2 ≤ c1e−c2/t.

Since∫|x|≤1/2 |Et(x)|dx → as t→ 0, it follows that∫

|η<|x|≤1/2|HS1

t (x)|dx → 0

as t→ 0 because Ht does.


16.3.6. Heisenberg’s uncertainty principle. In quantum mechanics, Heisen-berg’s uncertainty principle states that

(textunceetaintyo f position)× (uncertainty of momentum) ≥ h16π2 > 0

where h denotes the Planck’s constant.

Theorem 16.3.25. (Heisenberg’s uncertainty principle) Suppose f is a function inS(R) which satisfies the normalizing condition

∫ ∞−∞ | f (x)|2dx = 1. Then

(16.3.27)(∫ ∞

−∞x2| f (x)|2dx

)(∫ ∞

−∞ξ2| f (ξ)|2dξ

)≥ 1

16π2 ,

and equality holds if and only if f (x) = Ae−Bx2where B > 0 and A2 =

√2B/π. In

fact, we have

(16.3.28)[∫ ∞

−∞(x− x0)

2| f (x)|2dx] [∫ ∞

−∞(ξ − ξ0)

2| f (ξ)|2dξ

]≥ 1

16π2 .

PROOF. The second inequality (16.3.28) follows from (16.3.27) by replacingf (x) by e−2π

√−1xξ0 f (x + x0). We now prove (16.3.27) under the normalizing con-

dition. Compute

1 =∫ ∞

−∞| f (x)|2dx = −

∫ ∞

−∞

(x

ddx| f (x)|2

)dx

= −∫ ∞

−∞

[x f ′(x) f (x) + x f ′(x) f (x)

]dx

≤ 2∫ ∞

−∞|x|| f (x)|| f ′(x)|dx

≤ 2(∫ ∞

−∞x2| f (x)|2dx

)1/2 (∫ ∞

−∞| f ′(x)|2dx

)1/2.

Since f ′(ξ) = 2π√−1ξ f (ξ), it follows∫ ∞

−∞| f ′(x)|2dx =

∫ ∞

−∞| f ′(ξ)|2dξ = 4π2

∫ ∞

−∞ξ2| f (ξ)|2dξ.

These two inequalities now immediately imply (16.3.27).

The Heisenberg uncertainty principle can be formulated in terms of the Her-mite operator operator

(16.3.29) L := − d2

dx2 + x2,

which acts on S(R) by the formula

(16.3.30) L( f )(x) := − f ′′(x) + x2 f (x).

Consider the inner product on S(R) given by

( f , g) :=∫ ∞

−∞f (x)g(x)dx, f , g ∈ S(R).

In the proof of Theorem 16.3.25 we have actually proved

(16.3.31) (L f , f ) ≥ ( f , f ), f ∈ S(R).

This is usually denoted by L ≥ I.


Consider the annihilation and creation operators

(16.3.32) A :=d

dx+ x, A∗ := − d

dx+ x,

respectively acting on S(R).

Exercise 16.3.26. Prove that for all f , g ∈ S(R) we have

(A f , g) = ( f , A∗g), (A f , A f ) = (A∗A f , f ) ≥ 0, A∗A = L− I.

For each t ∈ R define

(16.3.33) At :=d

dx+ tx, A∗ := − d

dx+ tx.

We can think of (A∗t At f , f ) as a quadratic polynomial in t.

Exercise 16.3.27. Use the fact that (A∗t At f , f ) ≥ 0 to give another proof of theHeisenberg uncertainty principle.

The Hermite functions hk(x) are defined by

(16.3.34) ∑k≥0

hk(x)tk

k!= e−

x22 +2tx−t2

.


(16.3.35) hk(x) = (−1)kex2/2(

ddx

)ke−x2

.

For any f ∈ M(R) we see that∫ ∞

−∞f (y)e−y2

e2xydy = 0 for all x ∈ R =⇒ f ≡ 0.

Indeed, we can consider f ∗ e−x2.

We say that the family (hk(x))k≥0 is complete if ( f , hk) = 0 for all k ≥ 0 andf ∈ S(R).

Exercise 16.3.29. Verify that the family (hk(x))k≥0 is complete.

Define

(16.3.36) h∗k (x) := hk

(√2πx

).

Exercise 16.3.30. Show that h∗k (ξ) = (−√−1)kh∗k (ξ). Therefore, each h∗k is an

eigenfunction for the Fourier transform. Show also that Lhk = (2k + 1)hk. In par-ticular we conclude that the function hk are mutually orthogonal for the L2-innerproduct on the Schwartz space. Finally show that∫ ∞

−∞|hk(x)|2dx = π1/22kk!.


16.3.7. Differentiation and integration of the Fourier series. Let f be a peri-odic function with period 2π. Then

f (x) ∼ a0

2+ ∑

n∈N(an cos nx + bn sin nx) =: S( f )(x).


