Cálculo de $\int_0^{2\pi} \frac{x^2 \sin{x}}{1 - 2a\cos{x} + a^2}$

Question

Cómo calcular la siguiente integral. $$\int_0^{2\pi} \frac{x^2 \sin{x}}{1 - 2a\cos{x} + a^2}dx$$

Tengo la idea de transformar una parte de la función en serie de Fourier y obtuve $$ \frac1a \sum_{n=1}^{\infty}a^n\int_0^{2\pi}x^2\sin(nx)dx = \frac{4\pi^2 \ln(1-a)}{a}, \: |a|< 1$$ SOLUCIONADO

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Con $\ds{a \in \mathbb{R}\setminus\braces{-1,1}}$:

\begin{align} &\int_{0}^{2\pi}{x^{2}\sin\pars{x} \over 1 - 2a\cos\pars{x} + a^{2}} \,\dd x = \int_{-\pi}^{\pi}{\pars{x^{2} + 2\pi x + \pi^{2}}\bracks{-\sin\pars{x}} \over 1 + 2a\cos\pars{x} + a^{2}}\,\dd x \\[5mm] = &\ -4\pi\int_{0}^{\pi}{x\sin\pars{x} \over \pars{a + \expo{\ic x}}\pars{a + \expo{-\ic x}}}\,\dd x \\[5mm] = &\ -4\pi\int_{0}^{\pi}x\sin\pars{x}\, \pars{{1 \over a + \expo{-\ic x}} - {1 \over a + \expo{\ic x}}} \,{1 \over \expo{\ic x} - \expo{-\ic x}}\,\dd x \\[5mm] = &\ -4\pi\int_{0}^{\pi}x\sin\pars{x} \pars{{1 \over a + \expo{\ic x}} - {1 \over a + \expo{-\ic x}}} {1 \over 2\ic\sin\pars{x}}\,\dd x = -4\pi\,\Im\int_{0}^{\pi}{x \over a + \expo{\ic x}}\,\dd x \label{1}\tag{1} \end{align}

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$\ds{\Large a \to 0}.$ \begin{align} \lim_{a \to 0}\Im\int_{0}^{\pi}{x \over a + \expo{\ic x}}\,\dd x & = \bbx{\pi} \end{align}

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$\ds{\Large a \in \pars{-1,1}\setminus\braces{0}}.$ \begin{align} \Im\int_{0}^{\pi}{x \over a + \expo{\ic x}}\,\dd x & = \Im\int_{0}^{\pi}{x\expo{-\ic x} \over 1 + a\expo{-\ic x}}\,\dd x = \Im\sum_{n = 0}^{\infty}\pars{-a}^{n}\ \overbrace{\int_{0}^{\pi}x\expo{-\pars{n + 1}\ic x}\,\dd x} ^{\ds{1 + \pars{-1}^{n} + \pars{n + 1}\ic\pi \over \pars{n + 1}^{2}}} \\[5mm] & = \pi\sum_{n = 0}^{\infty}{\pars{-a}^{n} \over n + 1} = -\,{\pi \over a}\sum_{n = 1}^{\infty}{\pars{-a}^{n} \over n} = \bbx{\pi\,{\ln\pars{1 + a} \over a}}\label{2}\tag{2} \end{align}

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$\ds{\Large a \in \pars{-\infty,-1}\cup\pars{1,\infty}}.$ \begin{align} \Im\int_{0}^{\pi}{x \over a + \expo{\ic x}}\,\dd x & = {1 \over a}\,\Im\int_{0}^{\pi}{x \over 1 + \pars{1/a}\expo{\ic x}}\,\dd x = {1 \over a}\,\Im\int_{0}^{\pi} {x\expo{-\ic x} \over \pars{1/a} + \expo{-\ic x}}\,\dd x \\[5mm] & = -\,{1 \over a^{2}}\,\Im\int_{0}^{\pi}{x \over \pars{1/a} + \expo{-\ic x}}\,\dd x = {1 \over a^{2}}\,\Im\int_{0}^{\pi}{x \over \pars{1/a} + \expo{\ic x}}\,\dd x \\[5mm] & = {1 \over a^{2}}\,\pi\,{\ln\pars{1 + 1/a} \over 1/a} = \pi\,{\ln\pars{1 + 1/a} \over a}\label{3}\tag{3} \end{align} Aquí, he utilizado, el resultado anterior, \eqref{2}.

Con \eqref{1}, \eqref{2} y \eqref{3}:

$$ \left.\int_{0}^{2\pi}{x^{2}\sin\pars{x} \over 1 - 2a\cos\pars{x} + a^{2}} \,\dd x\,\right\vert_{\ un\ \en\ \mathbb{R}\setminus\llaves{-1,1}} = \left\{\begin{array}{lcl} \ds{-4\pi^{2}} & \mbox{if} & \ds{a = 0} \\[2mm] \ds{-4\pi^{2}\,{\ln\pars{1 + a} \over a}} & \mbox{if} & \ds{-1 < a < 1} \\[2mm] \ds{-4\pi^{2}\,{\ln\pars{1 + 1/a} \over a}} & \mbox{if} & \ds{\verts{a} > 1} \end{array}\right. $$

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La siguiente imagen es un gráfico de $\ds{a \in \mathbb{R}}$: