226 Singularities of Complex Functions
assuming that the integral on the left is well-defined. Then integration
is reduced to computing the residues inside the unit disk for the modified
integrand; the computations can be rather brutal, even for seemingly simple
functions H.
FOR EXAMPLE, let us evaluate the integral
-I
(^2) 2 + COSI/5 ,
•;—4 + 3sinv? ^ 7T~.—dip.
Step 1. We write z = eitfi, cosip = (z^2 + l)/(2z), simp = (z^2 - l)/(2iz),
dip = dz/(iz), and so
J\z\ = l
z^2 + 4z + 1
Ay
z(3z^2 + 8iz - 3)
Step 2. The function/(z) = (z^2 +4z+l)/(z(3z^2 +8iz-3)) we are integrating
has three simple poles: ZQ = 0, zi, and Z2, where zi,2 are the solutions of
3z^2 + 8iz - 3 = 0. By the quadratic formula, zi, 2 = (-4i ± V-16 + 9)/3 =
i(—4± V7)/3; only z\ = i(—4 + v/7)/3 has \zi\ < 1 and therefore lies inside
the unit circle. As a result,
I = 2ni Res/(z)+Res/(z) 1
z=0 z=zi
Step 3. We compute the residue at zero by formula (4.4.14):
1
Res/(z) - (z/(z))|z=0
z=zo O
We compute the residue at Z\ by formula (4.4.15). It is more convenient to
compute the derivative of the denominator by the product rule:
(z(3z^2 + 8iz - 3))' = (3z^2 + 8iz -3) + z(6z + 8i), because the first term on
the right vanishes at z. As a result,
z^2 + 4zi + 1 -4 + V7.
Res Hz) — -—; —, where z\ = 1.
=ziK ' 2zi(3zi+4i)' 3