1 Residues - Florida Atlantic Universitymath.fau.edu/schonbek/IntroCpxAn/Residues.pdf · such as...

The Theorem of Residues and Applications

1 Residues

Assume f is holomorphic in a deleted disc of positive radius centered at apoint z0. Precisely, assume f is holomorphic in D′R(z0). 0 < R ≤ ∞ and letf(z) =

∑∞n=−∞ an(z−z0)n be the Laurent expansion of f at z0. The residue of f

at z0 is the coefficient a−1 of the first negative power in this Laurent expansion;precisely

Res(f, z0) = a1.

A note on notation. The online text uses the notation Res(f(z), z = z0) orRes(f(z))z=z0 . I decided against this notation for typographical reasons, its alittle harder to implement in the software I use (Latex). Also the z in thatnotation is redundant.

Why is it called residue? Laurent series can be integrated term by termsover paths inside their region of convergence. So let Cr be a positively orientedcircle of radius r, center z0, 0 < r < R. Then∫

Cr

f(z) dz =

∞∑n=−∞

an

∫Cr

(z − z0)n dz.

Now if n 6= −1, then (z − z0)n =d

dz

(1

n+ 1(z − z0)n+1

)so that

∫Cr

(z − z0)n dz = 0

if n = 0, 1, 2,−2, 3,−3, . . . and all that remains is the term with (z − z0)−1, ifthere is such a term:∫

Cr

f(z) dz = a−1

∫Cr

(z − z0)−1 dz = Res(f, z0)

∫Cr

(z − z0)−1 dz.

We calculated this last integral before, but we will do it again. Parameterizethe circle as usual by z0 + reit, 0 ≤ t ≤ 2π, so dz = rieit dt; then∫

Cr

(z − z0)−1 dz =

∫ 2π

0

r−1e−itrieit dt = i

∫ 2π

0

dt = 2πi.

We thus have (let’s call it a lemma; why not?)

Lemma 1 Assume f is holomorphic in D′R(z0), 0 < R ≤ ∞. If o < r < R andCr denotes the positively oriented circle of radius r centered at z0, then∫

Cr

f(z) dz = 2πiRes(f, z0).

2 THE THEOREM OF RESIDUES 2

This is the reason for the name residue, up to a factor of 2πi it is all that re-mains when integrating a function around an isolated singularity. This Lemma,of course, implies a very particular case of Cauchy’s Theorem if the functionhappens to be holomorphic at z0; then a−1 = 0; that is, the residue is 0, so isthe integral.

2 The Theorem of Residues

Without any more ado, here it is. Some terms will be explained (or explainedagain) after the statement.

Theorem 2 Let f be holomorphic in the open set U except (possibly) for iso-lated singularities. Let γ be a closed, positively oriented, closed simple path inU such that the inside of γ is also in U . Assume no singularity of f is on γ.Let z1, z2, . . . , zn be the singularities inside γ. Then∫

γ

f(z) dz = 2πi

n∑j=1

Res(f, zj).

Remarks. The assumption is that the singularities are isolated. If at any pointthere is an accumulation point of singularities, that point would be a very bad,non isolated singularity, so we assume no such point exists in U . With thatone can show that in any bounded set that does not touch the boundary of U ,such as the inside of a simple closed curve, there can only be a finite number ofsingularities.

A closed simple curve, also called a Jordan curve, is a non self intersectingcurve. It can be parameterized by a continuous map γ : [a, b] → C such thatγ(t) 6= γ(s) if a ≤ s < t < b (i.e., γ(s) = γ(t) assuming s < t if and onlyif s = a and t = b). The famous Jordan Curve Theorem states that such acurve divides the plane into two connected regions; specifically, the complementof such a curve in the plane consists of two disjoint connected open sets, onebounded (called the inside of the curve) and one unbounded. We assume ourcurve is piecewise smooth, thus we call it a path.

A simple closed curve γ : [a, b] → C is positively oriented if as one movesalong the curve when the parameter goes from a to b, the interior is alwayson the left of where one is. For a circle, positive orientation is also calledcounterclockwise orientation, a definition that may become incomprehensible asall analog clocks get replaced by digital ones.

