Asymptotic Expansions for Integrals

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Gergo Nemes

Asymptotic Expansionsfor Integrals

M. Sc. Thesis

Lorand Eotvos University

May 23, 2012


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Lorand Eotvos University

Faculty of Science

M. Sc. Thesis

Asymptotic Expansionsfor Integrals

Author:Gergo NemesM. Sc. Pure Mathematics

Supervisor:Dr. Arpad Toth

associate professor

Examination Committee:

Committee Chairman: Prof. Dr. Laszlo Simon . . . . . . . . . . . . . . . . . . . . . . . .

Committee Members: Dr. Balazs Csikos . . . . . . . . . . . . . . . . . . . . . . . .

Dr. Alice Fialowski . . . . . . . . . . . . . . . . . . . . . . . .

Prof. Dr. Tibor Jordan . . . . . . . . . . . . . . . . . . . . . . . .

Dr. Tamas Mori . . . . . . . . . . . . . . . . . . . . . . . .

Dr. Peter Sziklai . . . . . . . . . . . . . . . . . . . . . . . .

Dr. Janos Toth . . . . . . . . . . . . . . . . . . . . . . . .

May 23, 2012


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Preface

The asymptotic theory of integrals is an important subject of applied mathematicsand physics. Although it is an old topic, with origins dating back to Laplace, newmethods, applications and formulations continue to appear in the literature. Thepurpose of this thesis is to review some of the classical procedures for obtainingasymptotic expansions of integrals, especially focusing on Laplace-type integrals.As a contribution to the topic, we give a new method for computing the coefficientsof these asymptotic series with several illustrating examples. This method is ageneralization of the one given in my paper about the Stirling Coefficients ( J.Integer Seqs. 13 (6), Article 10.6.6, pp. 5 (2010)).

Chapter 1 covers the asymptotic theory of real Laplace-type integrals. Thecomputation of the coefficients appearing in the asymptotic expansions are de-scribed completely in this chapter. In Chapter 2, we discuss two methods fromthe asymptotic theory of complex Laplace-type integrals: the Method of SteepestDescents and Perrons Method. Each chapter contains examples to demonstratethe application of the results. In Appendix A, we collect the basic properties ofsome combinatorial quantities we use in this thesis.

Budapest, Hungary, May 23, 2012 Gerg o Nemes

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Table of Contents

Preface i

Table of Contents ii

Acknowledgements iii

1 Real Laplace-type integrals 11.1 Fundamental concepts and results . . . . . . . . . . . . . . . . . . . 21.1.1 Asymptotic expansions . . . . . . . . . . . . . . . . . . . . . 21.1.2 Watsons Lemma . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Laplaces Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.1 Asymptotic expansion of Laplace-type integrals . . . . . . . 51.2.2 Generalization of Perrons Formula . . . . . . . . . . . . . . 71.2.3 The CFWW Formula . . . . . . . . . . . . . . . . . . . . . . 81.2.4 Another method . . . . . . . . . . . . . . . . . . . . . . . . 91.2.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Complex Laplace-type integrals 212.1 The Method of Steepest Descents . . . . . . . . . . . . . . . . . . . 212.2 Perrons Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

A Combinatorial objects 33

A.1 Ordinary Potential Polynomials . . . . . . . . . . . . . . . . . . . . 33A.2 The r-associated Stirling Numbers . . . . . . . . . . . . . . . . . . . 34

Bibliography 35

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Acknowledgements

I am deeply grateful to my supervisor, Professor Arpad Toth for his detailed andconstructive comments, and for his important support throughout this work. I owemy loving thanks to Dorottya Sziraki. Without her support it would have beenimpossible for me to finish this work. My special gratitude is due to my parentsand my brother for their loving support.

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Chapter 1

Real Laplace-type integrals

In this chapter, we investigate the asymptotic behavior of certain parametric inte-grals. The origins of the method date back to Pierre-Simon de Laplace (1749 1827), who studied the estimation of integrals arising in probability theory of

the formIn =

ba

enf(x)g (x) dx (n +) . (1.1)

Here the functions f and g are real continuous functions defined on the real (finiteor infinite) interval [a, b]. Hereinafter, we call integrals of the type (1.1), Laplace-type integrals. Laplace made the observation that the major contribution to theintegral In should come from the neighborhood of the point where f attains itssmallest value. If f has its minimum value only at the point x0 in (a, b) wheref (x0) = 0 and f (x0) > 0, then Laplaces result is

ba

enf(x)g (x) dx g (x0) enf(x0) 2nf (x0)

.

The sign is used to mean that the quotient of the left-hand side by the right-hand side approaches 1 as n +. This formula is now known as Laplacesapproximation. A heuristic proof of this formula may proceed as follows. First,we replace f and g by the leading terms in their Taylor series expansions aroundx = x0, and then extend the integration limits to and +. Hence,

b

a

enf(x)g (x) dx

b

a

en

f(x0)+

f(x0)2 (xx0)2

g (x0) dx

g (x0) enf(x0)+

enf(x0)

2 (xx0)2dx

= g (x0) enf(x0)

2

nf (x0).

We divided this chapter into two parts. In Section 1.1, we deal with the fundamen-tal theorem of the topic, called Watsons Lemma. Section 1.2 presents a generaltheorem on asymptotic expansions of Laplace-type integrals, and investigates sev-eral of their properties.

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1.1. Fundamental concepts and results

1.1 Fundamental concepts and results

This section is divided into two parts. In the first part, we collect all the basicconcepts which we will use hereinafter. In the second part we state and prove animportant theorem of the topic, known as Watsons Lemma.

1.1.1 Asymptotic expansions

Let f and g be two continuous complex functions defined on a subset H of thecomplex plane. Let z0 be a limit point of H.

Definition 1.1.1. We write f(z) = O(g(z)) (f is big-oh g), as z z0, tomean that there is a constant C > 0 and a neighborhood U of z0 such that |f(z)| C|g(z)| for all z U H.Definition 1.1.2. We writef(z) = o(g(z)) (f is little-ohg), asz z0, to meanthat for every > 0, there exists a neighborhood U of z0 such that

|f(z)

|

|g(z)

|for all z U H.To define the asymptotic expansion of a function we need the concept of an

asymptotic sequence.

Definition 1.1.3. Let {n}n0 be a sequence of continuous complex functionsdefined on a subset H of the complex plane. Let z0 be a limit point of H. Wesay that {n}n0 is an asymptotic sequence as z z0 in H if, for all n 0,n+1(z) = o(n(z)), as z z0.Definition 1.1.4. If

{n

}n

0 is an asymptotic sequence as z

z0, we say that

n=1

ann(z),

where the ans are constants, is an asymptotic expansion of the function f if foreach N 0

f(z) =N

n=1

ann(z) + o(N(z)) as z z0. (1.2)

If a function possesses an asymptotic expansion, we write

f(z) n=1

ann(z) as z z0.

Note that (1.2) may be written as

f(z) =N1n=1

ann(z) + O(N(z)),

which implies that the error is of the same order of magnitude as the first termomitted. Usually, we use

n(z) = zn with 0 < Re (

0) < Re (

1) < Re (

2) < . . .,

as z in some sector of the complex plane.

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1.1. Fundamental concepts and results

1.1.2 Watsons Lemma

Consider the integral

I() =

+0

ex

1 + xdx

where > 0. Using the well-known identity

1

1 + x=

N1k=0

(1)k xk + (1)N xN

1 + x(N 0) ,

term-by-term integration gives

I() =N1k=0

(1)k k!k+1

+ (1)N+0

xN

1 + xexdx,

where(1)N+0

xN

1 + xexdx

=+0

xN

1 + xexdx

+0

xNexdx =N!

N+1.

And hence, by definition,

I() k=0

(1)k k!k+1

, as +.

This is a special case of a much more general result, now known as WatsonsLemma. The theorem is due to George Neville Watson (1886 1965). It is a

significant result in the theory of asymptotic expansions of Laplace-type integrals.In view of its importance, the proof of the result is reproduced below.

Theorem 1.1.1 (Watsons Lemma). Letf(x) be a complex valued function of areal variable x such that

(i) f is continuous on (0, );(ii) as x 0+,

f(x)

k=0akx

k1

with 0 < Re (0) < Re (1) < Re (2) < . . ., limk+Re (k) = +; and(iii) for some fixed c > 0, f(x) = O (ecx) as x +.Then we have +

0

exf(x) dx k=0

ak(k)

k,

as +.

