TITLE: How to show that $\int_{0}^{\pi/4}{\ln(\cos x)\ln(\sin x)\over \sin(2x)\tan(2x)}\mathrm dx={\ln(2^2)-\ln^2(2)\over 8}?$
QUESTION [5 upvotes]: Given that:

$$\int_{0}^{\pi/4}{\ln(\cos x)\ln(\sin x)\over \sin(2x)\tan(2x)}\mathrm dx={\ln(2^2)-\ln^2(2)\over 8}\tag1$$

I am not sure how to go about to begin tackling this problem of proving $(1)$.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\ds{\int_{0}^{\pi/4}{\ln\pars{\cos\pars{x}}\ln\pars{\sin\pars{x}} \over \sin\pars{2x}\tan\pars{2x}}\,\dd x =
{\ln\pars{2^{2}}-\ln^{2}\pars{2} \over 8}:\ {\large ?}}$.

\begin{align}
&\int_{0}^{\pi/4}{\ln\pars{\cos\pars{x}}\ln\pars{\sin\pars{x}} \over \sin\pars{2x}\tan\pars{2x}}\,\dd x
=
\int_{0}^{\pi/4}{\ln\pars{\cos\pars{x}}\ln\pars{\sin\pars{x}} \over \sin^{2}\pars{2x}}\,\cos\pars{2x}\,\dd x
\\[5mm] \stackrel{2z\ \mapsto\ x}{=}\,\,\, &\
{1 \over 2}\int_{0}^{\pi/2}
\ln\pars{\cos\pars{x \over 2}}\ln\pars{\sin\pars{x \over 2}}\,
{\cos\pars{x} \over \sin^{2}\pars{x}}\,\dd x
\\[5mm] = &\
-\,{1 \over 2}\int_{x\ =\ 0}^{x\ =\ \pi/2}
\ln\pars{\cos\pars{x \over 2}}\ln\pars{\sin\pars{x \over 2}}\,\dd\bracks{1 \over \sin\pars{x}}
\\[1cm] \stackrel{\mbox{IBP}}{=}\,\,\, &\
-\,{1 \over 8}\,\ln^{2}\pars{2}
\\[3mm] &\ +
{1 \over 2}\int_{0}^{\pi/2}\!\!\!\!\!\!\!{1 \over \sin\pars{x}}\bracks{%
{-\sin\pars{x/2}/2 \over \cos\pars{x/2}}\,\ln\pars{\sin\pars{x \over 2}} +
\ln\pars{\cos\pars{x \over 2}}\,{\cos\pars{x/2}/2 \over \sin\pars{x/2}}}\dd x
\\[1cm] = &\
-\,{1 \over 8}\,\ln^{2}\pars{2}
\\[3mm] + &\
{1 \over 16}\int_{0}^{\pi/2}
\ln\pars{\cos^{2}\pars{x \over 2}}{\dd x \over \sin^{2}\pars{x/2}} -
{1 \over 16}\int_{0}^{\pi/2}\ln\pars{\sin^{2}\pars{x \over 2}}
{\dd x \over \cos^{2}\pars{x/2}}
\\[1cm] = &\
-\,{1 \over 8}\,\ln^{2}\pars{2}
\\[3mm] + &\
{1 \over 8}\int_{0}^{\pi/2}
\ln\pars{1 + \cos\pars{x} \over 2}{\dd x \over 1 - \cos\pars{x}} -
{1 \over 8}\int_{0}^{\pi/2}\ln\pars{1 - \cos\pars{x} \over 2}
{\dd x \over 1 + \cos\pars{x}}
\\[1cm] = &\
-\,{1 \over 8}\,\ln^{2}\pars{2}
\\[3mm] + &\
{1 \over 8}\ \underbrace{\int_{0}^{\pi/2}
\ln\pars{1 + \cos\pars{x} \over 2}{\dd x \over 1 - \cos\pars{x}}}
_{\ds{\equiv\ \,\mc{I}_{1}}}\ -\
{1 \over 8}\
\underbrace{\int_{\pi/2}^{\pi}\ln\pars{1 + \cos\pars{x} \over 2}
{\dd x \over 1 - \cos\pars{x}}}_{\ds{\equiv\ \,\mc{I}_{2}}}
\\[1cm] = &\
{\mc{I}_{1} - \mc{I}_{2} - \ln^{2}\pars{2} \over 8}\label{1}\tag{1}
\end{align}