The Theorem of Residues is of course valid also if f has no singularities inU . Then it reduces to Cauchy’s Theorem. In fact, the Theorem of Residuesgeneralizes both Cauchy’s Theorem and Cauchy’s formula.End of the Remarks.Proving the theorem. (Sort of) Suppose we have a situation as illustrated inthe figure below. The region U is the yellow region, we have a simple, positivelyoriented closed path γ and singularities z1, z2, . . . , zn inside γ. In the picture the


singularities in U are et the points z1, . . . , z10, but only z1, . . . , z5 are inside thepath; the others play no role in the integration process for the pictured path.

We can now draw a small circle around each singularity and give each oneof them a negative orientation (clockwise). Let Cj be the negatively orientedcircle around zj .

(As you can see the graphics software I have at my disposal is rather primi-tive.)

Next starting at any point P of the path γ we draw a polygonal line tothe circle around z1, from this circle a polygonal line to the circle around thesingularity z2, and so forth until we get to the circle around zn. These polygonallines are just segments in blue in the picture below.


We can now think of γ and the circles and the polygonals as forming one bigclosed curve that starts and ends at P . We start at P and get into the inside ofγ by the first polygonal, go around the part of C1 from where the first polygonalgoes in to where the second polygonal goes out. Follow the second polygonalto C2, and so forth until we get to Cn. We go completely around the last littlecircle, leave it where we came in and return to P going along the polygonallines and along the parts of the little circles not traversed yet. Back at P we goaround γ until we end again at P . Another way is to think of the polygonalsas double lines separated by a distance ε > 0, the circles having arcs cut out ofwidth ε, so that a detail of the picture showing two of the little circles reallylooks more or less as shown:

The point P on γ is where the first (upper) polygonal line goes in. Goingthrough the upper part of the circles and the upper polygonal lines first, thenaround the lower parts, then around γ beginning at where the last lower polyg-onal line goes out, finally ending again at P , we get a closed simple path thatcan be included in a simply connected region that avoids all the singularities off . By Cauchy’s Theorem, the integral over this curve is 0. Letting ε → 0 theintegrals over the polygonals cancel and we are left with∫

γ

f(z) dz +

n∑j=1

∫Cj

f(z) dz = 0. (1)

On the other hand, in the first way of looking at it, the path as pictured, inwhich one goes through the polygonals twice in opposite directions (so that the


integrals of f over these polygonals cancel out) is a closed path albeit not asimple path. But a path that has the property that one can shrink it to a pointwithout touching the singularities of f . By Cauchy’s Theorem, as the onlinetext has it, the integral over this path is zero, so we obtain (1) directly.

Appealing now to Lemma 1, since the circles C1, . . . , Cn are traversed in anegative direction, we get from equation (1) that∫

γ

f(z) dz − 2πi

n∑j=1

Res(f, zj) = 0

proving the theorem of residues.The theorem of residues is an extremely powerful integration rule, as I hope

we shall see. There are a few exercises of integration at the end of Chapter 9of the online text, but these are not the more interesting integrals that can beevaluated by residues. However, let’s do one for practice. Let’s do, for example,Exercise 9.8(e), compute ∫

C[0,3]

ez

(z + 2)2 sin zdz.

Computing this integral by the definition would be close to impossible. I am notsaying impossible because impossible is very difficult to prove; but I think thatthe only way someone could do it is by discovering the theorem of residues in theprocess. By residues, we ask first what are the singularities of the holomorphicintegrand inside the path of integration. There are poles at the zeros of sin zand of (z + 2). Of the zeros of sin z, only the zero at 0 is inside the circle (thenext two are at ±2π and 2π > 3. The zero of z + 2, namely −2 is inside. Thus∫C[0,3]

ez

(z + 2)2 sin zdz = 2πi

(Res(

ez

(z + 2)2 sin z, 0) +Res(

ez

(z + 2)2 sin z,−2)

).

