Compute
\[ \int _ { 0 } ^ { \pi } { \frac { d \theta } { a + \cos \theta } } \]
for \(a > 1\) using the method of residues.
Solution
Set \[ I\coloneqq\int_0^\pi\frac{d\theta}{a+\cos\theta}. \]
1. One has\[ 2I = \int_0^{2\pi}\frac{d\theta}{a+\cos\theta}. \]
With the substitution \(u=2\pi-\theta\), \[ \int_\pi^{2\pi}\frac{d\theta}{a+\cos\theta} = \int_0^\pi\frac{du}{a+\cos u} = I. \] Adding the integral over \([0,\pi]\) proves the claim.
2. The full-period integral satisfies\[ \int_0^{2\pi}\frac{d\theta}{a+\cos\theta} = -2i\int_{\abs z=1}\frac{dz}{z^2+2az+1}. \]
Put \(z=e^{i\theta}\). Then \[ dz=iz\,d\theta, \qquad \cos\theta=\frac{z+z^{-1}}{2}. \] As \(\theta\) runs from \(0\) to \(2\pi\), \(z\) traverses the unit circle once counterclockwise. Therefore \[ \begin{aligned} \int_0^{2\pi}\frac{d\theta}{a+\cos\theta} &= \int_{\abs z=1} \frac{dz/(iz)}{a+(z+z^{-1})/2}\\ &= \frac{2}{i} \int_{\abs z=1}\frac{dz}{z^2+2az+1}\\ &= -2i\int_{\abs z=1}\frac{dz}{z^2+2az+1}. \end{aligned} \]
3. Of the two poles\[ \alpha=-a+\sqrt{a^2-1}, \qquad \beta=-a-\sqrt{a^2-1}, \]only \(\alpha\) lies inside the unit circle.
The denominator factors as \[ z^2+2az+1=(z-\alpha)(z-\beta). \] Since \(a>1\), \[ \abs\alpha = a-\sqrt{a^2-1} = \frac{1}{a+\sqrt{a^2-1}} <1. \] Also \[ \abs\beta = a+\sqrt{a^2-1} >1. \]
4. The full-period integral equals\[ \frac{2\pi}{\sqrt{a^2-1}}. \]
By step 3, the residue theorem uses only the pole \(\alpha\). Its residue is \[ \operatorname{Res}_{z=\alpha} \frac{1}{z^2+2az+1} = \frac{1}{\alpha-\beta} = \frac{1}{2\sqrt{a^2-1}}. \] Thus step 2 gives \[ \begin{aligned} \int_0^{2\pi}\frac{d\theta}{a+\cos\theta} &= -2i\left( 2\pi i\frac{1}{2\sqrt{a^2-1}} \right)\\ &= \frac{2\pi}{\sqrt{a^2-1}}. \end{aligned} \]
5. The requested integral is\[ \boxed{ I=\frac{\pi}{\sqrt{a^2-1}} }. \]
Combine step 1 with step 4 and divide by \(2\).
6. Q.E.D.
Step 5 gives the requested value.