Evaluate $\int _0^{2\pi }\frac{\cos (n\theta) }{a+\cos\theta}\,d\theta$ with $a>1$, $n\in \mathbb{N}-\left\{0\right\}$

Question

Evaluate $$\int _0^{2\pi }\frac{\cos (n\theta) }{a+\cos\theta}\,d\theta,\quad\,a>1$$

I wrote $$f\left(z\right)=\frac{\frac{1}{2}\left(z^n+z^{-n}\right)}{\frac{iz^2}{2}+aiz+\frac{i}{2}}$$

The discriminant is $\Delta =a^2-1>0$

Poles are $z=-a\pm \sqrt{a^2-1}$ and $z=0$

$z=-a+\sqrt{a^2-1}$ and $z=0$ are within the unit circle.

Applying the residue theorem, we compute: $2i\pi \lim _{z\to -a+\sqrt{a^2-1}}\left(\frac{-\frac{i}{2}\left(z^n+z^{-n}\right)}{\frac{1}{2}\left(z+a+\sqrt{a^2-1}\right)}\right)$

The first residue is $\pi \frac{\left(-a+\sqrt{a^2-1}\right)^n+\left(-a+\sqrt{a^2-1}\right)^{-n}}{\sqrt{a^2-1}}$

Now $2i\pi \lim _{z\to 0}\left(\frac{-\frac{i}{2}\left(z^{n+1}+z^{-n+1}\right)}{\frac{1}{2}\left(z+a-\sqrt{a^2-1}\right)\left(z+a+\sqrt{a^2-1}\right)}\right)$ $=\lim _{z\to 0}\left(2\pi \left(z^{n+1}+z^{-n+1}\right)\right)\:$

Answer 1

To evaluate the integral $$I(a,n)=\int _0^{2\pi }\frac{\cos (n\theta) }{a+\cos\theta}\,d\theta=\operatorname{Re}\int _0^{2\pi }\frac{e^{in\theta} }{a+\cos\theta}\,d\theta,\quad\,a>1,\quad\,n\in\mathbb N $$ it is suggestive to use the substitution $z=e^{i\theta}$ so that $$ I(a,n)=\operatorname{Re}\left[\frac1i\oint\limits_{|z|=1}f(z)dz\right] =2\pi\operatorname{Re}\left[\sum_{z}^{|z|<1} \operatorname{Res}(f,z)\right]\tag1, $$ with $$ f(z)=\frac{2z^n}{z^2+2az+1}\tag2. $$ The only pole of $f(z)$ lying inside the circle $|z|=1$ is the simple pole at $z=-a+\sqrt{a^2-1}\equiv z_a$.

The residue at the pole can be easily evaluated as $$ \operatorname{Res}(f,z_a)=\lim_{z\to z_a} (z-z_a)f(z)=\frac{z_a^n}{\sqrt{a^2-1}}, $$ so that finally $$ I(a,n)=\frac {(\sqrt{a^2-1}-a)^n}{\sqrt {a^2-1}}2\pi\equiv (-1)^n\frac { e^{-n\operatorname{arccosh} a}}{\sqrt {a^2-1}}2\pi. $$

Answer 2

As cosine is an even function, integral from o to 2$\pi$ can be written as 2 times the the integral from 0 to $\pi$, So you would be left with the upper semicircle and hence take $\frac{-ai + i \sqrt{a^2 -1}}{2}$. This is because in the upper plane imaginary part $Im(z) >= 0$.