Complex Analysis – Using the Residue Theorem to Calculate a Definite Integral

Question

So I have some troubles for the following question:
Using the residue theorem, calculate :

$\int_{0}^{2\pi}\frac{1}{1-2a\cos{\theta}+a^2}d\theta$

with

i. $|a|$ < 1 ,
ii. |a| > 1

I guess that i. and ii. would give us different residues to calculate given that the surface isn't the same.

First I solved for a in the denominator. I get a= $\cos{\theta}\pm\sin{\theta}$. Now I could use the fact that $\cos{\theta}=\frac{e^{i\theta}+e^{-i\theta}}{2}$ but I don't really see what to do next. I have seen other questions like Evaluating $\int_0^{2 \pi} \frac {\cos 2 \theta}{1 -2a \cos \theta +a^2}$ but the conditions and the function aren't exactly the same and I'm still stuck.

Thanks for your help !

Answer 1

Make the change of variables $z=e^{i\theta}$ so $$\cos\theta=\frac{z+z^{-1}}{2}.$$ Then, the integral is $$\int_0^{2\pi} \frac{d\theta}{1-2a\cos\theta+a^2}=\int_{C(0;1)} \frac{dz}{1-a(z+z^{-1})+a^2}\frac{1}{iz}=-i\int_{C(0;1)} \frac{dz}{z-a(z^2+1)+a^2z}=i\int_{C(0;1)} \frac{dz}{(z-a)(z-1/a)a}$$

If $a\in(0,1)$, then $z=1/a >1$, so it doesn't belong to $C(0;1)$ and $I(C(0;1), z)=0$. For $z=a$ we have that $$\text{Res}(f,a)=(z-a)\left.\frac{i}{(z-a)(z-1/a)a} \right|_{z=a}=\frac{i}{a^2-1}$$ so $$\int_0^{2\pi} \frac{d\theta}{1-2a\cos\theta+a^2}=2\pi i \text{Res}(f,a) =\frac{-2\pi}{a^2-1}$$

If $a>1$, then $z=a >1$, so it doesn't belong to $C(0;1)$ and $I(C(0;1), z)=0$. For $z=1/a$ we have that $$\text{Res}(f,1/a)=(z-1/a)\left.\frac{i}{(z-a)(z-1/a)a} \right|_{z=1/a}=\frac{i}{1-a^2}$$ so $$\int_0^{2\pi} \frac{d\theta}{1-2a\cos\theta+a^2}=2\pi i \text{Res}(f,1/a) =\frac{2\pi}{a^2-1}$$