Evaluate $\int_{-\pi}^\pi \big|\sum^\infty_{n=1} \frac{1}{2^n} e^{inx}\big|^2 \operatorname d\!x$

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

$$\int_{-\pi}^{\pi} |g(x)|^2 dx = 2\pi \sum\limits_{n = 1}^{\infty} |c_n|^2 = 2\pi \sum_{n = 1}^{\infty} 4^{-n} = \frac{2\pi}{3}$$

\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 $$ \int_{-\pi}^\pi\frac{1}{2^{m+n}} \mathrm{e}^{i(m-n)x}\,dx=\frac{2\pi}{2^{m+n}}\,\delta_{m,n}. $$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}