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742 CHAPTER 22 Singularities and the Residue Theorem

so
1 πi π
∞

dx = (−2i) = .
6
−∞ x + 64 96 48
22.3.2 Rational Functions Times Cosine or Sine
Suppose p and q are polynomials with real coefficients, have no common zeros, and the degree
of q exceeds that of p by at least 2. Suppose also that q has no real zeros, and zeros z 1 ,··· , z m
are in the upper half-plane. We want to evaluate integrals of the form
p(x) p(x)

∞ ∞
cos(cx)dx and sin(cx)dx
−∞ q(x) −∞ q(x)
in which c can be any positive number.
The idea is to consider
p(z) icz

e dz
q(z)
R
with R as the curve of Section 22.3.1, consisting of the upper part of a semicircle and part of
the real axis joining the ends of the semicircle. We obtain
m
p(z)

icz
icz
e dz = 2πi Res(p(z)e /q(z), z j )
q(z)
R j=1

p(z) icz p(z) icz
= e dz + e dz
q(z) q(z)
γ R S R
p(z) p(x) p(x)
R R
icz
= e dz + cos(cx)dx + i sin(cx)dx.
q(z) −R q(x) −R q(x)
γ R
Take the limit as R →∞. As in Section 22.3.1, the integral over γ R tends to 0 and
∞ ∞
m
p(x) p(x) icz
cos(cx)dx + i sin(cx)dx = 2πi Res(p(z)e /q(z), z j ). (22.4)
−∞ q(x) −∞ q(x)
j=1
We actually obtain two real integrals in this calculation. After computing 2πi times the sum
of the residues, the real part of this number is the integral containing cos(cx), and the imaginary
part is the integral containing sin(cx).
EXAMPLE 22.18
We will evaluate
∞

cos(cx)
dx
2
2
2
2
−∞ (x + α )(x + β )
in which c,α and β are positive numbers and α = β.Let
e icz
f (z) = .
2
2
2
2
(z + α )(z + β )
The poles of f in the upper half-plane are αi and βi, and
e −cα
Res( f,αi) = ,
2αi(β − α )
2
2
and
e −cβ
Res( f,βi) = .
2
2βi(α − β )
2
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October 14, 2010 15:37 THM/NEIL Page-742 27410_22_ch22_p729-750