TITLE: Evaluate $\int_{-\pi}^\pi \big|\sum^\infty_{n=1} \frac{1}{2^n} e^{inx}\big|^2 \operatorname d\!x$
QUESTION [9 upvotes]: I am trying to solve exercises for the coming exam, and I am stuck on this exercise:

Evaluate the integral $$\int_{-\pi}^\pi \Big|\sum^\infty_{n=1} \frac{1}{2^n} \mathrm{e}^{inx}\,\Big|^2 \operatorname d\!x$$

A page before it intoduced the Parseval's identity, so I guess it is related to it.
I tried to solve it, but whan ever it try is bad.
Can you please give me some hints? Thanks!

REPLY [8 votes]: \begin{align}
\int_{-\pi}^\pi \Big|\sum^\infty_{n=1} \frac{1}{2^n} \mathrm{e}^{inx}\Big|^{\,2} 
\operatorname d\!x &=
\int_{-\pi}^\pi \sum^\infty_{n=1} \frac{1}{2^n} \mathrm{e}^{inx} 
\overline{\sum^\infty_{m=1} \frac{1}{2^m} \mathrm{e}^{imx}}
\operatorname d\!x=
\sum_{m,n=1}^\infty \int_{-\pi}^\pi\frac{1}{2^{m+n}} \mathrm{e}^{i(m-n)x}\,dx \\ &=\sum_{n=1}^\infty\frac{2\pi}{2^{2n}}=\frac{\frac{2\pi}{4}}{1-\frac{1}{4}}=\frac{2\pi}{3},
\end{align}
since
$$
\int_{-\pi}^\pi\frac{1}{2^{m+n}} \mathrm{e}^{i(m-n)x}\,dx=\frac{2\pi}{2^{m+n}}\,\delta_{m,n}.
$$