Mathematics of Physics and Engineering

Residue Integration 225

terms after (TV — l)!c_i disappear as well. As a result,

1 / dN~l \

In particular, if f(z) = g(z)(z — ZQ)~N, where the function g is analytic at
ZQ and g(zo) ^ 0, then

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/

^=(]^i)!^'

1)

^

)

(4

417)

FOR EXAMPLE, if /(z) = (z^6 + 3z^4 + 2z - \)/{z - l)^3 , then TV = 3 and

Res/(z) = -(6z=i 2 • 5z^4 + 3 • 4 • 3z^2 )|2=1 = 33.

If z — ZQ is an essential singularity, then the computation of the residue
usually requires the computation of the corresponding terms in the Laurent
series. FOR EXAMPLE, if f(z) = 2^2 sin(l/,z), then f(z) = z^2 (l/z-l/(6z^3 ) +
l/(120z^5 ) - ...) and so Res/(z) = -1/6.
z=0
EXERCISE 4.4.9.C Taking C = {z : \z - 5i| = 5}, verify that

£

-dz = 87T^2.

ic ez + 1

Hint: how many poles are enclosed by C?

One of the most elegant applications of residue integration is evaluation

of real integrals. In what follows, we describe several classes of real integrals

that can be evaluated using residues.

The easiest class is integrals involving rational expressions (sums, dif-

ferences, products, and ratios) of simp and cos<p, integrated from 0 to 2n:

j H(cos ip, sin (p) dip.

Jo

Such integrals are immediately reduced to complex integrals over the unit

circle. Indeed, writing z = el,p, we have z going around the unit circle

counterclockwise; dz = iet<pdip, so that dip = dz/(iz); also, by the Euler

formula, cos<p = (z + z~^1 )/2, simp = (z — z~^1 )/(2i). As a result,

/ H(cos ip, simp) dip = / H I

Jo J\z\=i \

z^2 + 1 z^2 - 1\ dz

2z 2iz I iz