The pole at 0 is of order 1, since the denominator of the fraction is not 0 at 0,and sin z has a zero of order 1 at 0. Thus

Res(ez

(z + 2)2 sin z, 0) = lim

z→0

zez

(z + 2)2 sin z= limz→0

ez

(z + 2)2· limz→0

z

sin z=

1

4·1 =

1

4.

The pole at -2 is of order 2. Higher order poles make the residues harder tocompute since derivatives are involved and in many cases functions get messierand messier on differentiation. If the singularity is essential, things can be evenworse; the best way to get the value of the residue could well be figuring outthe Laurent expansion of the function. Let us recall the formula for computinga residue if the pole is of order k. If f has a pole of order k at z0 then

Res(f, z0) =1

(k − 1)!

dk−1

dzk((z − z0)kf(z)

)∣∣∣∣z=z0

.

3 EVALUATION OF DEFINITE INTEGRALS 6

Frequently, evaluating the derivative above at z0 consists in finding the limit asz → z0 of the derivative away from z0. In our case, with the pole being of orderk = 2, we get

Res(ez

(z + 2)2 sin z,−2) =

d

dz

(ez

sin z

)∣∣∣∣−2

=ez(sin z + cos z)

sin2 z

∣∣∣∣−2

=e−2(cos 2− sin 2)

sin2 2

(using the fact that cos is even and sin is odd). Thus∫C[0,3]

ez

(z + 2)2 sin zdz = 2πi

(1

4+e−2(cos 2− sin 2)

sin2 2

).

3 Evaluation of Definite Integrals

A number of integrals coming up in applications bowed their at first proud headsto the theorem of residues. In this section I present a very reduced list of suchintegrals and the needed techniques, the list that can be found in most textbooksbut that our online text seems to ignore. As many things in mathematics,using residues to evaluate integrals gets better with practice.

3.1 Rational functions of sine and cosine

These are integrals of the form∫ 2π

0

R(cos θ, sin θ) dθ,

whereR(x, y) is a rational function (quotient of polynomials) in x, y. That is, theintegrand is a quotient of two functions, both the numerator and denominatorcontain only a combination of products (including powers) of sin θ, cos θ. If thepowers are high things can get very hairy, so the example I will demonstrate willcontain very low powers of these trigonometric functions. The integral goes from0 to 2π; it could go over any interval of length 2π. Frequently (sometimes?) onecan also evaluate integrals in which the path of integration is an integer fractionof 2π. The technique consists in transforming this integral to an integral overthe unit circle which would be parameterized by z = eiθ. Noticing that for|z| = 1 one has z̄ = 1/z, one makes the following replacements in the integrand:

cos θ =1

2

(z +

1

z

)=z2 + 1

2z,

sin θ =1

2i

(z − 1

z

)=z2 − 1

2iz,

dθ =dz

iz.


With these substitutions R(cos θ, sin θ) dθ =P (z)

Q(z)dz, where P,Q are polynomi-

als and ∫ 2π

0

R(cos θ, sin θ) dθ =

∫|z|=1

P (z)

Q(z)dz.

And now one applies residues.Here are a several examples.

• Example 1. Compute ∫ 2π

0

1 + cos θ

2− sin θdθ.

Isn’t this exciting? Have you ever calculated as complicated an integralas this one? With the indicated changes∫ 2π

0

1 + cos θ

2− sin θdθ =

∫C[0,1]

1 + z2+12z

2− z2−12iz

dz

iz

=

∫C[0,1]

z2 + 2z + 1

z(4iz − z2 + 1)dz

The denominator has a simple zero at z = 0, thus the integrand has asimple pole at 0. Applying the quadratic formula, we see the denominatorhas two further simple zeros at z = (2 ±

√3)i so that the denominator

factors as

z(4iz − z2 + 1) = −z(z2 − 4iz − 1) = −z(z − (2 +√

3)i)(z − (2−√

3)i).

Of the zeros of the quadratic factor only (2−√

3)i is inside the unit circle,so it is the only one we need to consider. So the only singularities withinthe unit circle are the simple poles at 0 and at (2−

√3)i. Let us calculate

the residues at these poles.

