COMPLEX VARIABLES
formula (24.56) withm= 2. Choosing the latter method and denoting the integrand by
f(z), we have
d
dz[z^2 f(z)] =d
dz[
z^4 +1
(z−a/b)(z−b/a)]
=
(z−a/b)(z−b/a)4z^3 −(z^4 +1)[(z−a/b)+(z−b/a)]
(z−a/b)^2 (z−b/a)^2.
Now settingz= 0 and applying (24.56), we find
R(0) =a
b+
b
a.
For the simple pole atz=a/b, equation (24.57) gives the residue asR(a/b) = lim
z→(a/b)[
(z−a/b)f(z)]
=
(a/b)^4 +1
(a/b)^2 (a/b−b/a)=−a^4 +b^4
ab(b^2 −a^2 ).
Therefore by the residue theorem
I=2πi×i
2 ab[
a^2 +b^2
ab−
a^4 +b^4
ab(b^2 −a^2 )]
=
2 πa^2
b^2 (b^2 −a^2 ).
24.13.2 Some infinite integralsWe next consider the evaluation of an integral of the form
∫∞
−∞f(x)dx,wheref(z) has the following properties:
(i)f(z) is analytic in the upper half-plane, Imz≥0, except for a finite number
of poles, none of which is on the real axis;
(ii) on a semicircle Γ of radiusR(figure 24.14),Rtimes the maximum of|f|
on Γ tends to zero asR→∞(a sufficient condition is thatzf(z)→0as
|z|→∞);
(iii)∫ 0
−∞f(x)dxand∫∞
0 f(x)dxboth exist.Since ∣
∣
∣
∣
∫Γf(z)dz∣
∣
∣
∣≤^2 πR×(maximum of|f|on Γ),condition (ii) ensures that the integral along Γ tends to zero asR→∞, after
which it is obvious from the residue theorem that the required integral is given
by
∫∞
−∞f(x)dx=2πi×(sum of the residues at poles with Imz>0).
(24.68)