(16.3.37) limn→∞

an = 0 = limn→∞

bn, f ∈ R[−π, π].

Theorem 16.3.31. (A) If f ∈ R[−π, π] then

(16.3.38)∫ x

cf (t)dt =

∫ x

c

a0

2dt + ∑

n∈N

∫ x

c(an cos nt + bn sin nt) dt

for any c, x ∈ [−π, π].(B) If f ′′ ∈ R[−π, π], then

(16.3.39) f ′(x) = ∑n∈N

(−nan sin nx + nbn cos nx)

for any x ∈ [−π, π].(C) If f ′ ∈ R[−π, π] then S( f )(x) uniformly converges f (x) in [−π, π].(D) (Weierstrass) If f is a continuous and periodic function with period 2π, then

there exists a sequence of trigonometric polynomials (Tn)n∈N, where Tn(x) =αn

02 +

∑1≤k≤n(αnk cos kx + βn

k sin kx), such that for any ε > 0 there exists an integer N ∈ Nso that

supx∈R| f (x)− Tn(x)| ≤ ε

whenever n ≥ N.(E) (Parseval) If f ∈ R[−π, π], then

(16.3.40)a2

02+ ∑

n∈N

(a2

n + b2n

)=

1π

∫ π

−πf 2(x)dx.

PROOF. (D) Recall that

Sn( f )(x) :=a0

2+ ∑

1≤k≤n(an cos kx + bk sin kx) = ( f ∗ Dn)(x),

whereDn(x) := ∑

|k|≤ne√−1kx.

The Fejer kernel is given by

σn( f ) :=1n ∑

0≤i≤n−1Si( f ) = f ∗ Fn, Fn(x) =

1N

sin2(Nx/2)sin2(x/2)

.

For any x ∈ [−π, π] one has

| f (x)− σn( f )(x)| =

∣∣∣∣ 12π

∫ π

−πf (x)Fn(y)dy− 1

2π

∫ π

−πf (y)Fn(x− y)dy

∣∣∣∣≤ 1

2π

∫ π

−πFn(y)| f (x)− f (x− y)|dy.


Lemma 16.3.32. (Best approximation) Suppose f ∈ R[−π, π]. For any trigonometricpolynomial Tn(x) = α0

2 + ∑1≤k≤n(αn cos nx + βn sin nx) we have

(16.3.41)1

2π

∫ π

−π| f (x)− Sn( f )(x)|2dx ≤ 1

2π

∫ π

−π| f (x)− Tn(x)|2dx.

PROOF. Since f − Tn ∈ R[−π, π], we can define

∆n :=(

12π

∫ π

−π| f (x)− Tn(x)|2dx

)1/2.

Compute

∆2n =

12π

∫ π

−π| f (x)− Rn(x)|2dx

=1

2π

[∫ π

−πf 2(x)dx− 2

∫ π

−πf (x)Tn(x)dx +

∫ π

−πT2

n(x)dx]

=1

2π

∫ π

−πf 2(x)dx− 1

π

∫ π

−πf (x)

[α2

2+ ∑

1≤k≤n(αk cos kx + βk sin kx)

]dx

+1

2π

∫ π

−π

[α0

2+ ∑

1≤k≤n(αk cos kx + βk sin kx)

]2

dx

=1

2π

∫ π

−πf 2(x)dx− α0a0

2− ∑

1≤k≤n(αkak + βkbk) +

α20

2+

12 ∑

1≤k≤n

(α2

k + β2k

)=

12π

∫ π

−πf 2(x)dx +

12

[(α0 − a0)

2

2+ ∑

1≤k≤n(αk − ak)

2 + (βk − bk)2

]

−a2

04− 1

2 ∑1≤k≤n

(a2

k + b2k

).

On the other hand,

12π

∫ π

−π| f (x)− Sn( f )(x)|2dx =

12π

∫ π

−πf 2(x)dx−

a20

4− 1

2 ∑1≤k≤n

(a2

k + b2k

).

Hence

∆2n −

12π

∫ π

−π| f (x)− Sn( f )(x)|2dx

=12

[(α0 − a0)

2

2+ ∑

1≤k≤n(αk − ak)

2 + (βk − bk)2

]≥ 0.


Lemma 16.3.33. If f ∈ R[−π, π], then Sn( f )→ f in L2[−π, π]. That is,

(16.3.42) limn→∞

∫ π

−π| f (x)− Sn( f )(x)|2dx = 0.

PROOF. For any m < n we have∫ π

−π| f (x)− Sn( f )(x)|2dx ≤

∫ π

−π| f (x)− Sm( f )(x)|2dx


(B) Since f ′′ ∈ R[−π, π], we can apply part (C) to f ′.(A) Let

g(x) :=

π2 , t = 0 or t = x,π, 0 < t < x,0, x < t < 2π.

Then

g(x) ∼ x2+ ∑

n∈N

(sin nx

xcos nt +

1− cos nxx

sin nt)

.

Using (16.3.43) we get

1π

∫ x

0π f (t)dt =

a0x2

+ ∑n∈N

(an

nsin nx +

bn

n(1− cos nx)

)for any 0 ≤ x ≤ 2π. We then proved (16.3.38) for c = 0.

Remark 16.3.34. We can find a trigonometric series T(x) = α02 + ∑n∈N(αn cos nx +

βn sin nx) such that for any integrable and periodic function f (x) we have S( f )(x) 6=T(x). In fact, consider

T(x) := ∑n∈N

cos(nx)√n

.

If T(x) = S( f )(x), then(16.3.40) implies that

∞ = ∑n∈N

(1√n

)2=

1π

∫ π

−πf 2(x)dx < ∞

a contradiction.

In general, if f , g ∈ R[−π, π], then

(16.3.43)a0α0

2+ ∑

n∈N(anαn + bnβn) =

1π

∫ π

−πf (x)g(x)dx,

where

f (x) ∼ a0

2+ ∑

n∈N(an cos nx + bn sin nx) , g(x) ∼ α0

2+ ∑

n∈N(αn cos nx + βn sin nx) .


Example 16.3.36. Consider f (x) = x2

2 − πx for 0 ≤ x ≤ 2π. Since

π − x2∼ ∑

n∈N

sin mxn

it follows that

f (x) =12

∫ x

0

π − t2

dt =12 ∑

n∈N

∫ x

0

sin ntn

dt =12

[− ∑

n∈N

1n2 + ∑

n∈N

cosnxn2

].

Hence

(16.3.44)x2

2− πx = −π2

3+ ∑

n∈N

2 cos nxn2 , x ∈ [0, 2π].

16.4. ISOPERIMETRIC INEQUALITY 207

16.4. Isoperimetric inequality

A parameter curve γ is a mapping γ : [a, b] → R2. The image of γ is a set ofpoints in the plane which we call a curve and denote by Γ. The curve is simple ifit does not interest itself, and closed if the two endpoints coincide. We can extendγ to a periodic function on R of period b− a and think of γ as a function on thecircle. Moreover the parametrization γ induces an orientation on Γ as the param-eter s travels from a to b.

Assumption: We also always assume that γ is of class C1, and that its deriv-ative γ satisfies γ(s) 6= 0.

Any C1 bijective mapping s : [c, d] → [a, b] gives rise to another parametriza-tion of Γ by the formula

η(t) := γ(s(t)).

Observe that the conditions that γ be closed and simple are independent of thechosen parametrization. We say that the two parametrizations γ and η are equiv-alent if s(t) > 0 for all t; this means that η and γ induce the same orientation onthe curve Γ. If s(t) < 0 then η reverses the orientation.

16.4.1. Length and area. If γ is parametrized by γ(s) = (x(s), y(s)) then thelength of the curve Γ is defined by

(16.4.1) L(γ) :=∫ b

a|γ(s)|ds =

∫ b

a

√x(s)2 + y(s)2ds.

The length of γ does not depend on its parametrization. Indeed, if γ(s(t)) = η(t)for some C1 bijective mapping s : [c, d]→ [a, b], then∫ b

a|γ(s)|ds =

∫ d

c|γ(s(t))||s(t)|dt =

∫ d

c|η(t)|dt.

We say that γ is a parametrization by arc-length if |γ(s)| = 1 for all s. Thismeans that γ(s) trvels at a constant speed and hence the length of Γ is b− a. There-fore, after a possible additional translation, a parametrization by arc-length will bedefined on [0, `], where ` := L(γ).

Exercise 16.4.1. Let γ : [a, b]→ R2 be a parametrization for the closed curve Γ.

(i) Prove that γ is a parametrizxation by arc-length if and only if the lengthof the curve from γ(a) to γ(s) is s− a, that is,∫ s

a|γ(t)|dt = s− a.

(ii) Prove that any curve Γ admits a parametriztion by arc-length.(iii) If γ(t) = (x(t), y(t)), then

12

∫ b

a[x(s)y(s)− y(s)x(s)]ds =

∫ b

ax(s)y(s)ds = −

∫ b

ay(s)x(s)ds.

(iv) Define the reverse parametrization of γ by γ− : [a, b]→ R2 with γ−(t) :=γ(b + a− t). The image of γ− is Γ, except that the points γ−(t) and γ(t)


travel in opposite directions. Thus γ− “reverses” the orientation of thecurve. Prove that∫

γ(xdy + ydx) = −

∫γ−

(xdy− ydx).

16.4.2. Isoperimetric inequality.

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