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1.2. Laplaces Method

Proof. By the conditions (i), (ii) and (iii); the integral+0

exf(x) dx

converges for > c. Conditions (ii) and (iii) imply thatf(x) N1k=0

akxk1

KNecx xN1 for x > 0,for every N 0 with some constant KN. Thus we have+0

exf(x) dx N1k=0

ak

+0

exxk1dx

KN+0

e(c)xxN1 dx.

Note that we have for > 0+0

exxk1dx =1

k

+0

ettk1dt =(n)

k.

Hence we have+0

exf(x) dx N1k=0

ak (n)

k

KN+0

e(c)xxN1 dx

= KN (Re (N))

|( c)N| ,

that is +0

exf(x) dx =N1k=0

ak(n)

k+ O

1

N

as +, which proves Watsons Lemma.

1.2 Laplaces Method

In this section, we revisit the classical Laplace Method. In the first subsection, weprove the fundamental theorem on asymptotic expansion of Laplace-type integrals,an extension of the formula mentioned in the introduction of the chapter. Whenthe functions appearing in the integral are holomorphic, the coefficients of theasymptotic expansion can be given explicitly by Perrons Formula. In the secondpart, we extend this formula to the case when the power series of the functions arenot necessarily convergent but asymptotic. In the third subsection, we reproducethe proof of the CFWW Formula which gives another explicit form of the coeffi-cients in the asymptotic expansion. The fourth part of the section is about a newand simpler explicit formula for these coefficients. Finally, in the fifth subsection,we give three examples to demonstrate the application of the results.

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1.2.1 Asymptotic expansion of Laplace-type integrals

Laplaces Method is one of the best-known techniques developing asymptotic ex-pansions for integrals. Here we present an extension of the formula mentioned inthe introduction of this chapter. Consider the integral of the form

I() = ba

ef(x)g (x) dx,

where (a, b) is a real (finite or infinite) interval, is a large positive parameterand the functions f and g are continuous. We observe that by subdividing therange of integration at the minima and maxima of f, and by reversing the sign ofx whenever necessary, we may assume, without loss of generality, that f has onlyone minimum in [a, b] which occurs at x = a. Next, we assume that, as x a+,

f(x) f(a) +

k=0ak (x a)k+, (1.3)

and

g (x) k=0

bk (x a)k+1 (1.4)

with > 0, Re () > 0; and that the expansion of f can be term wise differenti-ated, that is,

f (x) k=0

ak (k + ) (x a)k+1 (1.5)

as x

a+. Also, we suppose, without loss of generality, that a0= 0 and b0

= 0.

The following theorem is due to Arthur Erdelyi (1908 1977) (see, e.g., [8, p.85], [14, p. 57]).

Theorem 1.2.1. For the integral

I() =

ba

ef(x)g (x) dx,

we assume that

(i) f(x) > f(a) for allx

(a, b), and for every > 0 the infimum off(x)

f(a)

in [a + , b) is positive;

(ii) f(x) and g(x) are continuous in a neighborhood of x = a, expect possibly ata;

(iii) the expansions (1.3), (1.4) and (1.5) hold; and

(iv) the integral I() converges absolutely for all sufficiently large .

Then

I() ef(a)

n=0 n +

cn

(n+)/, (1.6)

as +, where the coefficients cn are expressible in terms of an and bn.

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The first three coefficients cn are given explicitly by

c0 =b0

a/0

, c1 =1

a(+1)/0

b1

(+ 1) a1b02a0

,

and

c2 =1

a(+2)/0

b2

(+ 2) a1b12a0

+

(+ + 2) a21 2a0a2 (+ 1) b0

22a20

.

Proof. By conditions (ii) and (iii) there exists a number c (a, b) such thatf(x) and g(x) are continuous in (a, c], and f(x) is also positive there. Let T =f(c) f(a), and introduce the variable t as

t = f(x) f(a).

Since f(x) is increasing in (a, c), we can write

ef(a)ca

ef(x)g (x) dx =T0

eth (t) dt (1.7)

with h(t) being the continuous function in (0, T] given by

h(t) = g(x)dx

dt=

g(x)

f(x). (1.8)

By assumption

t

k=0 ak (x a)

k+

as x a+

,

and thus, by series reversion we obtain

x a k=1

dktk/ as t 0+.

Substituting this into (1.8) yields

h (t)

k=0 ckt(k+)/1

as t 0+, where the coefficients ck are expressible in terms of ak and bk. We nowapply Watsons Lemma to the integral on the right-hand side of (1.7) to obtainc

a

ef(x)g (x) dx ef(a)n=0

n +

cn

(n+)/,

as +. All that remains is to show that the integral on the remaining range(c, b) is negligible. Define

def= inf cx


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This is positive by condition (i). Let 0 be a value of for which I() is absolutelyconvergent. Assume that 0, then

(f(x) f(a)) = ( 0) (f(x) f(a)) + 0 (f(x) f(a)) ( 0) + 0 (f(x) f(a))

and ef(a) bc

ef(x)g (x) dx Ke,

where

K = e0(+f(a))bc

e0f(x) |g (x)| dxis a constant.

1.2.2 Generalization of Perrons Formula

In this subsection, we derive an explicit formula for the coefficients cn appearingin the asymptotic expansion (1.6). The notations are the same as in the previoussubsection.

Theorem 1.2.2. The coefficients cn appearing in (1.6) are given explicitly by

cn =1

n!

dn

dxn

G (x)

(x a)

f(x) f(a)(n+)/

x=a

=1

n

k=0bnk

k!

dk

dxk

(x a)

f(x)

f(a)

(n+)/x=a

,

(1.9)

where G (x) is the (formal) power series

k=0 bk (x a)k andf(x) should be iden-tified with its (formal) expansion (1.3).

Remark. When (f(x) f(a)) (x a) and g (x) (x a)1 are holomorphicfunctions, the representation (1.9) is known as Perrons Formula [10] [14, p. 103](see also Section 2.2). But here we show that this formula holds also when theexpansions (1.3), (1.4) are merely asymptotic. An alternative proof of essentiallythe same formula has been given previously by Wojdylo [13].

Proof. Let > 0 be the index of the first nonvanishing coefficient in the expansion(1.3) of f apart from a0. Define

f (x)def=

f(x) f(a) a0 (x a)(x a)+

k=0

ak+ (x a)k as x a+.

Then

I() =

ba

ef(x)g (x) dx = ef(a)ba

ea0(xa)

e(xa)+f(x)g (x) h(,x)

dx

= ef(a)ba

ea0(xa)

h (, x) dx

= ef(a)

1/1/(ba)0

ea0th , 1/t + adt.(1.10)

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Set s = 1/t; then the function h

, 1/t + a

reads

h (, s + a) = etsf(s+a)g (s + a) .

We have

etsf(s+a)

k=0

1k!

dkdwk

etwf(w+a)

w=0

sk as s 0+

and hence,

h (, s + a) n=0

n

k=0

bnkk!

dk

dwket

wf(w+a)

w=0

sn+1 as s 0+.

Introducing this expansion in the last integral of (1.10) we find

I() = ef(a)

1/1/(ba)0

ea0th , 1/t + a dt e

f(a)

1/

n=0

n

k=0

bnkk!

dk

dwk

1/(ba)0

e(a0+wf(w+a))tsn+1dt

w=0

= ef(a)n=0

n

k=0

bnkk!

dk

dwk

1/(ba)0

e(a0+wf(w+a))ttn+1dt

w=0

1

(n+)/

ef(a)

n=0

n

k=0bnk

k!

dk

dwk

+0

e(a0+wf(w+a))ttn+1dt

w=0

1

(n+)/

= ef(a)n=0

n+

nk=0

bnkk!

dkdwk

1(a0 + wf (w + a))

(n+)/

w=0

1(n+)/

and using the definition of f (w + a):

I() ef(a)n=0

n+

nk=0

bnkk!

dk

dwk

w

f(w + a) f(a)(n+)/

w=0

1

(n+)/

= ef(a)

n=0

n+

n

k=0bnk

k!

dk

dxk (x a)

f(x)

f(a)

(n+)/x=a

1

(n+)/

as +. Formula (1.9) now follows from the asymptotic expansion (1.6) andthe uniqueness theorem on asymptotic series.