Res

(z2 + 2z + 1

z(4iz − z2 + 1), 0

)= lim

z→0

z(z2 + 2z + 1)

z(4iz − z2 + 1)

= lim +z → 0z2 + 2z + 1

4iz − z2 + 1= 1,

Res

(z2 + 2z + 1

z(4iz − z2 + 1), (2−

√3)i

)= − lim

z→(2−√

3)i

(z − (2−√

3)i)(z2 + 2z + 1)

z(z − (2 +√

3)i)(z − (2−√

3)i)

= − limz→(2−

√3)i

z2 + 2z + 1

z(z − (2 +√

3)i)

= −1−√

3

3i.


You might want to check these computations. Assuming they are right∗

we get that the sum of residues is

1 +

(−1−

√3

3i

)= −√

3

3i,

so that ∫ 2π

0

1 + cos θ

2− sin θdθ = 2πi

(−√

3

3i

)=

2π√

3

3.

Note When computing a definite integral there are some minimum checksone should run on the final result to avoid really stupid mistakes. Forexample, in computing this integral we went into the complex field butthe original integrand is real so the integral has to be real. For example iffor some reason we had forgotten to use both residues, or had one of theresidues wrong, we could have ended with a complex, non real number.If our final result contained an i we know that we made a mistake. Theintegrand is also never negative, so the integral has to be positive. Theresult we obtained passes the minimum tests. It could still be wrong†,but at least not stupidly wrong, so wrong that even Martians looking atit from Mars can see that it is wrong.

• Example 2. The following integral plays a big role in harmonic analysis.Evaluate ∫ 2π

0

1

1− 2r cos θ + r2dθ,

where 0 ≤ r < 1. This time I will go much faster. Can you follow it? Ifyes, good for you! If no, why not? Go back and reread the instructions.Notice that because 0 < r < 1, we have z = 1/r outside of the unit circle.∫ 2π

0

1

1− 2r cos θ + r2dθ = −1

i

∫C[0,1]

1

rz2 − (1 + r2)z + rdz

= − 1

ir

∫1

(z − r)(z − 1r )dz

= −2π

rRes(

1

(z − r)(z − 1r ), r) = − 2π

r2 − 1=

2π

1− r2,

• Example 3. Some integrals over intervals that are a fraction of an intervalof length 2π can be transformed into an integral over an interval of length2π by symmetry considerations, and thus integrated by residues. Forexample, compute ∫ π

0

(1 + cos θ) dθ.

∗They are.†It isn’t.


Of course nobody in her or his right mind would use residues here, butI’ll do it just for illustration purposes. Because the integrand is even, wehave∫ π

0

(1 + cos θ) dθ =1

2

∫ π

−π(1 + cos θ) dθ =

1

2i

∫C[0,1]

(z + 1)2

z2dz

=1

4i

∫C[0,1]

(1 +

2

z+

1

z2

)dz.

The integrand has a pole of order 2 at 0; the residue is obviously 2 so that∫ π

0

(1 + cos θ) dθ1

4i· 4πi = π.

• Example 4. In general, if there is a pole of higher order things are a littleharder. If the order is high enough they can get very hard. The followingexample is a bit artificial, but it should show what to do. We will evaluate∫ π

−π

(cos θ − i sin θ)2 cos2 θ

(2− cos θ + i sin θ)2δθ.

Looks sort of scary, but it really isn’t. One simply has to be carefulwhen operating; one has to pay attention to what one does and checkoccasionally one’s calculations. If one substitutes cos θ, sin θ, dθ by theirexpressions in terms of z one gets∫ π

−π


(2− cos θ + i sin θ)2δθ =

1

4i

∫C[01,]

(z2 + 1)2

z3(2z − 1)2dz.

The integrand has a pole of order 3 at 0 and one of order 2 at 1/2. Let us

denote the integrand by f(z), so f(z) =(z2 + 1)2

z3(2z − 1)2.

– Computing the residue at 0. Since the pole is of order 3, we will use

Res(f, 0) =1

2

d2

dz2(z3f(z))

∣∣∣∣0

.

