TITLE: A difficult integral evaluation problem
QUESTION [8 upvotes]: How do I compute the integration for $a>0$, 
$$
\int_0^\pi \frac{x\sin x}{1-2a\cos x+a^2}dx?
$$
I want to find a complex function and integrate by the residue theorem.

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\half}{{1 \over 2}}
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$\ds{\int_{0}^{\pi}{x\sin\pars{x} \over 1-2a\cos\pars{x} + a^{2}}\,\dd x:\
     {\large ?}.\qquad{\large a > 0}}$.

\begin{align}&\color{#c00000}{\int_{0}^{\pi}
{x\sin\pars{x} \over 1-2a\cos\pars{x} + a^{2}}\,\dd x}
=\half\int_{-\pi}^{\pi}{x\sin\pars{x} \over \pars{a - \expo{\ic x}}\pars{a - \expo{-\ic x}}}\,\dd x
\\[3mm]&=\half\int_{-\pi}^{\pi}{x\sin\pars{x} \over 2\ic\sin\pars{x}}
\,\pars{{1 \over a - \expo{\ic x}} -{1 \over a - \expo{-\ic x}}}\,\dd x
=\half\,\Im\int_{-\pi}^{\pi}{x \over a - \expo{\ic x}}\,\dd x
\\[3mm]&=\half\Im
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
{-\ic\ln\pars{z} \over a - z}\,{\dd z \over \ic z}
\\[3mm]&={1 \over 2a}\,\Im\bracks{%
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
{\ln\pars{z} \over z - a}\,\dd z\ -\ \overbrace{%
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
{\ln\pars{z} \over z}\,\dd z}^{\ds{=\ 0}}}
\end{align}

$$
\color{#c00000}{\int_{0}^{\pi}
{x\sin\pars{x} \over 1-2a\cos\pars{x} + a^{2}}\,\dd x}
={1 \over 2a}\,\Im
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
{\ln\pars{z} \over z - a}\,\dd z
$$

\begin{align}&\color{#c00000}{\int_{0}^{\pi}
{x\sin\pars{x} \over 1-2a\cos\pars{x} + a^{2}}\,\dd x}
\\[3mm]&={1 \over 2a}\Im\bracks{%
2\pi\ic\ln\pars{a}\Theta\pars{1 - a}
-\int_{-1}^{0}{\ln\pars{-x} + \ic\pi \over x - a}\,\dd x
-\int_{0}^{-1}{\ln\pars{-x} - \ic\pi \over x - a}\,\dd x}
\\[3mm]&={\pi \over a}\bracks{\ln\pars{a}\Theta\pars{1 - a}
-\int_{-1}^{0}{\dd x \over x - a}}
={\pi \over a}\bracks{\ln\pars{a}\Theta\pars{1 - a} + \ln\pars{1 + a \over a}}
\end{align}

$$\color{#66f}{\large%
\left.\int_{0}^{\pi}{x\sin\pars{x}\,\dd x \over 1-2a\cos\pars{x} + a^{2}}
\right\vert_{\ds{\color{#c00000}{\,a > 0}}}}
=\color{#66f}{\large%
\left\lbrace\begin{array}{lcrcl}
{\pi \over a}\,\ln\pars{1 + a} & \mbox{if} & a & < & 1
\\[3mm]
{\pi \over a}\,\ln\pars{1 + a \over a} & \mbox{if} & a & > & 1
\end{array}\right.}
$$