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

Question

How to compute the following integral. $$\int_0^{2\pi} \frac{x^2 \sin{x}}{1 - 2a\cos{x} + a^2}dx$$

I have got idea to transform a part of function in Fourier series and I obtained $$ \frac1a \sum_{n=1}^{\infty}a^n\int_0^{2\pi}x^2\sin(nx)dx = \frac{4\pi^2 \ln(1-a)}{a}, \: |a|< 1$$ SOLVED

I think it is an almost right approach. Assume that $|a| < 1.$ Expand $\dfrac{1}{a-e^{ix}}$ in a geometric series and observe that ${\rm Im}\dfrac{x^2}{a-e^{ix}} = \dfrac{x^2\sin x}{1-2a\cos x +a^2}$. In this case the value of the integral will be $4\pi^2\dfrac{\ln(1-a)}{a}.$ If $|a|>1$ you have to make some modifications. — JanG, Jun 14 '17 at 20:24
look : https://math.stackexchange.com/questions/2312096/evaluate-int-02-pi-frac-cos-2x1-2a-cos-xa2-with-a21 — Almot1960, Jun 14 '17 at 20:32

Felix Marin · Accepted Answer · 2017-06-27T07:09:26.797

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\mbox{Integrand} = {x^{2}\sin\pars{x} \over 1 - 2a\cos\pars{x} + a^{2}}}$. $$ \substack{Integrand\\ Behaviour}:\quad \left\{\begin{array}{l} \ds{\mrm{As}\ a \to -1\,,\ \sim {1 \over 2}\,{x^{2}\sin^{2}\pars{x} \over 1 + \cos\pars{x}} = {1 \over 4}\,{x^{2}\sin^{2}\pars{x} \over \cos^{2}\pars{x/2}} = x^{2}\sin^{2}\pars{x \over 2}} \\[3mm] \ds{\mrm{As}\ a \to \phantom{-\,}1\,,\ \sim {1 \over 2}\,{x^{2}\sin^{2}\pars{x} \over 1 - \cos\pars{x}} = {1 \over 4}\,{x^{2}\sin^{2}\pars{x} \over \sin^{2}\pars{x/2}} = x^{2}\cos^{2}\pars{x \over 2}} \end{array}\right. $$

As $\ds{a \to -1}$, the integrand has an integrable singularity at $\ds{x = \pi}$.
As $\ds{a \to +1}$, the integrand has an integrable singularity at $\ds{x = 0}$ and at $\ds{x = 2\pi}$.

With $\ds{a \in \mathbb{R}\setminus\braces{1}}$:

\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 1 + 2a\cos\pars{x} + a^{2}}\,\dd x = -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}

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

$\ds{\Large\verts{a} > 1}.$ \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} Here, I used, the previous result, \eqref{2}.

With \eqref{1}, \eqref{2} and \eqref{3}:

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

The cases $\ds{a \to 0}$ and $\ds{a \to -1}$ are given by $\ds{-4\pi^{2}}$ and $\ds{-4\pi^{2}\ln\pars{2}}$, respectively. The integral diverges logarithmically when $\ds{a \to 1}$ $\ds{\pars{~\mbox{as}\ 4\pi^{2}\ln\pars{1 - a}~}}$.

The following picture is a plot for $\ds{a \in \pars{-4,4}\setminus\braces{1}}$:

Running an NIntegrate in Mathematica, and comparing, it looks like the answer of @JanG in the comments is the correct one. I guess that you have a sign error somewhere. (Perhaps in the first part) — mickep, Jun 26 '17 at 07:47
@mickep Thanks. I'll check everything. I'm loosing my confident in $\texttt{Mathematica}$ somehow. — Felix Marin, Jun 26 '17 at 20:41
Well, in your case $|a|<1$, what you really get (there was a $(-1)^{n+1}$ missing in the calculation of the integral) is $$-\pi\sum_{n=0}^{+\infty}\frac{a^n}{n+1}=\pi\frac{\ln(1-a)}{a}.$$ This seems to be it. Now it is the same as JanG and Mathematica get. — mickep, Jun 27 '17 at 06:19
@mickep Fixed. Thanks. I evaluated that integral with $\texttt{Mathematica}$. Then, I $\texttt{Expand}$ed the $\texttt{Numerator}$. Now, I see that $\texttt{Mathematica}$ didn't include the prefactor $\left(,-1,\right)^{n + 1}$ when the $\texttt{Numerator}$ was expanded. Thanks a lot: You really spent your time in my answer. — Felix Marin, Jun 27 '17 at 07:07

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

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