Now finding the second derivative of this function gets quite messy.It might be a good exercise to find it on your own, so I only give herethe result of evaluating it at 0. It works out to 28. Dividing by 2 weget

Res(f, 0) = 14.

The pole at 1/2 is of order 2 so that

Res(f,1

2) =

d

dz((z − 1

2)2f(z))

∣∣∣∣12

=d

dz

z2 + 1

4z3

∣∣∣∣12

= −z2 + 3

z4

∣∣∣∣12

= −13.


The sum of residues is thus 14 − 13 = 1 and our integral equals(remember the factor of 1/4i in front of the complex integral)∫ π

−π


(2− cos θ + i sin θ)2δθ =

1

4i2πi =

π

2.

Exercise 1 Compute the following integral.∫ π

0

1

(5 + 3 cos θ)2dθ

Notice the integrand is even.

3.2 Integrals from −∞ to ∞ of rational functions

One of the simplest applications of residues is to compute∫ ∞∞

P (x)

Q(x)dx

where P,Q are polynomials in x and the degree of Q is greater then or equalthan the degree of P plus 2; degQ ≥ degP + 2, and Q has no real zeroes.The last condition can be slightly lessened; simple zeros on the real axis canbe treated. Under these circumstances, let z1, . . . , zn be the zeros of Q in theupper half plane. Then∫ ∞

∞

P (x)

Q(x)dx = 2πi

n∑j=1

Res

(P (z)

Q(z), zj

). (2)

Here are some examples, before we explain why it works.

• Compute ∫ ∞−∞

1

x2 + 1dx.

The denominator of the integrand has simple zeros at ±i; the one in theupper half plane is i, thus∫ ∞

−∞

1

x2 + 1dx = 2πiRes

(1

z2 + 1, i

)= 2πi lim

z→i

z − iz2 + 1

= 2πi1

2i= π.

• Compute ∫ ∞−∞

x4

(x4 + 1)(x2 + 4)2dx.

The denominator of the integrand has simple zeros at ±1 + i√2,±1− i√

2and

double poles at ±2i. The ones in the upper plane are

1 + i√2,−1 + i√

2, 2i.


The residues are:

Res

(z4

(z4 + 1)(z2 + 4)2,

1 + i√2

)=

√2(23 + 7i)

2312.

Res

(z4

(z4 + 1)(z2 + 4)2,−1 + i√

2

)= −√

2(23− 7i)

2312.

Res

(z4

(z4 + 1)(z2 + 4)2,−1 + i√

2

)= − 13

578i.

Nobody says that things have to be always pretty!

Exercise 2 Compute these residues, showing work, to verify that they arecomputed correctly.

The sum of residues works out to (7√

2− 26)i/1156, thus multiplying by2πi ∫ ∞

−∞

x4

(x4 + 1)(x2 + 4)2dx =

(26− 7√

2)π

578.

Exercise 3 Evaluate ∫ ∞−∞

x2 + 1

x2 + x+ 4dx.

Exercise 4 Evaluate ∫ ∞0

1

x4 + 16dx.

(Use the fact that the integrand is even.)

Let us see now why (2) works. The proof of the formula is based onthree facts that are satisfied by P (z)/Q(z) if P,Q are polynomials withQ(x) 6= 0 for x real and deg ≥ degP + 2:

Fact 1. P (z)/Q(z) is holomorphic at all points of the real line.

Fact 2. P (z)/Q(z) has only a finite number of singularities in the upperhalf plane.

Fact 3. |P (z)/Q(z)| ≤ C/|z|2 for large values of |z|, some constant C sothat one has (and this is really all one needs; actually more than oneneeds): If for R > 0

MR = max|z|=R,Im (R)>0

∣∣∣∣ P (z)

Q(Z)

∣∣∣∣ ,then limR→∞RMR = 0.


Any other function that is holomorphic in an open set containing theupper half plane and the real line except for isolated singularities, andsatisfies these three facts could be integrated over the real line by thesame formula.

