1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 465

where '"'ft is a circle with center ak and sufficiently small radius rk described

once in the positive direction. Suppose that lf(z)I <Mk in the punctured

disk 0 < lz -ak I ::::; r'k· Then for any given f > 0, we have

(7.14-2)

by taking r'k < e:/27r Mk, if r'k does not already satisfy this inequality. Since

f was arbitrary and the integral in (7.14-2) does not depend on rk, we have

jt(z)dz=O

•t

which implies that fc f(z) dz = 0 in view of (7.14-1).

Theorem 7.21 will have some significant implications later.

7.15 APPLICATION OF THE CAUCHY-GOURSAT

THEOREM TO THE EVALUATION OF SOME

REAL IMPROPER INTEGRALS

As an illustration we propose to evaluate the integral

1

00

e -x

2

cos 2ax dx (a f:. 0 real)

assuming as known the result J 000 e-x
2
dx =^1 / 2 ,.fif (see, e.g., Widder [40],
p. 371).
Consider the function f( z) = e-z
2
which is analytic in C, and so analytic
in a region containing the rectangle

R = {(x,y): -u::; x::; u,O::::; y::::; a}

for a> 0 fixed and any u > 0 (Fig. 7.17). By the Cauchy-Goursat theorem

we have

j e-z

2

dz= 0

c+

(7.15-1)

where c+ denotes the boundary of the rectangle described once in the

positive direction.