1.2.3 The CFWW Formula

In the previous subsection, we introduced formula (1.9) for the coefficients cnapperaing in the asymptotic expansion of certain Laplace-type integrals. In thissubsection, we shall extract the higher order derivatives in formula (1.9) by means

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of the Partial Ordinary Bell Polynomials (see Appendix A.1). Using the seriesexpansion (1.3) of f(x) we find

cn =1

nk=0

bnkk!

dk

dxk

(x a)

f(x) f(a)(n+)/

x=a

=1

a(n+)/0

nk=0

bnkk!

dkdxk

a0 (x a)f(x) f(a)

(n+)/x=a

=1

a(n+)/0

nk=0

bnkk!

dk

dxk

f(x) f(a)a0 (x a)

(n+)/x=a

=1

a(n+)/0

nk=0

bnkk!

dk

dxk

1 +

k=1

aka0

(x a)k(n+)/

x=a

.

From the definition of the Ordinary Potential Polynomials and the Partial Ordi-

nary Bell Polynomials we obtain

cn =1

a(n+)/0

nk=0

bnkA(n+)/,k

a1a0

,a2a0

, . . . ,aka0

=1

a(n+)/0

nk=0

bnkk

j=0

n+

j

1

aj0Bk,j (a1, a2, . . . , akj+1)

=1

n+

nk=0

bnkk

j=0

(1)ja(n+)/+j0

Bk,j (a1, a2, . . . , akj+1)j!

n +

+ j

.

(1.11)

This is the CampbellFromanWallesWojdylo Formula (from now on the CFWWFormula) [1, 12, 13]. Sometimes it is possible to find the Partial Ordinary BellPolynomials explicitly, in other cases the recurrence (A.4) could be used.

1.2.4 Another method

In the previous subsection, we showed an explicit formula for the coefficients cnappearing in the asymptotic expansion of Laplace-type integrals. This formula,that we called the CFWW Formula, followed from the generalization of PerronsFormula (1.9) using the Partial Ordinary Bell Polynomials. In this subsection,we give a new explicit formula using Lagrange Interpolation and the OrdinaryPotential Polynomials. We write formula (1.9) in the form

cn =1

a(n+)/0

nk=0

bnkk!

dk

dxk

a0 (x a)

f(x) f(a)(n+)/

x=a

. (1.12)

Define the sequence of functions (Fk (z))k0 by the generating function

a0 (x a)

f(x) f(a)z

=

k=0

Fk (z) (x a)k

.

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Now, equation (1.12) can be written as

cn =1

a(n+)/0

nk=0

bnkFk

n +

. (1.13)

Sincea0 (x a)

f(x) f(a)z

=

1 +

k=1

aka0

(x a)kz

=k=0

k

j=0

zj

1

aj0Bk,j (a1, a2, . . . , akj+1)

(x a)k,

the Fk is a polynomial of degree at most k. By Lagrange Interpolation

Fk (z) =

(z+ k + 1)

k!(z)

k

j=0 (1)

j kjFk (j)z+ j , (1.14)where the Fk (j)s can be computed as follows

k=0

Fk (j) (x a)k =

f(x) f(a)a0 (x a)

j=

1 +

k=1

aka0

(x a)kj

=k=0

Aj,k

a1a0

,a2a0

, . . . ,aka0

(x a)k,

that isFk (j) = Aj,k

a1a0

,a2a0

, . . . ,aka0

. (1.15)

From (1.14) and (1.15) we deduce

Fk

n +

=

n+

+ k + 1

k!n+

kj=0

(1)jn+

+ j

k

j

Aj,k

a1a0

,a2a0

, . . . ,aka0

.

Plugging this into (1.13) yields

Theorem 1.2.3. The coefficients cn appearing in (1.6) are given explicitly by

cn =1

n+

nk=0

n+

+ k + 1

bnk

k!a(n+)/0

kj=0

(1)jn+

+ j

k

j

Aj,k

a1a0

,a2a0

, . . . ,aka0

.

(1.16)

As we will see in the following subsection, it is possible, in several cases, toderive an explicit formula for the Ordinary Potential Polynomials due to the some-what simple generating function

1 + k=1

aka0

(x a)kj = f(x) f(a)a0 (x a)

j = k=0

Aj,ka1

a0, a2

a0, . . . , ak

a0 (x a)k,

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whereas in the case of the CFWW formula, the corresponding Partial OrdinaryBell Polynomials have more complicated generating functions, such as

exp

y

k=1

ak (x a)k

= exp

y

f(x) f(a) a0 (x a)(x a)

=k=0

kj=0

Bk,j (a1, a2, . . . , akj+1)j!

yj

(x a)k.

1.2.5 Examples

Example 1.2.1. As a first example we take the Gamma Function

( + 1) =

+0

ettdt (1.17)

for > 0. If we put t = (1 + x), we obtain

( + 1) = +1e+1

e(xlog(1+x))dx

and hence,

()

e=

+0

e(xlog(1+x))dx +10

e(xlog(1x))dx.

Using Theorem 1.2.1 with f(x) = x log (1 + x) (and f(x) = x log (1 x)),g(x) 1, = 2 and = 1; one finds that+

0

e(xlog(1+x))dx n=0

n + 1

2

cn

(n+1)/2

and 10

e(xlog(1x))dx n=0

n + 1

2

(1)n cn(n+1)/2

,

as +, where, by Theorem 1.2.2,

cn =1

2n! dn

dxn x

2

x log (1 + x)(n+1)/2

x=0

.

Finally,

()

21/2en=0

nn

, (1.18)

as +, where

n =

2

n +

1

2

c2n (1.19)

=1

2nn! d2n

dx2n1

2

x2

x log (1 + x)n+1/2

x=0

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are the so-called Stirling Coefficients. The first few are 0 = 1 and

1 =1

12, 2 =

1

288, 3 = 139

51840, 4 = 571

2488320.

We shall give a more explicit formula for the ns using the CFWW Formula (1.11).

Sincef(x) = x log (1 + x) =

k=0

(1)kk + 2

xk+2, g(x) 1,

we have to compute the Partial Ordinary Bell Polynomials Bk,j

1

3, 14

, . . . , (1)kj+1

kj+3

.

We have the exponential generating function

exp

y

k=1

(1)kk + 2

xk

= exp

y

x2

x x

2

2 log (1 + x)

=

k=0

kj=0

Bk,j13 , 14 , . . . , (1)

kj+1

k j + 3 yj

j! xk.

From the generating functions of the Hermite Polynomials and the Stirling Num-bers of the First Kind we have

exp

t

x x

2

2

=

k=0

k/2

j=0

(1)jj! (k 2j)!2j t

kj

xk

and

exp(t log (1 + x)) = k=0

(1)k kj=0

s (k, j) tj xkk!

.

Performing the Cauchy product of the series and rearranging the terms in thecoefficients yields

exp

t

x x

2

2 log (1 + x)

=

k=0k/3

j=0kj

i=0(1)i

2ii!

ji

r=0(1)k+r s (j i r, k 2i r)

r! (k

2i

r)!

tj

xk.

Plugging t = yx2

gives

exp

y

x2

x x

2

2 log (1 + x)

=k=0

k/3

j=0

kji=0

(1)i2ii!

jir=0

(1)k+r s (j i r, k 2i r)r! (k 2i r)!

yj

x2j

xk

=

k=0

k

j=0

j

i=0

(1)i

2ii!

ji

r=0

(1)r s (k + 2j 2i r, j i r)r! (k + 2j 2i r)!

yjxk,

12


18/40


that is

Bk,j

1

3,

1

4, . . . ,

(1)kj+1k j + 3

= j!

j

i=0

(

1)i

2ii!

ji

r=0

(

1)r s (k + 2j

2i

r, j

i

r)

r! (k + 2j 2i r)! .Finally, formula (1.11) produces

cn =n

j=0

2(n1)/2+jn+12

+ j

n+12

ji=0

1

2ii!

jir=0

(1)j+i+r s (n + 2j 2i r, j i r)r! (n + 2j 2i r)! .

From this and the expression (1.19) we deduce

n =2n

j=02n+j

n + j + 1

2

j

i=01

2ii!

ji

r=0(1)j+i+r s (2n + 2j 2i r, j i r)

r! (2n + 2j

2i

r)!

.

(1.20)This is the result of Lopez, Pagola and Sinusa [6].

Now we show how our new method gives a much simpler result than (1.20) withless calculation. To use formula (1.16) we have to compute the Ordinary Potential

Polynomials Aj,k

2

3, 12

, . . . , (1)k 2k+2

. This time the generating function is

1 +k=1

(1)k 2k + 2

xk

j=

2

x log (1 + x)x2

j

=

k=0

Aj,k23 , 12 , . . . , (1)k 2k + 2 xk.

