Mathematics of Physics and Engineering

228 Singularities of Complex Functions

Such integrals are evaluated by integrating the function P(z)/Q(z) over
the curve CR on Figure 4.4.1(a), and then passing to the limit R —> oo. The
integral over CR is determined by the residues of P(z)/Q(z) at the roots of
Q that have positive imaginary part, and the difference of degrees of P and
Q ensures that the integral over the semi-circle tends to zero as R increases.
If R is sufficiently large, then the curve CR encloses all the roots of Q with
positive real part.
Compared to the integration of the trigonometric expressions, we have
more complicated theoretical considerations, such as passing to the limit as
R —> oo, but often much easier computations of the residues.
FOR EXAMPLE, consider the integral

/= r ±

J_ 00 (2 + 2x + x*)Z-

Step 1. We consider the complex integral

IR = dz

Jc "CR(2 + 2Z + Z2\2'^2 )

where the curve CR is from Figure 4.4.1(a), and R is sufficiently large.

Step 2. The function f(z) = (2 + 2z + z^2 )~^2 we are integrating has

two second-order poles zit2 at the roots of the polynomial z^2 + 2z + 2 =

(z + l)^2 + 1, and these roots are z\ = — 1 + i, z 2 = —\ — i. The root in the

upper half-plane is z\ = — 1 + i, and z\ is also inside the domain enclosed

by CR if R > y/2. As a result,

IR = 2ni Res f(z), z\ — — 1 + i.

Step 3. We notice that f(z) = (z - z{)~^2 {z - z<i)~^2 and then find the

residue at z\ by formula (4.4.17) with N = 2 :

d(z - z 2 )-^2

Res f(z)

dz

= -2(zi - z 2 )~^3 = -2(2i)~A = -t/4;

recall that i_1 = —i. Accordingly, IR = n/2 for all R > y/2.

Step 4- We have IR = lR,r + lR,c, where IR^ is the integral over the real

axis from —R to R, and lRth is the integral over the circular arc. Then / =

limfi-,00 lRtT. As for i#iC, we note that \z\ = Ron the arc, and, for R > 10,

we have \z^2 + 2z + 2| > \z\^2 - 2\z\ -2 = R^2 -2R-2> R{R - 4) > R^2 /2.

Since the length of the arc is nR, inequality (4.2.8) on page 196 implies

ATTR

\IRA < -^4- -» 0, R - oo.