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

Question

Estoy tratando de resolver los ejercicios para el próximo examen, y yo estoy atrapado en este ejercicio:

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

Una página antes de que se intoduced la Parseval's identity, así que supongo que está relacionado con él.

He intentado solucionarlo, pero whan cada vez que intento es malo.

Puede usted por favor darme algunos consejos? Gracias!

Answer 1

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

Answer 2

$\newcommand{\+}{^{\daga}}% \newcommand{\ángulos}[1]{\left\langle #1 \right\rangle}% \newcommand{\llaves}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \cima {= \cima \vphantom{\enorme}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\mitad}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\a la derecha\vert\,}% \newcommand{\cy}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$ \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}

Answer 3

Sugerencia:

$$\sum^\infty_{n=1} 2^{-n} e^{inx} = \frac{e^{ix}/2}{1-e^{ix}/2}$$

Después elevarlo al cuadrado y simplificando, el contorno de la integración en torno a $|z|=1$ debe hacer el truco.