TITLE: Evaluating $\int^{\pi}_0\arctan\left(\frac{p\sin x}{1-p\cos x}\right)\sin(nx) dx$ by differentiation under integral?
QUESTION [7 upvotes]: I saw that 
$$
\int^{\pi}_{0}\arctan \left(\frac{p \sin x}{1-p \cos x}\right) \sin(nx) dx=\frac{\pi}{2n} p^n  
$$
for $$p^2 <1$$
I tried to prove using differentiation under integral but got stuck at this step
$$
I^{\prime} (p)=\int^{\pi}_{0} \frac{\sin x \sin (nx)}{1+p^2-2p \cos x}dx 
$$ 
What to do next?

REPLY [7 votes]: For $p^{2} <1$, $$\frac{\sin x}{1+p^{2}-2p \cos (x)} = \sum_{k=1}^{\infty} p^{k-1} \sin(kx). $$
This can be derived by evaluating the geometric series $\sum_{k=0}^{\infty}(p e^{ix})^{k} $.
So assuming $n$ is a positive integer, we have
$$ \begin{align} I'(p) &= \int_{0}^{\pi} \sin (nx) \sum_{k=1}^{\infty} p^{k-1} \sin(kx) \, dx \\ &= \sum_{k=1}^{\infty}p^{k-1} \int_{0}^{\pi} \sin(nx) \sin(kx) \, dx \\ &= p^{n-1} \int_{0}^{\pi} \sin^{2}(nx) \, dx \tag{1} \\ &= \frac{p^{n-1}}{2} \left(\int_{0}^{\pi} \, dx - \int_{0}^{\pi} \cos(2nx) \, dx \right) \\&= \frac{p^{n-1}}{2} (\pi-0) \\ &= \frac{\pi}{2}p^{n-1}. \end{align}$$
$(1)$ The functions $\sin(nx)$ and $\sin(kx)$ are orthogonal on $[0, \pi]$ unless $k=n$.