Using the generating function of the Stirling Numbers of the First Kind yields

(x log (1 + x))j =ji=0

j

i

xji ( log (1 + x))i

=

ji=0

j

i

xji

k=0

(1)k i!s (k, i) xk

k!

=

k=0

ji=0

ji (1)k i!s (k, i) xk+jik!

=k=0

j

i=0

j

i

(1)kj+i i!s (k j + i, i)

(k j + i)!

xk,

which gives2

x log (1 + x)x2

j=

k=0

2j

ji=0

j

i

(1)kj+i i! s (k j + i, i)

(k j + i)!

xk2j

=

k=0

2jji=0

ji (1)kj+i i! s (k + j + i, i)(k + j + i)! xk,

13


19/40


that is

Aj,k

2

3,

1

2, . . . , (1)k 2

k + 2

= 2j

ji=0

j

i

(1)kj+i i! s (k + j + i, i)

(k + j + i)!.

Finally, by formula (1.16) we find

cn =3n+12

+ 1

n+12

nj=0

2n/2+j1/2n+12

+ j

(n j)!ji=0

(1)n+i s (n + j + i, i)(j i)! (n + j + i)!

=3n2

+ 32

n+12

nj=0

2n/2+j+1/2

(n + 2j + 1) (n j)!ji=0

(1)n+ji s (n + 2j i, j i)i! (n + 2j i)! .

From this and the expression (1.19) it follows that

n =

2nj=0

2n+j+13n + 32 (2n + 2j + 1) (2n j)!ji=0

(

1)ji s (2n + 2j

i, j

i)

i! (2n + 2j i)!

=3n

j=n

(1)n 2j+13n + 32

(2j + 1) (3n j)!jni=0

(1)ji s (2j i, j n i)i! (2j i)! ,

which is much simpler than (1.20). Remark. The substitution x = log(t/) in (1.17) leads to the form

()

e=

+

0

e(exx1)dx +

+

0

e(ex+x1)dx.

Similar procedures to that described above give (1.18) with

n =1

2nn!

d2n

dx2n

1

2

x2

ex x 1n+1/2

x=0

=2n

j=0

2n+j

n + j + 12

ji=0

1

2ii!

jir=0

(1)j+i+r S(2n + 2j 2i r, j i r)r! (2n + 2j 2i r)!

=

3nj=n

(

1)n 2j+13n + 32 (2j + 1) (3n j)!

jn

i=0

(

1)ji S(2j

i, j

n

i)

i! (2j i)! ,

using formula (1.9), (1.11) and (1.16) respectively. Here S(n, k) denotes the StirlingNumbers of the Second Kind.

Example 1.2.2. Our second example is the family of the Modified Bessel Func-tions of the Second Kind K. Suppose that ,t > 0, then we have the integralrepresentation

K(t) =1

2

+

e(t cosh ss)ds.

14


20/40


The substitution s = sinh1 (1/t) + x gives

K (t) =1

2

t2 + 1 + 1

t

+

e(t2+1coshx+sinhxx)dx

=

1

2

t2 + 1 + 1

t

et2+1 + e(

t2+1(coshx

1)+sinhx

x)

dx,

and by splitting up the integral

2

t2 + 1 + 1

t

et2+1K (t)

=

+0

e(t2+1(coshx1)+sinhxx)dx +

+0

e(t2+1(coshx1)sinhx+x)dx.

Using Theorem 1.2.1 with f(x) =

t2 + 1 (cosh x

1) + sinh x

x (and f(x) =t2 + 1 (cosh x 1) sinh x + x), g(x) 1, = 2 and = 1; one finds that+

0

e(t2+1(coshx1)+sinhxx)dx

n=0

n + 1

2

cn

(n+1)/2

and +0

e(t2+1(coshx1)sinh x+x)dx

n=0

n + 1

2

(1)n cn(n+1)/2

,

as +, where, by Theorem 1.2.2,

cn =1

2n!

dndxn

x2t2 + 1 (cosh x 1) sinh x + x

(n+1)/2x=0

.

Finally,

K(t)

t2 + 1 + 1

t

e

t2+1

2

t2 + 1

n=0

(1)nUn()n

, (1.21)

as +, where

Un() = (1)n2t2 + 1n + 12 c2n (1.22)= (1)n (t

2 + 1)1/4

22n+1/2n!

d2n

dx2n

x2

t2 + 1 (cosh x 1) sinh x + x

n+1/2x=0

=(1)n

22n+1/2n!1/2

d2n

dx2n

x2

1 (cosh x 1) sinh x + xn+1/2

x=0

,

and = (t2 + 1)1/2. It is known that the Un()s are polynomials in of degree3n. The first few are given by U0() = 1 and

U1 () = 524

3 +18

,

15


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U2 () =385

11526 77

1924 +

9

1282,

U3 () = 8508582944

9 +17017

92167 4563

51205 +

75

10243.

Thus, the expansion (1.21) holds uniformly for t > 0. Since

t2 + 1 (cosh x 1) sinh x + x = k=1

1(2k)!

x2k k=1

1(2k + 1)!

x2k+1,

we have

a2k =1

(2k + 2)!, a2k+1 = 1

(2k + 3)!for k 0.

From the CFWW Formula (1.11) we obtain

cn =1

n+12

nj=0

(1)j (2)(n+1)/2+jj!

Bn,j

1

6,

1

24, . . . , anj+1

n + 1

2+ j

,

from which it follows by (1.22) that

Un () =2n

j=0

(1)n+j 2n+j+1n+j

j!

n + j +

1

2

B2n,j.

Here Bn,j = Bn,j1

6, 124

, . . . , anj+1

, and from the recurrence of the PartialOrdinary Bell Polynomials, Bn,0 = 0, Bn,1 = an and

Bn,j = 16Bn1,j1 +

1

24Bn2,j1 + + anj+1Bj1,j1.

In this case, it is quite complicated to give any explicit formula for the PartialOrdinary Bell Polynomials. Nevertheless, formula (1.16) enables us to derive afairly explicit expression for the polynomials Un (). To use formula (1.16) wehave to compute the Ordinary Potential Polynomials Aj,k

3

, 112

, . . . , 2 ak

. Thegenerating function is

21 (cosh x 1) sinh x + x

x2

j=

k=0

Aj,k

3,

1

12, . . . , 2 ak

xk.

Algebraic manipulation and the generating function of the 1-associated StirlingNumbers of the Second Kind (see expression (A.5)) yields

1 (cosh x 1) sinh x + xj=

(ex x 1)

1

2 1

2

+

ex + x 1 12

+1

2

j

=

ji=0

j

i

1

2 1

2

i 12

+1

2

ji(ex x 1)i ex + x 1ji

= j!

ji=0

1

2 1

2

i1

2+

1

2

ji k=0

S1 (k, i)xk

k!

m=0

(1)m S1 (m, j i) xm

m!

= j!

k=0

ji=0

k2j+2im=2i

(1)km 12 12i 1

2+ 1

2ji S1 (m, i) S1 (k m, j i)m! (k m)! xk,

16


22/40


which gives2

1 (cosh x 1) sinh x + xx2

j

= j!

k=0

j

i=0

k2j+2i

m=2i

(

1)km (1

)i (1 + )jiS1 (m, i) S1 (k m, j i)

m! (k m)! xk2j= j!

k=0

j

i=0

k+2im=2i

(1)km (1 )i (1 + )ji S1 (m, i) S1 (k + 2j m, j i)m! (k + 2j m)!

xk,

that is

Aj,k

3,

1

12, . . . , 2 ak

= j!

j

i=0

k+2i

m=2i

(

1)km (1

)i (1 + )jiS1 (m, i) S1 (k + 2j m, j i)

m! (k + 2j m)!.

Finally, from formula (1.16), we find

cn = (2)(n+1)/2

3n+12

+ 1

2n+12

nj=0

(1)jn+12

+ j Aj,n 3 , 112 , . . . , 2 an

j! (n j)!

= (2)(n+1)/23n2

+ 32

n+12

nj=0

(1)j(n + 2j + 1)

Aj,n

3

, 112

, . . . , 2 an

j! (n j)! ,

and thus by (1.22),

Un () = (1)n2 (2)n

3n + 3

2

2nj=0

(1)jj

i=0 un (i, j) (1 )i (1 + )ji(2n + 2j + 1) (2n j)! ,

where

un (i, j)def=

2n+2im=2i

(1)m S1 (m, i) S1 (2n + 2j m, j i)m! (2n + 2j m)! .

