TITLE: Calculate $\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx$
QUESTION [6 upvotes]: Calculate:$$\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx$$

One can prove $\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}$ converges uniformly by Dirichlet's test, integrate term-by-term, and since $\int_{-\pi}^\pi\frac{\sin(nx)}{2^n}dx=0,$ we get series of $0$'s and the final result would be $0.$
Thing is I'm not sure how to deal with the square.
Any help appreciated.

REPLY [2 votes]: Hint:
First,
$$\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}= Im\bigg( \sum_{n=1}^\infty\bigg(\frac{e^{ix}}{2}\bigg)^n\bigg) =  Im\bigg(\frac{1}{1-\frac{e^{ix}}{2}}\bigg) =  Im\bigg(2\frac{2-\cos x+i\sin x}{(2-\cos x)^2+\sin^2 x}\bigg)= \frac{2\sin(x)}{5-4\cos(x)}$$
Thus, 
$$\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx = 2\int^{\pi}_0\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx = 8\int^{\pi}_0\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx$$
From here you can use standard rule for integration of trigonometries functions.
See here: What is tha simpliest way to compute :$\int^{\pi}_0\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx$