Evaluate $\int_{-\pi}^\pi \big|\sum^\infty_{n=1} \frac{1}{2^n} e^{inx}\big|^2 \operatorname d\!x$ 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!
 A: Define $$g(x) = \sum\limits_{n = 1}^{\infty} \frac{e^{inx}}{2^n}$$
This is absolutely convergent everywhere, and defines a continuous function (by, for example, the Weierstrass $M$-test). The Fourier coefficients of $g$ are easy to compute: Define $$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} g(x) e^{-inx} dx$$ 
By orthogonality of the functions $\{e^{inx}\}$, we find that
$$c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} \sum_{k = 1}^{\infty} \frac{e^{i(k - n)x}}{2^k} dx = \frac{1}{2\pi} \frac{2\pi}{2^n} = \frac{1}{2^n}$$
Then Parseval's identity shows that
$$\int_{-\pi}^{\pi} |g(x)|^2 dx = 2\pi \sum\limits_{n = 1}^{\infty} |c_n|^2 = 2\pi \sum_{n = 1}^{\infty} 4^{-n} = \frac{2\pi}{3}$$
A: \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
$$
\int_{-\pi}^\pi\frac{1}{2^{m+n}} \mathrm{e}^{i(m-n)x}\,dx=\frac{2\pi}{2^{m+n}}\,\delta_{m,n}.
$$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\color{#00f}{\large\int_{-\pi}^{\pi}\verts{\sum^{\infty}_{n=1}{1 \over 2^{n}}\,\expo{inx}}^{2}\,\dd x}
=\int_{-\pi}^{\pi}\verts{\expo{\ic x}/2 \over 1 - \expo{\ic x}/2}^{2}\,\dd x
=\int_{-\pi}^{\pi}{\dd x \over \bracks{2 - \cos\pars{x}}^{2} + \sin^{2}\pars{x}}
\\[3mm]&=2\int_{0}^{\pi}{\dd x \over 5 - 4\cos\pars{x}}
=2\int_{0}^{\infty}{1 \over 5 - 4\pars{1 - t^{2}}/\pars{1 + t^{2}}}\,{2\,\dd t \over 1 + t^{2}}
=4\int_{0}^{\infty}{1 \over 9t^{2} + 1}\,\dd t
\\[3mm]&={4 \over 3}\int_{0}^{\infty}{1 \over t^{2} + 1}\,\dd t=
{4 \over 3}\lim_{t \to \infty}\arctan\pars{t} = {4 \over 3}\,{\pi \over 2}
=\color{#00f}{\large{2 \over 3}\,\pi}
\end{align}
A: Hint:
$$\sum^\infty_{n=1} 2^{-n} e^{inx} = \frac{e^{ix}/2}{1-e^{ix}/2}$$
After squaring and simplifying, contour integration around $|z|=1$ should do the trick.
