1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications 695

(9.11-14)

Hence by using the residue theorem and the estimates (9.11-12) to

(9.11-14), we have

If we first let Ra ~ oo, then R2 ~ oo, and finally Ri ~ oo (independently

of each other), we obtain

1

+00 m

eiax J(x) dx = 27l"i L Res eiaz J(z)

-oo k=l z=b1c

(9.11-15)

In particular, the result applies to rational functions f(z) = P(z)/Q(z)

with the degree of Q exceeding the degree of P by at least one.

After applying (9.11-15) it remains to equate real and imaginary parts

in that equation to evaluate the given integrals.

Example To evaluate f 0
00
[(xsinax)/(x^2 + b^2 )]dx, a> 0, b > 0.
Consider the complex integral

J

zeiaz

2 b2 dz

z +

c+

where c+ = [-R,R] + r+, with r+: z = Rei^9 , 0 $ (} $ 71", R > b. Inside

c+ the integrand has a simple pole at z = ib, and

zeiaz ibe-ab 1

Res = --= -e-ab

z=ib z2 + b^2 2ib 2

The rational function f(z) = z/(z^2 + b^2 ) is continuous on r+ for R > b,
and the degree of the denominator exceeds the degree of the numerator
by one, so that

izi R

lf(z)I $ lzl2 -b2 = --R2,,......_-b=2 ~ 0