Let R > 0 and consider the closed path ΓR consisting of the line segmentfrom −R to R, followed by the half circle of center 0, radius R in the upperhalf plane. Let us write ΓR = λR + γR, where

λR(t) = t, −R ≤ t ≤ R

is the line segment and

γR(t) = Reit, 0 ≤ t ≤ π

is the half circle.

If we take R large enough all the poles of the integrand will be inside thesimple closed path ΓR so that∫

ΓR

P (z)

Q(z)dz = 2πi

n∑j=1

Res(p(z)/Q(z), zj) (3)

where z1, . . . , zj are the zeros ofQ in the upper half plane. Since∫λRf(z) dz =∫ R

−R f(t)dt, we may write (3 in the form∫ R

−R

P (x)

Q(x)dx+

∫γR

P (z)

Q(z)dz = 2πi

n∑j=1

Res(p(z)/Q(z), zj). (4)

With MR as defined above we have that∣∣∣∣∫γR

P (z)

Q(z)dz

∣∣∣∣ ≤MRΛ(γR) = πRMR

so that by Fact 3,

limR→∞

∫γR

P (z)

Q(z)dz = 0.


Letting R → ∞ in (4), since the right hand side does not depend on Ronce all the poles are within the half disc, we get∫ ∞

−∞

P (x)

Q(x)dx = 2πi

n∑j=1

Res(p(z)/Q(z), zj).

as claimed.

It should be mentioned that there is nothing special about the upper halfplane, one might as well have used the lower half plane integrating over the linefrom −R to R and then over the half circle from −R to R in the lower halfplane. For example, suppose we want to evaluate∫ ∞

−∞

1

(x− i)2(x+ i)dx.

The integrand has a double pole in the upper half plane, a simple pole in thelower half plane; we may decide to use the lower half plane. There is, however,a proviso. The path we will be following, pictured below, is negatively oriented.

The formula for a rational function P/Q if degQ ≥ degP + 2, Q(x) 6= 0 if x isreal, and if z1, . . . , zm are the poles in the lower half plane. is∫ ∞

−∞

P (x)

Q(x)dx = −2πi

m∑k=1

Res(P/Q, zk).

Returning to our integral,∫ ∞−∞

1

(x− i)2(x+ i)dx = −2πiRes(

1

(z − i)2(z + i),−i)

= −2πi limz→−i

(z + i)

(z − i)2(z + i)= −2πi · 1

−4=πi

2.

If we had integrated over the upper hallf plane the result would have been∫ ∞−∞

1

(x− i)2(x+ i)dx = 2πiRes(

1

(z − i)2(z + i), i)

= 2πid

dz

((z − i)2

(z − i)2(z + i)

)∣∣∣∣i

= 2πid

dz

(1

(z + i)

)∣∣∣∣i

= − 2πi(z + i)−2∣∣i

=πi

2.


Exercise 5 Assume f has a simple pole at x0 ∈ R, For ε > 0 let γε be the halfcircle in the upper halfplane from x0 + ε to x0 − ε. Show that

limε→0

∫γε

f(z) dz = πiRes(f, x0).

Hints: You can write f(z) =a

z+g(z), where g is analytic in some disc centered

at x0 (and discs are about as simply connected as they get) and a = Res(f, z0).Using the fact that g has an antiderivative in (at least) some disc around x0 (why

does it have an antiderivative?) it should be easy to prove that

∫γε

g(z) dz → 0

as ε → 0. The integral of a/z overthe curve should be easy to calculate; justparameterize and compute it, and let ε→ 0.

This exercise allows for some calculations of improper integrals. For example,let us compute ∫ ∞

−∞

1

x3 + 1dx.

The integrand has simple poles at −1, (1 + i√

3)/2, (1 − i√

3)/2. One of thepoles is on the real axis. If you consult your Calculus texts you will learn thatan integral where the integrand has a problem at a point x0 on the axis mayexist in the improper sense. That is, suppose f has a discontinuity at x0 anda < x0 < b. Then∫ b

a

f(x) dx = limε→0+

∫ x0−ε

a

f(x) dx+ limδ→0+

∫ b

x0+δ

f(x) dx,

and the two limits must exist independently of each other (and be finite). In thissense, the integral we are planning to compute does not exist. But existence is arelative concept in mathematics, occasionally it is convenient to relax conditionsa bit. Consider, for example ∫ 4

−3

1

xdx.