Remark. It can be shown that the Modified Bessel Functions of the First KindI have a similar asymptotic expansion:

I (t)

tt2 + 1 + 1

et2+12

t2 + 1

n=0

Un ()

n,

as +, uniformly for t > 0. The coefficients Un() are the same as above. Itis known that a recurrence for the polynomials Un() is given as follows

Un+1 () =1

22

1 2

Un ()

1

8

0 5x2 1

Un (x) dx,

with U0() = 1 [8, p. 376].

17


23/40


Example 1.2.3. As a last example we take the Legendre Polynomials Pm. Sup-pose that t > 1, then we have the integral representation

Pm (t) =1

0

t + cos x

t2 1

mdx.

The substitution t = cosh ( > 0) and a little algebraic manipulation gives

Pm (cosh ) =em

0

em( log(1sin2(x2 )(1e2)))dx.

Using Theorem 1.2.1 with f(x) = log 1 sin2 x2

1 e2, g(x) 1, = 2

and = 1; one finds that0

em( log(1sin2(x2 )(1e2)))dx

n=0

n + 1

2

cn

m(n+1)/2

as m +, where, by Theorem 1.2.2,

cn =1

2n!

dn

dxn

x2

log 1 sin2 x2

(1 e2)

(n+1)/2x=0

.

It is not hard to see that cn = (1)ncn, thus

Pm (cosh ) e(m+1)

m (e2 1)

n=0

n ()

mn, (1.23)

as m +, where

n () =1

1 e2

n +

1

2

c2n (1.24)

=

1 e2

22n+1n!

d2n

dx2n

x2

log 1 sin2 x2

(1 e2)

n+1/2x=0

.

The first few are 0 () = 1 and

1 () = e2

3

8 (e2 1) , 2 () =e4 + 10e2 + 25

128 (e2 1)2 , 3 () =5e6 + 35e4 + 455e2 + 105

1024(e2 1)3 .Since

log

1 sin2x

2

1 e2 =

k=1

1 e2k

ksin2k

x2

=k=1

1 e2k

k

j=0

(1)k+j

22k1 (2k + 2j)!

ki=1

(1)i

2k

k i

i2j+2k

x2k+j

=

k=1

k1j=0

(

1)k 1 e

2j+122j+1 (j + 1) (2k)!j+1i=1

(1)i 2j + 2j i + 1i2k x2k,

18


24/40


we have

a2k =k

j=0

(1)k+1 1 e2j+122j+1 (j + 1) (2k + 2)!

j+1i=1

(1)i

2j + 2

j i + 1

i2k+2

and a2k+1 = 0 for k

0. From the CFWW Formula (1.11) we obtain

cn =1

2n+12

nj=0

(1)j 4(n+1)/2+jj! (1 e2)(n+1)/2+j

n + 1

2+ j

Bn,j,

with Bn,j = Bn,j

0, 3e

44e2+196

, . . . , anj+1

. From this and (1.24) it follows that

n () =2n

j=0

(1)j 4n+j

j!

(1 e2)n+j

n + j +1

2

B2n,j.

By the recurrence of the Partial Ordinary Bell Polynomials, B2n,0 = 0, B2n,1 = a2n

andB2n,2j =

3e4 4e2 + 196

B2n2,2j1 + + a2n2jB2j,2j,

B2n,2j+1 =3e4 4e2 + 1

96B2n2,2j + + a2n2jB2j,2j

for j 0. If one thinks of the generating function of these Partial Ordinary BellPolynomials, it is clear that giving an explicit formula for them would be quitecomplicated. On the other hand, formula (1.16) enables us to give an explicitexpression for the coefficients n () in terms of the Stirling Numbers of the FirstKind. To use formula (1.16) we have to compute the Ordinary Potential Polyno-

mials Aj,k 0, 13e224 , . . . , 4ak1e2. The generating function is4 log 1 sin2 x

2

1 e2

x2 (1 e2)

j=

k=0

Aj,k

0,

1 3e224

, . . . ,4ak

1 e2

xk.

To compute the corresponding Ordinary Potential Polynomials, note that

log

1 sin2

x2

1 e2j =

k=1

j!s (k, j)

1 e2k

k!sin2k

x2

=

k=1

j!s (k, j) 1 e2kk!

m=0

(1)k+m22k1 (2k + 2m)!k

i=1

(1)i 2kk ii2k+2mx2k+m=

k=1

k

m=0

(1)kj!s (m, j) 1 e2m22m1m! (2k)!

mi=1

(1)i

2m

m i

i2k

x2k,

for j 1, which gives

Aj,2k

0,

1 3e224

, . . . ,4a2k

1 e2

=

k+jm=0

(

1)k+j 22j2m+1j!s (m, j)

(1 e2)jm m! (2k + 2j)!mi=1

(1)i 2m

m ii2k+2j ,19


25/40


for k, j 1. Finally, formula (1.16) yields

cn =1

2n+12

2n+1n+12 + n + 1n! (1 e2)(n+1)/2

nj=0

(1)jn+12

+ jn

j

Aj,n

=

1

n+12 2n+1

3n2

+ 32(1 e2)(n+1)/2

n

j=0

(

1)j

(n + 2j + 1)

Aj,n

(n j)!j! ,

for n 1 with Aj,n = Aj,n

0, 13e2

24, . . . , 4an

1e2

. From this and the expression

(1.24) we deduce

n () =

3n + 32

2nj=0

(1)n(2n j)!

n+jm=0

22n+2j2m+2s (m, j)

(1 e2)n+jm m! (2n + 2j + 1)!mi=1

(1)i

2m

m i

i2n+2j

=

3n + 3

2

3n

j=n(1)n

(3n

j)!

j

m=04jm+1s (m, j n)

(1

e2)jm m! (2j + 1)!

m

i=1(1)i

2m

m

ii2j ,

(1.25)

for n 1, with 0() = 1. Remark. Another possible way to derive the asymptotic expansion (1.23) is touse the generating function

11 2x cosh + x2 =

m=0

Pm (cosh ) xm, > 0.

A simple algebraic manipulation yields

11 z

ee2 z =

m=0

emPm (cosh ) zm.

This generating function has an algebraic singularity at z = 1, the full asymptoticexpansion can now be carried out by Darbouxs Method (see, e.g., [8, p. 309], [14, p.116]). The coefficients n () this time take the form

n () =n

k=0

(1)k (2n 4k + 1)24n4k (2n 2k + 1) (e2 1)nk

2n 2k

n k2

n k + 1/2k

B(nk+1/2)k ,

where B()n denote the Generalized Bernoulli Numbers. Using the explicit formula[5]

B()k =k

j=0

k+kj

kk+j

kk

S(k + j, j),we get the following expression involving the Stirling Numbers of the Second Kind:

n () =n

k=0

(1)k (2n 4k + 1)24n4k (e2 1)nk

2n 2k

n k2

n + 1/2

2k

kj=0

2k

k j

(1)j S(k + j, j)2n 2k + 2j + 1 ,

which is even simpler than (1.25).

20


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Chapter 2

Complex Laplace-type integrals

In this chapter, we investigate the asymptotic behavior of Laplace-type integralswith complex parameter. The integrals under consideration are of the form

I() = C

ef(z)g (z) dz, (2.1)

where the path of integration C is a contour in the complex plane and is a largereal parameter. The functions f and g are independent of and holomorphic in adomain containing the path of integration.

Chapter 2 is organized as follows. In the first section, we revisit the Method ofSteepest Descents, a well-known procedure in the asymptotic theory of complexLaplace-type integrals. In the second part, we present Perrons Method that avoidsthe computation of the path of steepest descent. Finally, in the third section, wegive three examples to demonstrate the application of these methods.

2.1 The Method of Steepest Descents

Consider the integral (2.1) with the assumptions we made. The basic idea of themethod is to deform the contour of integration C into a new path of integrationC so that the following conditions hold:

(i) C passes through one or more zeros of f;

(ii) the function Im(f) is constant on C.

The choice of a path with Im(f) = constant has two major advantages. It removesthe rapid oscillations of the integrand and on such paths Re(f) changes the mostrapidly (see, e.g., [9, p. 5]). Thus, the dominant contribution will arise from aneighborhood of the point where Re(f) is the greatest.

