\[ \int_{0}^{2 \pi} \frac{d \theta}{1+a^{2}-2 a \cos (\theta)} = \begin{cases}\frac{2 \pi}{a^{2}-1} & \text { if }|a|>1 \\ \frac{2 \pi}{1-a^{2}} & \text { if }|a|<1\end{cases} .\]
Solution
The usual substitution: \(z=e^{i\theta}, \dz = (iz)\dtheta\). \[ I\da \int_{[0, 2\pi]} \inverseof{\qty{a^2 - 2a\cos(\theta) + 1}} \dtheta &= \oint \inverseof{\qty{a^2-2(z+\inverseof{z}) + 1}} \inverseof{(iz)} \dz \\ &= -i\oint \inverseof{\qty{za^2 - a(z^2+1) +z}} \dz \\ &= -i \oint\inverseof{\qty{-az^2 + (a^2+1)z - a}}\dz \\ &= {i\over a}\oint \inverseof{\qty{z^2 - \qty{a^2+a\over a}z + 1}} \dz \\ &= {i\over a}\oint \inverseof{(z-a)} \inverseof{(z-\inverseof{a})} \dz ,\] noting that \({a^2+a\over a} = a+\inverseof{a}\). Now there are two cases:
\(\abs{a} < 1\): then \(a\in \DD,\inverseof{a}\in \DD^c\), so there is a simple pole at \(a\). Then \[ I &= {i\over a}\, 2\pi i \Res_{z=a} \inverseof{(z-a)}\inverseof{(z-\inverseof{a})} \\ &= -{2\pi \over a} \inverseof{(z-\inverseof{a})} \evalfrom_{z=a} \\ &= -{2\pi \over a(a-\inverseof{a})} \\ &= {2\pi \over 1 - a^2} .\]
\(\abs{a}> 1\): then \(\inverseof{a} \in \DD, a\in \DD^c\) so there is a simple pole at \(\inverseof{a}\). Then \[ I &= {i\over a}\, 2\pi i \Res_{z=\inverseof{a} } \inverseof{(z-a)}\inverseof{(z-\inverseof{a})} \\ &= -{2\pi \over a} \inverseof{(z-a)} \evalfrom_{z=\inverseof{a}} \\ &= -{2\pi \over a(\inverseof{a} - a)} \\ &= {2\pi \over a^2 - 1} .\]