We have ∫ −ε−3

1

xdx = log |x|

∣∣∣∣−ε−3

= log ε− log 3→∞ as ε→ 0+,∫ 4

δ

1

xdx = log |x|

∣∣∣∣4δ

= log 4− log δ → −∞ as δ → 0 + .

The integral does not exist in the Calculus sense. But it so happens the infinitiescoming from the left and from the right balance each other perfectly. If we letδ = ε so that both go to 0 at the same rate we get∫ 4

−3

1

xdx “ =′′ lim

ε→0+

∫ −ε−3

1

xdx+ lim

ε→0+

∫ 4

ε

1

xdx

= limε→0+

(log ε− log 3 + log 4− log ε) = log 4 = log 3 = log(4/3).


I put the first equal sign above in quotation marks because one does not considerit really a full blown equality. One says the integral exists in the principal valuesense and sometimes will write

p.v.

∫ 4

−3

1

xdx = log(4/3).

Because the pole on the real line of 1/(z3 + 1) is simple, the integral exists in aprincipal value sense. What I am about to do here should NOT be attemptedwith poles on the real line of order 2 or higher. I will compute the integral andthen state the formula in a theorem summarizing the results of this subsection.We will integrate the mostly holomorphic function f(z) = 1/(z3 + 1) over aclosed path

ΓR,ε = λR,ε + γε + λ̃R,ε + γR,

where 0 < ε < R and

λR,ε(t) = t, −R ≤ t ≤ −1− ε, (interval [−R,−1− ε],)

γε(t) = −1 + εei(π−t) 0 ≤ t ≤ π (halfcircle −1− ε to 1 + ε,

clockwise),

λ̃R,ε(t) = t, −1 + ε ≤ t ≤ R, (interval [−1 + ε, R],)

γR(t) = Reit 0 ≤ t ≤ π (halfcircle R to −R,counterclockwise).

The parameterizations play no big role, we don’t compute integrals anymore byparameterizing. The only relevant facts are that

limR→∞

∫γR

1

z3 + 1dz = 0, (5)

limR→∞

∫λR,ε

=

∫ −1−ε

−∞

1

x3 + 1dx, (6)

limR→∞

∫λ̃R,ε

=

∫ ∞−1+ε

1

x3 + 1dx, (7)

limε→0

∫γe

1

z3 + 1dz = −πiRes

(1

z3 + 1,−1

). (8)

The reason why (5) holds is the same as before, because the degree of thedenominator is 2 or more above the degree of the numerator; (6) and (7) hold,I hope, for obvious reasons. That is, I hope the reasons for their holding areobvious. Concerning (8), it is due to Exercise 5; the minus being due to thepath going in a clockwise direction. If we take R large enough (in our case largerthan 1 will do) all the poles in the upper half plane of the integrand are in theinside of the curve ΓR and we can apply the theorem of residues. Notice that


the pole at −1 is outside of this path. We thus have∫ΓR

1

z3 + 1dz = 2πiRes

(1

z3 + 1,

1 + i√

3

2

)

= 2πi limz→ 1+i

√3

2

z + 1−i√

32

(z − 1+i√

32 )(z − 1−i

√3

2 )(z + 1)

=4πi

i√

3(3 + 3i√

3)=π(√

3− i)3

.

If we let R→∞ and use (5), (6) and (7) we get∫ −1−ε

−∞

1

x3 + 1dx+

∫ ∞−1+ε

1

x3 + 1dx+

∫γε

1

z3 + 1dz =

π(√

3− i)3

Letting ε→ 0 and using (8) we get that

p.v.

∫ ∞−∞

1

x3 + 1dx− πiRes

(1

z3 + 1,−1

)=π(√

3− i)3

. (9)

Now

Res

(1

z3 + 1,−1

)= limz→−1

z + 1

z3 + 1=

1

3,

so that (9) becomes

p.v.