In order to obtain a geometrical interpretation of the method and the new pathof integration C, we introduce the notations

f(z) = u(x, y) + iv(x, y)

with z = x + iy, and the functions u and v are real. Suppose that z0 = x0 + iy0is a zero of f. It follows easily that (x0, y0) is a critical point of u, and since u is

harmonic, it must be a saddle point ofu. For this reason, we call z0 a saddle pointof f.

21


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2.1. The Method of Steepest Descents

Consider the surface F in the (x,y,u) space defined by u = u(x, y). The shapeof the surface F can also be represented on the (x, y) plane by drawing the levelcurves on which u is constant. From the CauchyRiemann equations it followsthat the families of curves corresponding to constant values of u(x, y) and v(x, y)are orthogonal at all their points of intersection [9, p. 6]. The regions whereu(x, y) > u(x0, y0) are called hills and those where u(x, y) < u(x0, y0) are calledvalleys. The level curve through the saddle, u(x, y) = u(x0, y0), separates theneighborhood of the saddle point (x0, y0) into a series of hills and valleys.

Suppose that z0 is a saddle point of order m 1, m 2, i.e.,f (z0) = f (z0) = = f(m1) (z0) = 0,

with f(m) (z0) = aei, a > 0. Then, if z = z0 + re

i, r > 0, we have

f(z) = f(z0) +rm

m!aei(m+) +

and hence, near the saddle point z0 the level curves and steepest paths are roughlythe same as

u (x, y) = u (x0, y0) +rm

m!a cos(m + ) ,

v (x, y) = v (x0, y0) +rm

m!a sin(m + ) .

The directions of the level curves where u is constant, are given by the solutionsof the equation cos (m + ) = 0, i.e.,

= m

+

k +

1

2

m, k = 0, 1, . . . , 2m 1.

Similarly, the directions of the steepest paths satisfy sin (m + ) = 0, i.e.,

= m

+ k

m, k = 0, 1, . . . , 2m 1.

Therefore, there are 2m equally spaced steepest directions from z0: m directionsof steepest descent and m directions of steepest ascent. In the neighborhood ofz0,the level curves u = u(x0, y0) form the boundaries of m valleys surrounding thesaddle point, in which cos (m + ) < 0, and m hills on which cos (m + ) > 0.The valleys and hills are situated respectively entirely below and above the saddlepoint, and each has angular width equal to /m.

Now suppose that the path of integration C in (2.1) is a steepest path throughthe saddle point z0 of order m 1. On this path we have

f(z) = f(z0) ,where is non-negative and monotonically increasing as one progresses down thepath of steepest descent. Then (2.1) becomes

I() = ef(z0)Tz0

eg (z) dz = ef(z0)T0

eg (z)dz

dd , (2.2)

where T denotes some point on the steepest descent path and T > 0 is the mapof T in the -plane. Since, for large positive , the factor e decays rapidly, the

22


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2.1. The Method of Steepest Descents

main contribution comes from the neighborhood of = 0. Thus, we can applyWatsons Lemma to the integral in the right-hand side of (2.2). To do this, wefirst require a series expansion for g (z) dz

din ascending powers of . Near z0,

f(z) = f(z0)

k=0 ak (z z0)m+k, a0

= 0,

that is

=k=0

ak (z z0)m+k

as z z0. If we put = m, then

= a1/m0 (z z0) (z) ,

where is analytic around z0 with (z0) = 0 and a1/m0 takes its principal value.Hence, is a single-valued analytic function ofz in a neighborhood of z0 and that(z0) = 0. Therefore, by the inverse function theorem (see [2, p. 121])

z z0 =k=1

kk =

k=1

kk/m,

where

1 =1

a1/m0

, 2 = a1ma

1+2/m0

, 3 =(m + 3) a21 2ma0a2

2m2a2+3/m0

, . . . .

Furthermore, for small , g (z) dzd has a convergent expansion of the form

g (z)dz

d=

n=0

cn(nm+1)/m. (2.3)

The first three coefficients cn are given explicitly by

c0 =b0

ma1/m0

, c1 =1

a2/m0

b1m

2a1b0m2a0

,

and

c2 =1

a3/m0

b2m

3a1b1m2a0

+

(m + 3) a21 2ma0a2 b0

m2a20

.

Here n!bn = g(n) (z0). We are now in a position to derive the asymptotic expansion

of the integral (2.2) taken from the saddle point z0 of order m 1 down one ofthe m valleys. Depending on the path C, we choose one of the m valleys, labelledby l, say. We replace by e2il in the expansion (2.3) and substitute it into theintegral (2.2). The asymptotic expansion of I() can now be obtained directlyfrom Watsons Lemma:

I() ef(z0)

n=0

n + 1m cne2il(n+1)/m

(n+1)/m , (2.4)

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2.2. Perrons Method

as +. To obtain an explicit formula for the coefficients cn we observe thatwith = m, so that dz

d=

1m

mdzd

, we find from (2.3)

g (z)dz

d= m

n=0

cnn.

Cauchys formula then shows that

cn =1

2im

g (z)dz

d

d

n+1=

1

2im

g (z)

(f(z0) f(z))(n+1)/mdz,

with an appropriate branch of (f(z0) f(z))1/m. Here and are simple closedcontours with positive orientation that enclose the points = 0 and z = z0,respectively. This is Dingles Formula [3, p. 119]. It follows that

cn =

1

mn! dn

dzn g (z) (z z0)m

f(z0) f(z)(n+1)/m

z=z0

=1

m

nk=0

bnkk!

dk

dzk

(z z0)m

f(z0) f(z)(n+1)/m

z=z0

,

(2.5)

which is Perrons Formula. In the next subsection we derive this formula in moregeneral conditions.

Remark. It can be shown that the asymptotic expansion (2.4) is also valid when

|arg + arg a0 + m 2l|

2 0 for all z on the portion ofC joining c to b.(iii) In a neighborhood of a,

f(a) f(z) =k=0

ak (z a)k+, (2.7)

where a0 = 0 and > 0. Since is not necessarily a positive integer, f isneed not be analytic at a. We make (z a) single-valued by introducing acut in the z-plane from a to infinity along a convenient radial line.

(iv) The contour C must not cross this cut. Furthermore, suppose that thereexists a point c = a on C such that for any c on C with |ac| < |ac|, theportion ofC from a to c can be deformed into the straight line arg(z a) =arg(c a).

Let = lim

Czaarg(z a)

be the angle of slope ofC at a. In order to condition (i) holds, the set throughwhich || + must be contained in the sector

2l 1

2 + arg + arg a0 + 2l +1

2 (2.8)for some fixed in 0 <

2, where l is a fixed integer.

Theorem 2.2.1. Consider the integral (2.6), and assume that it exists absolutelyfor every fixed satisfying the inequalities in (2.8). Furthermore, assume that ina neighborhood of a,

g (z) =k=0

bk (z a)k+1, b0 = 0,

with some fixed complex number , Re() > 0. Then, under the conditions (i) to(iv), we have

I() ef(a) n=0

n +

cne2li(n+)/(n+)/

,

as || +, uniformly with respect to arg confined to the sector (2.8). Thecoefficients cn are given by

cn =1

a(n+)/0 n!

dn

dzn

G (z)

a0 (z a)

f(a) f(z)(n+)/

z=a

=1

a(n+)/0

n

k=0bnk

k!

dk

dzk a0 (z a)

f(a)

f(z)

(n+)/z=a

,

(2.9)

where G (z)def=

k=0 bk (z a)k.