$\ds{\Large\mc{I}_{1}\ \mbox{and}\ \,\mc{I}_{2}:\ ?}$
\begin{align}
\mc{I}_{1} & \equiv
\int_{0}^{\pi/2}\ln\pars{1 + \cos\pars{x} \over 2}{\dd x \over 1 - \cos\pars{x}} \\[5mm] = &\
\left.-\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\ln\pars{1 + \bracks{z^{2} + 1}/\bracks{2z} \over 2}
{1 \over \pars{z^{2} + 1}/\pars{2z} - 1}\,{\dd z \over \ic z}\,
\right\vert_{\ z\ \equiv\ \exp\pars{\ic x}}
\\[5mm] = &\
\left.-2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\ln\pars{\bracks{z + 1}^{2} \over 4z}
{1 \over \pars{z - 1}^{2}}\,\dd z\,
\right\vert_{\ z\ \equiv\ \exp\pars{\ic x}}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, &\
2\,\Im\int_{1}^{\epsilon}\ln\pars{\bracks{\ic y + 1}^{2} \over 4\ic y}
{1 \over \pars{\ic y - 1}^{2}}\,\ic\,\dd y
\end{align}
The $\ds{\ln}$ function has a branch-cut along the 'negative real axis' with
$\ds{-\pi < \,\mrm{arg}\pars{z} < \pi}$. The integration was performed along a quarter circumference in the first quadrant with an indent 'around
$\ds{z = 0}$'. The $\ds{\,\mc{I}_{1}}$ whole contribution comes from an integral which 'runs' along the vertical axis and from the above mentioned indent. By simplicity I omitted 'contributions' from terms which are reals or goes to zero as $\ds{\epsilon \to 0^{+}}$. Then, as $\ds{\epsilon \to 0^{+}}$
\begin{align}
\mc{I}_{1} & =
-2\,\Re\int_{0}^{1}\braces{\ln\pars{1 + y^{2} \over 4y} +
\bracks{2\arctan\pars{y} - {\pi \over 2}}\ic}
{\exp\pars{2\arctan\pars{y}\ic} \over 1 + y^{2}}\,\dd y
\\[5mm] \stackrel{y\ =\ \tan\pars{x}}{=}\,\,\, &\
-2\,\Re\int_{0}^{\pi/4}\braces{-\ln\pars{2\sin\pars{2x}} +
\bracks{2x - {\pi \over 2}}\ic}\exp\pars{2\ic x}\,\dd x
\\[5mm] = &\
2\
\overbrace{\int_{0}^{\pi/4}\ln\pars{2\sin\pars{2x}}\cos\pars{2x}\,\dd x}
^{\ds{{1 \over 2}\,\ln\pars{2} - {1 \over 2}}}\ +\
2\
\overbrace{\int_{0}^{\pi/4}\pars{2x - {\pi \over 2}}\sin\pars{2x}\,\dd x}
^{\ds{{1 \over 2} - {\pi \over 4}}}
\\[5mm] &\ \implies
\bbx{\mc{I}_{1} = \ln\pars{2} - {\pi \over 2}}.\quad
\mbox{Similarly,}\quad\bbx{\mc{I}_{2} = -\ln\pars{2} - {\pi \over 2}}.
\end{align}

Expression \eqref{1} becomes
\begin{align}
\int_{0}^{\pi/4}{\ln\pars{\cos\pars{x}}\ln\pars{\sin\pars{x}} \over \sin\pars{2x}\tan\pars{2x}}\,\dd x & =
{\bracks{\ln\pars{2} - \pi/2} - \bracks{-\ln\pars{2} - \pi/2} - \ln^{2}\pars{2}\over 8}
\\[5mm] & = 
\bbx{\ln\pars{2^{2}} - \ln^{2}\pars{2}\over 8}
\end{align}