∫ ∞−∞

1

x3 + 1dx− πi

3=π(√

3− i)3

.

so that finally we obtain

p.v.

∫ ∞−∞

1

x3 + 1dx =

π(√

3)

3.

Here is the promised theorem.

Theorem 3 Assume P,Q are polynomials and assume that degQ ≥ degP + 2.Assume Q only has simple zeros (or no zeros at all) on the real axis. Letz1, . . . , zn be all the zeros of Q inthe upper halfplane and let x1, . . . , xm be thezeros of Q on the real axis. Then

p.v.

∫ ∞−∞

P (x)

Q(x)dx = 2πi

n∑j=1

Res(P/Q, zj) + πi

m∑k=1

Res(P/Q, xk).

If Q has no zeros on the x-axis, then the second sum above is void and theintegral exists in the conventional sense.


3.3 Integrals of the type (P (z)eαz)/Q(z), and related ones

Most of the techniques of the previous subsection apply in a much more generalway. For example, assume that f(z) is analytic at least in an open set containingthe upper half plane and the real axis, except for a finite number of singularitiesin the upper halfplane. If

limR→∞

∫γR

f(z) dz = 0, (10)

where γR is the half circle centered at 0 in the upper half plane from R to −R,then ∫ ∞

−∞f(z) dz =

n∑j=1

Res(f, zj) (11)

where z1, . . . , zn are the residues of f in the upper half plane.

Let us use this to try to compute∫ ∞−∞

cos ax

x2 + 1dx

where a > 0. We could think of doing the obvious, try to apply the theoremwith f(z) = cos az/(z2 + 1). This would result in a massive failure becausecos az grows exponentially in the upper half plane and (10) is absolutely false.We resort to a common trick. We will use that∫ ∞

−∞

cos ax

x2 + 1dx = Re

∫ ∞−∞

eiax

x2 + 1dx

and calculate ∫ ∞−∞

eiax

x2 + 1dx

first. Notice that if z = x+ iy is in the upper half plane so y > 0, then

|eiaz| = |eiax−ay| = e−ay ≤ 1

and (10) holds for the same reason it held for plain old 1/(z2 + 1). Thus∫ ∞−∞

eiax

x2 + 1dx = 2πiRes

(eiaz

z2 + 1, i

)= lim

z→i

(z − i)eiaz

z2 + 1= 2πi

e−a

2i= πe−a.

Taking real parts, we get ∫ ∞−∞

cos ax

x2 + 1dx = πe−a.


If we had taken imaginary parts we would get∫ ∞−∞

sin ax

x2 + 1dx = 0,

a fact that is sort of obvious since the integrand is odd and the interval ofintegration is symmetric about 0. Here is another example, done a bit faster.We want to compute ∫ ∞

0

x sin 4x

(x2 + 4)2dx.

Because the integrand is even, we have∫ ∞0

x sin 4x

(x2 + 4)2dx =

1

2

∫ ∞−∞

x sin 4x

(x2 + 4)2dx =

1

2Im

∫ ∞−∞

xe4ix

(x2 + 4)2dx.

Now∫ ∞−∞

xe4ix

(x2 + 4)2dx = 2πiRes

(ze4iz

(z2 + 4)2, 2i

)=

d

dz

(z − 2i)2e4iz

(z2 + 4)2

∣∣∣∣2i

=d

dz

e4iz

(z − 2i)2

∣∣∣∣2i

=πe−8(8 + i)

16.

Taking imaginary parts and dividing by 2,∫ ∞0

x sin 4x

(x2 + 4)2dx =

πe−8

32.

These are just a few, very few, of the applications of the theorem of residues.But our residue story will probably have to end here.

1 Residues - Florida Atlantic Universitymath.fau.edu/schonbek/IntroCpxAn/Residues.pdf · such as...

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Transcript of 1 Residues - Florida Atlantic Universitymath.fau.edu/schonbek/IntroCpxAn/Residues.pdf · such as...