Calculate the integral for $\left|a\right|<1$
$$\int_{-\pi}^{\pi} \dfrac{\cos(n\theta)}{1-2a\cos(\theta)+a^2}d\theta$$
I'm supposed to evaluate this using method of residues, but the parameter a is confusing me. I'm not even sure what contour I should use. Any help would be very appreciated.
Best Answer
Look at the integral $$ I_n = \int_{-\pi}^{\pi} \frac{e^{in\theta}}{(1-ae^{i\theta})(1-ae^{-i\theta})} \, d\theta. $$ It is easy to see the denominator expands to $1-2a\cos{\theta}+a^2$.
Since the interval of integration is symmetric, we can add the integral with $\theta \mapsto -\theta$ to $I$, and we find the integrand is $$ \frac{2\cos{n\theta}}{1-2a\cos{\theta}+a^2}, $$ so $I_n$ is equal to your integral. Now set $e^{i\theta}=z$, so the interval of integration transforms to the circle $|z|=1$, and $dz/z = i d\theta$. Hence $$ I_n = \frac{1}{i}\int_{|z|=1} \frac{z^{n}}{(1-az)(z-a)} \, dz. $$ Now you do the residues bit: since $|a|<1$, the only pole inside the unit circle is at $z=a$, where the residue is just $$ \left. \frac{z^n}{1-az} \right\rvert_{z=a} = \frac{a^n}{1-a^2} $$ Now we do $I_n = 2\pi i \sum \text{Res}$, and so $$ I_n = \frac{2\pi i}{i} \frac{a^n}{1-a^2} = \frac{2\pi a^n}{1-a^2}. $$