- Calculate the integral \(\textstyle \int _ { 0 } ^ { 2 \pi } { \frac { d \theta } { 1 - 2 a \cos \theta + a ^ { 2 } } }\) , where \(0 < a < 1\)
Solution
The integral is \(\boxed{2\pi/(1-a^2)}\).
1. The unit-circle substitution produces a rational contour integral.
Since \(0<a<1\), the real denominator is \[ 1-2a\cos\theta+a^2=|e^{i\theta}-a|^2\geq(1-a)^2>0, \] so the integral is an ordinary integral of a continuous function. Put \(z=e^{i\theta}\), which traverses the unit circle counterclockwise, and use \(d\theta=dz/(iz)\). The denominator becomes \(1-a(z+z^{-1})+a^2=(z-a)(1-az)/z\). Therefore \[ I=\int_{|z|=1}\frac{dz}{i(z-a)(1-az)}. \]
2. The pole at \(a\) gives the value.
The two poles are \(a\) and \(1/a\). Only \(a\) lies inside the unit circle, and neither lies on it. The enclosed pole is simple, with residue \(1/(i(1-a^2))\). The residue theorem consequently gives \[ I=2\pi i\frac1{i(1-a^2)}=\frac{2\pi}{1-a^2} \] [@SS03], as required.