TITLE: What is $\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x$?
QUESTION [16 upvotes]: Could you prove that: 
\begin{align} \displaystyle 2\left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{7} + & 53 \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{5} \\ + \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{4} + & \quad \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{3} \\ + & 19\int_{0}^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x = 2014 \end{align}
Just for the fun of it. 
Observe that $2\times53\times1\times1\times19=2014.$ 
EDIT: A different proof of
$$
\int_{0}^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x = 2
$$
may be found here. Thanks.

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\pi}{x \over x^{2} + \ln^{2}\pars{2\sin\pars{x}}}\,\dd x = 2:
     \ {\large ?}}$

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

We close the contour with the 'line segment'
$\ds{\braces{\pars{x,0}\ \mid\ x \in \pars{-1,1}}}$ with 'indented points' at $\ds{z = -1,0,1}$. I t turns out that the whole contribution to the integral arises from the 'indented point' at $\ds{z = 0}$: See my detailed calculation in one of  my previous answers which is very close to the present one. Then,

\begin{align}&\color{#66f}{\large%
\int_{0}^{\pi}{x \over x^{2} + \ln^{2}\pars{2\sin\pars{x}}}\,\dd x}
=\left. -\lim_{\epsilon\ \to\ 0^{+}}\Re\int_{\pi}^{0}
{1 \over \ln\pars{\bracks{1 - z^{2}}\ic}}\,{\dd z \over z}
\right\vert_{\,z\ \equiv\ \epsilon\expo{\ic\theta}}
\\[3mm]&=-\,\Re\int_{\pi}^{0}{1 \over \ln\pars{\ic}}\,\ic\,\dd\theta
=-\,\Re\int_{\pi}^{0}{1 \over \pi\,\ic/2}\,\ic\,\dd\theta
=-\,{2 \over \pi}\int_{\pi}^{0}\dd\theta
=\color{#66f}{\Large 2}
\end{align}