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2.2. Perrons Method

Proof. Let F be the function

F(z)def=

k=1

aka0

(z a)k

which is analytic at z = a and satisfies

f(z) = f(a) a0 (z a) (1 F(z)) (2.10)in a neighborhood ofa. Since the function G (z)exp(wF (z)) is, for each fixed w,analytic at z = a, we have

G (z)exp(wF (z)) =k=0

Pk (w) (z a)k, |z a| < ,

for sufficiently small , where Pk (w) is a polynomial in w whose degree does notexceed k, and

Pk (w) = 1k!

dkdzk

(G (z)exp(wF (z)))z=a

. (2.11)

Using (2.10), (2.11) can be written as

ewPk (w) =1

k!

dk

dzk

G (z)exp

w

f(z) f(a)a0 (z a)

z=a

. (2.12)

By Cauchys inequality

|Pk (w)| 1rk

max|za|=r

|G (z)exp(wF (z))| ,

for 0 < r < . Since F vanishes at a, for every 0 < r < ,

|Pk (w)| 1rk

M1 exp(M2 |w| r) ,

where M1 and M2 are fixed numbers. Thus, for any fixed N > 0,

G (z)exp(wF (z)) =Nk=0

Pk (w) (z a)k + RN, (2.13)

where, for|z

a|

r < ,

|RN| M3 |z a|N+1 exp(M2 |w| r) ,where M3 is a fixed number depending on N. Now, we put w = a0(z a) in(2.13). It follows that

g (z)exp(f(z) f(a)) =N

k=0

ewPk (w) (z a)k+1 + ew (z a)1 RN(2.14)

and, for |z a| r < ,ew (z a)1 RN M3 (z a)N+ exp(Re (w) + M2 |w| r) . (2.15)26


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2.2. Perrons Method

We write the integral (2.6) in the form

I() =

ca

ef(z)g (z) dz+

bc

ef(z)g (z) dz, (2.16)

where c is a point on C such that the portion ofC form a to c can be deformed

into the radial line arg(z a) = arg(c a) (see condition (iv)). Furthermore, werequire |c a| < r so that the expansion (2.14)-(2.15) holds for all z on the newpath of integration from a to c. Inserting (2.14) in the first integral on the rightof (2.16) yields

ca

ef(z)g (z) dz = ef(a)

N

k=0

Ik () + EN ()

, (2.17)

where

Ik ()def=

c

a

ewPk (w) (z

a)k+1 dz (2.18)

and

EN ()def=

ca

ew (z a)1 RNdz.

By choosing c closer to a if necessary, we also have from (2.8)2l 1

2

+ 0 arg + arg a0 + arg(c a)

2l +

1

2

0

for some 0 < 0 < . This implies that

2l 1

2

+ 0 arg w

2l + 1

2

0, (2.19)

and thus Re(w) |w| sin 0. Thus we obtain from (2.15)ew (z a)1 RN M3 (z a)N+ exp(|w| sin 0 + M2 |w| r)for the points on the path of integration. If we choose r such that r < sin 0/M2,then there will exist a constant M4 > 0 such that for all z with arg(z a) =arg(c a) and |z a| r < ,ew (z a)1 RN M3 (z a)N+ exp(M4 |w|) ,and therefore,

|EN ()| M3ca

(z a)N+ exp(M4 |a0 (z a)|) |dz|.If we let = arg(c a) and z a = ei, then

|EN ()

| M3

+

0 +N exp(M4 |a0| ) d = O (+N+1)/ . (2.20)

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2.2. Perrons Method

We now turn our attention to the integrals Ik() in (2.18). It is not hard toshow that a+ei

c

ewPk (w) (z a)k+1 dz = O

e||

for some > 0. Next we note, from (2.19), that w = a0(z a) gives

z a = w1/e2li/a1/0 1/

and(z a)k+1 = w(k+1)/e2li(k+1)/a(k+1)/0 (k+1)/.

Hence,

Ik () =1

a(k+)/0

(k+)/e2li(k+)/ei0

ewPk (w) w(k+)/1dw+O

e||

,

where, for fixed ,

|

|<

2. Now, we deform the path of integration into the

positive real axis. Therefore

Ik () =1

a(k+)/0

(k+)/e2li(k+)/+0

ewPk (w) w(k+)/1dw +O

e||

.

We now insert (2.12) into the integral, and then interchange the order of integrationand differentiation. This leads to

Ik () =

k +

cke

2li(k+)/

(k+)/+ O e|| .

Using this and (2.20) in (2.17) yieldsca

ef(z)g (z) dz = ef(a)

N

k=0

k +

cke

2li(k+)/

(k+)/+ O

1

(N++1)/

.

(2.21)What remains is to consider the second integral on the right-hand side of (2.16).

We choose a point 0 that satisfies the inequalities in (2.8) and writebc

ef(z)g (z) dz =

bc

e(0)f(z)e0f(z)g (z) dz,

whence bc

ef(z)g (z) dz

max e(0)f(z)bc

e0f(z)g (z) |dz|,where the maximum is taken over all points on the portion ofC joining c to b. Byassumption, the integral on the right exist, thus

bc

ef(z)g (z) dz

Kmax e(0)f(z)for some constant K > 0. It follows that

bc

ef(z)g (z) dz Ke0f(a) ef(a) max e(0)(f(a)f(z)) . (2.22)

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2.3. Examples

Then

Re (( 0) (f(a) f(z)))= |f(a) f(z)| {|| cos[arg( (f(a) f(z)))] |0| cos [arg (0 (f(a) f(z)))]}= || |f(a) f(z)| {cos[arg( (f(a) f(z)))] + o (1)} ,

as || +. Thus, by conditions (i) and (ii),Re (( 0) (f(a) f(z))) || (sin + o (1)) ,

as || + in the sector (2.8), and consequently,Re (( 0) (f(a) f(z))) 1 ||

for some 1 > 0, uniformly in arg for all satisfying (2.8) and uniformly in z forall z on C from c to b. This together with (2.22) implies that

bc

ef(z)g (z) dz = O ef(a)1|| . (2.23)The result now follows from (2.16), (2.21) and (2.23).

Remark. From (2.7) and (2.9) it can be seen that the representations (1.11) and(1.16) hold in the complex case too.

2.3 Examples

Example 2.3.1. Our first example is the Reciprocal Gamma Function

1

()=

1

2i

H

ettdt,

where the path of integration starts at ei, goes round the origin once and endsat ei. Suppose that is real an positive, then the substitution t = z gives

1

()=

1

2i1

H

e(zlog z)dz,

with the same path as before. This formula holds also whenRe

() > 0, providedthat means e log where log has its principal value. Using the notations ofSection 2.1, f(z) = z log z and g(z) 1. The function f has only one saddlepoint, namely z0 = 1. The steepest paths through z0 are described by the equation

y tan1y

x

= 0.

The equation y = 0 represents the path of steepest ascent whereas the equationx = y cot y (or rather the branch through (1, 0)) gives the path of steepest descent.

Then H can be deformed into Cdef= {(x, y) : x = y cot y, |y| < }. On the path

of steepest descent, we have

f(z) f(1) = 1 + z log z = . (2.24)

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2.3. Examples

As z 1,f(z) f(1) = (z 1)

2

2 (z 1)

3

3+ ,

thus

z

= 1 +

n=0 (1)n n

n/2, (2.25)

dzd

=n=0

(1)n+1 cn(n1)/2, cn = n + 12

n+1, (2.26)

where z denotes the upper and lower path ofC through the saddle point z0 = 1.Equation (2.24) defines a many valued function z() of a complex variable withbranch points at = 2in with n Z \ {0}. Hence, the expansions (2.25)-(2.26)are converget in the disc || < 2. Since

dz

d

=z

1 zis bounded when > 0, Watsons Lemma is applicable and gives

1

()=

e

2i1

+0

e

dz+d

dzd

d

e

2

n=0

2

n +

1

2

ic2nn

,

as +. From Perrons Formula (2.5) we find

cn =1

2n!

dndzn

(z 1)2

1 z+ log z(n+1)/2

z=1

=in+1

2n!

dn

dzn

z2

z log (1 + z)(n+1)/2

z=0

,

and hence2

n +

1

2

(ic2n) =

2

n +

1

2

(1)n2 (2n)!

d2n

dz2n

z2

z log (1 + z)n+1/2

z=0

= (1)n n.

Here n denotes the Stirling Coefficients given in Example 1.2.1. Therefore,

1

() e

2

n=0

(1)n nn

,

as +. This expansion is also valid when |arg | 2

< 2

, with fixed0 <

2.

Example 2.3.2. As a second example we take the Airy Function

Ai 2 = 12i L e2t t33 dt,

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2.3. Examples

where > 0 and the path L consists of two rays, one with end points e 43i and0, and the other with end points 0 and e 23i. The substitution t = z yields

Ai

2

=

2i

L

e3z z33

dz,

where L can be taken as the same as in the original integral. Using the notationsof Section 2.1, f(z) = z z3

3, g(z) 1 and = 3. The saddle points of f are

z0 = 1 and z1 = 1, we choose the former one. The steepest paths through z0 aredescribed by the equation

y

y2 3x2 + 3 = 0.The equation y = 0 represents the path of steepest ascent whereas the equationy2 3x2 + 3 = 0 (left branch of a hyperbola) gives the path of steepest descent. Itis clear that L can be deformed into C

def= {(x, y) : y2 3x2 + 3 = 0, x < 0}. We

put

f(z) f(1) = z z3

3+

2

3= .

