TITLE: Challenging integral: $\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\sin x}dx$
QUESTION [6 upvotes]: How to tackle
$$I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\sin x}dx\ ?$$
This integral popped up in my solution ( see the integral $\mathcal{I_3}\ $ at the end of the solution.)
My attempt:  By Weierstrass substitution we have
$$I=2\int_0^1\frac{\arctan^2(x)}{x}\ln\left(\frac{2x}{1+x^2}\right)dx$$
$$=2\int_0^1\frac{\ln(2)+\ln x}{x}\arctan^2(x)dx-2\int_0^1\frac{\ln(1+x^2)}{x}\arctan^2(x)dx$$
The first integral simplifies to known harmonic series using the identity
$$\arctan^2(x)=\frac12\sum_{n=1}^\infty\frac{(-1)^n\left(H_n-2H_{2n}\right)}{n}x^{2n}$$
But using this series expansion in the second integral yields very complicated harmonic series. Also integrating by parts, yields the integrand $\frac{\text{Li}_2(-x^2)\arctan(x)}{1+x^2}$ which complicates the problem. Any thought how to approach any of these two integrals?
Thank you.

REPLY [5 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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)}
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\begin{align}
I & \equiv \bbox[5px,#ffd]{\int_{0}^{\pi/2}x^{2}{\ln\pars{\sin\pars{x}} \over \sin\pars{x}}\,\dd x}
\\[5mm] = &\
\left. \Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\bracks{-\ic\ln\pars{z}}^{\, 2}\,{\ln\pars{\bracks{z - 1/z}/\bracks{2\ic}} \over \pars{z - 1/z}/\pars{2\ic}}\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left. 2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\ln^{2}\pars{z}
\ln\pars{{1 - z^{2} \over 2z}\,\ic}\,{\dd z \over 1 - z^{2}}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &
-2\,\Re\int_{1}^{0}\bracks{\ln\pars{y} + {\pi \over 2}\,\ic}^{2}
\ln\pars{1 + y^{2} \over 2y}\,{\ic\,\dd y \over 1 + y^{2}}
\\[2mm] &\
-\! 2\,\Re\int_{0}^{1}\ln^{2}\pars{x}
\ln\pars{{1 - x^{2} \over 2x}\,\ic}\,{\dd x \over 1 - x^{2}}
\\[5mm] = &
-2\pi\int_{0}^{1}\ln\pars{y}\ln\pars{1 + y^{2} \over 2y}
\,{\dd y \over 1 + y^{2}} 
\\[2mm] & \,\, -2\int_{0}^{1}\ln^{2}\pars{x}
\ln\pars{1 - x^{2} \over 2x}\,{\dd x \over 1 - x^{2}}
\\[5mm] = &
-2\pi\
\overbrace{\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + y^{2}} \over 1 + y^{2}}\,\dd y}^{\ds{I_{1}}}\ +\
2\pi\ln\pars{2}\
\overbrace{\int_{0}^{1}{\ln\pars{y} \over 1 + y^{2}}\,\dd y}^{\ds{I_{2}}}
\\[2mm] &\
+2\pi\
\underbrace{\int_{0}^{1}{\ln^{2}\pars{y} \over 1 + y^{2}}\,\dd y}
_{\ds{I_{3}}}
\\[2mm] & \,\, -2\ \overbrace{\int_{0}^{1}{\ln^{2}\pars{x}
\ln\pars{1 - x^{2}} \over 1 - x^{2}}\,\dd x}^{\ds{I_{4}}}\ +\
2\ln\pars{2}\
\overbrace{\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}
\,\dd x}^{\ds{I_{5}}}
\\[2mm] &\
+ 2\ \underbrace{\int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x^{2}}\,\dd x}
_{\ds{I_{6}}}
\\[5mm] = &\
-2\pi I_{1} + 2\pi\ln\pars{2}I_{2} + 2\pi I_{3} - 2I_{4} +
2\ln\pars{2}I_{5} + 2I_{6}\label{1}\tag{1}
\end{align}

\begin{equation}
\!\!\!\!\!\!\!\!\!\!\!\!\left\{\begin{array}{rcl}
\ds{\LARGE\bullet} && \ds{I_{2}, I_{3}, I_{5}}\ \mbox{and}\ \ds{I_{6}}\
\mbox{are rather trivial ones or/and amenable to standard} 
\\ && \mbox{techniques}\ (~\mbox{IBP, Polylogarithms, rescaling, etc$\ldots$}~)
\\[1mm]
\ds{\LARGE\bullet} && \ds{I_{4}}\ \mbox{can be evaluated via the Beta Function, after the rescaling}\ \ds{x^{2} \mapsto x}.
\\[1mm]
\ds{\LARGE\bullet} && \mbox{After the rescaling}\ \ds{y^{2} \mapsto y},\ \ds{I_{1}}\ \mbox{can be written as a sum that involves} 
\\ && \mbox{the}\ harmonic\ number\ \mbox{because}\
\ds{{\ln\pars{1 + x} \over 1 + x} =
-\sum_{k = 1}^{\infty}H_{k}\,\pars{-1}^{k}x^{k}}.
\\ && \mbox{It turns out that}\ \ds{I_{1} =
\sum_{k = 1}^{\infty}\pars{-1}^{k}\,{H_{k} \over \pars{2k + 1}^{2}}}
\\[3mm] && \mbox{Indeed,}\ \ds{I_{1}}\ \mbox{was}\
\underline{evaluated}\
\mbox{in a previous post by user}\ {\tt @user97357329}.
\\ && \mbox{See the link at the very end.}  
\\ &&
\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S
\\[2mm]
\ds{I_{1}} & \ds{=} &
\ds{-\,{\pi^{3} \over 64} - \ln\pars{2}G - {\pi\ln^{2}\pars{2} \over 16} +
2\,\Im\mrm{Li}_{3}\pars{1 + \ic \over 2}}
\\[2mm]
\ds{I_{2}} & \ds{=} & \ds{-G\,\qquad\pars{~G:\ Catalan\ Constant~}}
\\[2mm]
\ds{I_{3}} & \ds{=} & \ds{\phantom{-}{\pi^{3} \over 16}}
\\[2mm]
\ds{I_{4}} & \ds{=} & \ds{-\,{\pi^{4} \over 32}
+ {7\ln\pars{2}\zeta\pars{3} \over 2}}
\\[2mm]
\ds{I_{5}} & \ds{=} & \ds{\phantom{-}{7\zeta\pars{3} \over 4}}
\\[2mm]
\ds{I_{6}} & \ds{=} & \ds{-\,{\pi^{4} \over 16}}
\end{array}\right. \label{2}\tag{2}
\end{equation}

Finally, with (\ref{1}) and (\ref{2}):
\begin{align}
I & \equiv \bbox[5px,#ffd]{\int_{0}^{\pi/2}x^{2}{\ln\pars{\sin\pars{x}} \over \sin\pars{x}}\,\dd x}
\\[5mm] & =
\bbx{{3\pi^{4} \over 32} + {\pi^{2}\ln^{2}\pars{2} \over 8} -
4\pi\,\Im\mrm{Li}_{3}\pars{1 + \ic \over 2} -
{7\ln\pars{2}\zeta\pars{3} \over 2}} \\ &
\end{align}

Thanks to user ${\tt @Ali Shather}$ who calls my attention to a link where $\ds{I_{1}}$ is evaluated.