We have

f(z) f(1) = (z+ 1)2 (z+ 1)3

3

as z 1, thusz = 1 +

n=0

(1)n nn/2, (2.27)

dzd =

n=0

(1)n+1

cn(n1)/2

, cn =

n + 1

2 n+1, (2.28)

where z denotes the upper and lower path ofC through the saddle point z0 = 1.Since

dzd

=1

z2 1,

and (1) = 43

, the expansions (2.27)-(2.28) are converget in the disc || < 43

.Watsons Lemma then gives

Ai 2 =

2i +

0

e23

33dz+

d dz

d d e

233

2

n=0

n +

1

2

2ic2n3n

,

as +. From Perrons Formula (2.5) we find

cn =1

2n!

dn

dzn

3

z 2(n+1)/2

z=1=

(i)n+12n!

dn

dzn

1 z

3

(n+1)/2z=0

=(i)n+1

2

3n+12 3nn!n+12 ,

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2.3. Examples

and thus

Ai

2 e 233

2

n=0

(1)n

3n + 12

9n (2n)!

1

3nas +.

This expansion is also valid when

|arg 3

| 2

<

2, with fixed 0 <

2

. The

substitution z = 2 leads to the final form

Ai (z) e 23z3/2

2

z1/4

n=0

(1)n

3n + 12

9n (2n)!

1

z3n/2,

as |z| + in the sector |arg z| 3

< 3

, with fixed 0 < 3

. Example 2.3.3. Our last example is the integral

I(, ) =1

+

0

e(2zz2)zdz, Re() > 1.

We shall use Perrons Method to obtain the asymptotic expansion of I(, ) forfixed and || + in an appropriate sector of the complex plane. We split upthe integral into two parts as follows

I(, ) =1

+0

e(1z2) (1 + z)dz+

1

10

e(1z2) (1 z)dz.

Using the notations of Section 2.2, f(z) = 1 z2, g+(z) = (1 + z), g(z) =(1 z). The only saddle point of f is z0 = 0. We have f(0) f(z) = z2 and

g (z) =k=0

(1)k

k

zk,

as z 0. We can apply Theorem 2.2.1 with = 2, = 1:+0

e(1z2) (1 + z)dz e

n=0

n + 1

2

cn

(n+1)/2,

1

0

e(1z2) (1 z)dz e

n=0

n + 1

2 (1)n cn(n+1)/2

,

as || + in the sector |arg | 2

< 2

, with fixed 0 < 2

. Here

cn =1

2n!

dn

dzn(1 + z)

z=0

=1

2

n

.

After simplification, the final result is

I(, ) e

n=0

2n

(2n)!

22nn!

1

n,

as || + and |arg | 2 < 2 , with fixed 0 < 2 .

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Appendix A

Combinatorial objects

A.1 Ordinary Potential Polynomials

Let F(x) = 1 + n=1 fnxn be a formal power series. For any complex number ,we define the Ordinary Potential PolynomialA,n (f1, f2, . . . , f n) (associated to F)by the generating function

(F(x)) =

1 +

n=1

fnxn

=

n=0

A,n (f1, f2, . . . , f n) xn.

The first few are A,0 = 1, A,1 = f1, A,2 = f2 +2

f21 , and in general

A,n (f1, f2, . . . , f n) =

k

k!

k1!k2! kn! fk11 f

k22 fknn , (A.1)

where the sum extending over all sequences k1, k2, . . . , kn of non-negative integerssuch that k1 + 2k2 + + nkn = n and k1 + k2 + + kn = k. We write

A,n (f1, f2, . . . , f n) =n

k=1

k

Bn,k (f1, f2, . . . , f nk+1),

where the Bn,k (f1, f2, . . . , f nk+1)s are called the Partial Ordinary Bell Polyno-mials. From (A.1) it follows that

Bn,k (f1, f2, . . . , f nk+1) = k!

k1!k2! knk+1!fk11 f

k22

fknk+1n

k+1 . (A.2)

Here the sum runs over all sequences k1, k2, . . . , kn of non-negative integers suchthat k1 + 2k2 + + (n k + 1) knk+1 = n and k1 + k2 + + knk+1 = k.

Since (F(x)) = (F(x))1 F(x), we have the recurrence

A,n (f1, f2, . . . , f n) = A1,n (f1, f2, . . . , f nk) +n

k=1

fkA1,nk (f1, f2, . . . , f nk).

Taking into account formula (A.2), one finds that

(F(x) 1)k = n=1

fnxnk =

n=k

Bn,k (f1, f2, . . . , f nk+1) xn, (A.3)

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A.2. The r-associated Stirling Numbers

and therefore

Bn,k+1 (f1, f2, . . . , f nk) =nkk=1

fkBnk,k (f1, f2, . . . , f n2k+1). (A.4)

If G (x) = n=0 gnxn is a formal power series, then by (A.3), we haveG (y (F(x) 1)) =

n=0

n

k=0

gkBn,k (f1, f2, . . . , f nk+1) yk

xn.

Specially,

exp

y

n=1

fnxn

=

n=0

n

k=0

Bn,k (f1, f2, . . . , f nk+1)k!

yk

xn.

For more details see, e.g., Riordans book [11, p. 189].

A.2 The r-associated Stirling Numbers

For every nonnegative integers r and k we define the r-associated Stirling Numbersof the First and Second Kind by the generating functions

1

k!

log (1 x)

rm=1

xm

m

k=

n=(r+1)k

sr (n, k)xn

n!,

1

k!ex

rm=0

xm

m!k =

n=(r+1)kSr (n, k)

xn

n! . (A.5)

If r = 0 then s (n, k)def= s0 (n, k) and S(n, k)

def= S0 (n, k) are the Stirling Numbers

of the First and Second Kind, respectively.It follows that

exp

y

log (1 x)

rm=1

xm

m

=

n=0

n/r+1

j=0

sr (n, j) yj

xn

n!,

exp

y

ex r

m=0

xmm!

= n=0

n/r+1j=0

Sr (n, j) yj xn

n!.

It is known (see [4]) that

sr (n, k) = n!k

j=0

(1)j sr1 (n rj,k j)rjj! (n rj)! ,

Sr (n, k) = n!k

j=0(1)j Sr1 (n rj,k j)

(r!)jj! (n

rj)!

.

For further identities see, e.g., Howards paper [4].

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Bibliography

[1] J. A. Campbell, P. O. Froman, E. Walles, Explicit series formulae forthe evaluation of integrals by the method of steepest descent, Studies in AppliedMathematics 77 (2) (1987), 151172.

[2] E. T. Copson, Theory of Functions of a Complex Variable, Oxford UniversityPress, London, 1935.

[3]R. B. Dingle

, Asymptotic Expansions: Their Derivation and Interpretation,Academic Press, New York, 1973.

[4] F. T. Howard, Associated Stirling Numbers, Fibonacci Quart. 18 (1980),303315.

[5] F. T. Howard, A theorem relating potential and Bell polynomials, DiscreteMathematics 39 (2) (1982), 129143.

[6] J. L. Lopez, P. Pagola, E. P. Sinusa, A simplification of Laplacesmethod: Applications to the Gamma function and Gauss hypergeometric func-tion, Journal of Approximation Theory 161 (1) (2009), 280291.

[7] G. Nemes, On the coefficients of the asymptotic expansion of n!. J. IntegerSeqs. 13 (6) (2010), Article 10.6.6, pp. 5.

[8] F. W. J. Olver, Asymptotics and Special Functions, Academic Press, NewYork, 1974.

[9] R. B. Paris, Hadamard Expansions and Hyperasymptotic Evaluation: AnExtension of the Method of Steepest Descents, Cambridge University Press,2011.

[10] O. Perron, Uber die naherungsweise Berechnung von Funktionen groer

Zahlen, Sitzungsberichte der Koniglich Bayerischen Akademie der Wis-senschaften (1917), 191219.

[11] J. Riordan, Combinatorial Identities. Wiley, New York, 1979.

[12] J. Wojdylo, Computing the coefficients in Laplaces method, SIAM Review48 (1) (2006), 7696.

[13] J. Wojdylo, On the coefficients that arise from Laplaces method, Journalof Computational and Applied Mathematics 196 (1) (2006), 241266.

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Asymptotic Expansions for Integrals

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