TITLE: Integral ${\large\int}_0^{\pi/2}\frac{x\,\log\tan x}{\sin x}\,dx$
QUESTION [9 upvotes]: Could you please help me to find closed form expressions for the following definite integrals:
$$I_1=\int_0^{\pi/2}\frac{x\,\log\tan x}{\sin x}\,dx\approx0.3606065973884796896...$$
$$I_2=\int_0^{\pi/3}\frac{x\,\log\tan x}{\sin x}\,dx\approx-0.845708026471324676...$$

REPLY [7 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\,\mathcal{#1}}
 \newcommand{\mrm}[1]{\,\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, the
  $\ds{\ln}$ branch-cut 'sits' along the 'negative $\ds{x}$ axis': Namely,
  $\ds{-\pi < \mrm{arg}\pars{z} < \pi}$.

\begin{align}
\color{#f00}{I_{1}} & \equiv
\color{#f00}{\int_0^{\pi/2}{x\ \ln\pars{\tan\pars{x}} \over \sin\pars{x}}
\,\dd x} =
\Re\int_{\verts{z}\ =\ 1 \atop
{\vphantom{\huge A}0\ <\ \mrm{arg}\pars{z}\ <\ \pi/2}}
{-\ic\ln\pars{z}\ln\pars{-\ic\bracks{z^{2} - 1}/\bracks{z^{2} + 1}} \over
\pars{z^{2} - 1}/\pars{2\ic z}}\,{\dd z \over \ic z}
\\[5mm] & =
-2\,\Im\int_{\verts{z}\ =\ 1 \atop
{\vphantom{\huge A}0\ <\ \mrm{arg}\pars{z}\ <\ \pi/2}}\quad\quad
\ln\pars{z}\ln\pars{{1 - z^{2} \over 1 + z^{2}}\,\ic}\,{\dd z \over 1 - z^{2}}
\\[5mm] & =
2\,\Im\int_{1}^{0}
\ln\pars{y\ic}\ln\pars{{1 + y^{2} \over 1 - y^{2}}\,\ic}
\,{\ic\,\dd y \over 1 + y^{2}} +
2\,\Im\int_{0}^{1}
\ln\pars{x}\ln\pars{{1 - x^{2} \over 1 + x^{2}}\,\ic}
\,{\dd x \over 1 - x^{2}}
\\[5mm] & =
-2\int_{0}^{1}\ln\pars{y}\ln\pars{1 + y^{2} \over 1 - y^{2}}
\,{\dd y \over 1 + y^{2}} -
2\int_{0}^{1}{-\pars{\pi/2}^{2} \over 1 + y^{2}}\,\dd y +
2\int_{0}^{1}{\ln\pars{x}\pars{\pi/2} \over 1 - x^{2}}\,\dd x
\tag{1}
\end{align}

Since
  $\ds{\int_{0}^{1}{\dd y \over 1 + y^{2}} = {\pi \over 4}}$ and
  $\ds{\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x = -\,{\pi^{2} \over 8}}$, the last two terms don't yield any
  
  contribution to the final result. The whole contribution is coming from the first integral in $\ds{\pars{1}}$. Namely,


\begin{align}
\color{#f00}{I_{1}} & \equiv
\color{#f00}{\int_0^{\pi/2}{x\ \ln\pars{\tan\pars{x}} \over \sin\pars{x}}
\,\dd x} =
-2\int_{0}^{1}\ln\pars{y}\ln\pars{1 + y^{2} \over 1 - y^{2}}
\,{\dd y  \over  1 + y^{2}}
\\[5mm] & =
2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + y} \over 1 + y^{2}}\,\dd y +
2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y} \over 1 + y^{2}}\,\dd y
-4\,\Re\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + y\ic} \over 1 + y^{2}}\,\dd y
\\[5mm] & =
\mrm{f}\pars{1} + \mrm{f}\pars{-1} - 2\mrm{f}\pars{\ic}\tag{2}
\end{align}
where
\begin{equation}
\mrm{f}\pars{a} \equiv
2\,\Re\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + ay} \over 1 + y^{2}}\,\dd y -
\int_{0}^{1}{\ln^{2}\pars{y} \over 1 + y^{2}}\,\dd y
\end{equation}

$\ds{\mrm{f}\pars{a}}$ can be rewritten in the following form:
\begin{align}
\mrm{f}\pars{a} & =
\Re\int_{0}^{1}{\ln^{2}\pars{1 + ay} \over 1 + y^{2}}\,\dd y -
\Re\int_{0}^{1}\ln^{2}\pars{y \over 1 + ay}\,{\dd y \over 1 + y^{2}}
\end{align}
In the first integral we make the substitution
$\ds{\ {1 \over 1 + ay}\ \mapsto\ y\ }$ while in the second
$\ds{\ {y \over 1 + ay}\ \mapsto\ y.\ }$ $\ds{\mrm{f}\pars{a}}$ is reduced to:
\begin{align}
\mrm{f}\pars{a} & =
-\Re\int_{1}^{1/\pars{1 + a}}
{a\ln^{2}\pars{y} \over \pars{1 + a^{2}}y^{2} - 2y + 1}\,\dd y -
\Re\int_{0}^{1/\pars{1 + a}}{\ln^{2}\pars{y} \over
\pars{1 + a^{2}}y^{2} - 2ay + 1 }\,\dd y
\end{align}
Those integral are easily reduced to the form
$\ds{\pars{~y_{0}\ \mbox{is a constant}~}}$
$\ds{\int{\ln^{2}\pars{y} \over y_{0} - y}\,\dd y}$ which involves $\ds{\Li{\mathrm{s}}}$ functions after integration by parts. 'Partial' results
are straightforward given by:
\begin{equation}
\left\lbrace\begin{array}{rcl}
\ds{\mrm{f}\pars{1}} & \ds{=} & \ds{%
-\int_{1}^{1/2}
{\ln^{2}\pars{y} \over 2y^{2} - 2y + 1}\,\dd y -
\int_{0}^{1/2}{\ln^{2}\pars{y} \over 2y^{2} - 2y + 1 }\,\dd y}
\\[5mm] & \ds{=} &
\ds{\int_{1/2}^{1}{\ln^{2}\pars{y} \over 2y^{2} - 2y + 1 }\,\dd y -
\int_{0}^{1/2}{\ln^{2}\pars{y} \over 2y^{2} - 2y + 1 }\,\dd y}
\\[5mm] & \ds{=} & \ds{%
{7 \over 64}\,\pi^{3} - 6\,\Im\Li{3}\pars{1 + \ic \over 2} - 4G\ln\pars{2} +
{3 \over 16}\,\pi\ln\pars{2}^{2}}
\\[1mm]&& G\ \mbox{is the}\ Catalan\ Constant.
\\[1cm]
\ds{\mrm{f}\pars{-1}} & \ds{=} & \ds{%
\int_{1}^{\infty}
{\ln^{2}\pars{y} \over 2y^{2} - 2y + 1}\,\dd y -
\int_{0}^{\infty}{\ln^{2}\pars{y} \over 2y^{2} + 2y + 1 }\,\dd y}
\\[5mm] & \ds{=} & \ds{%
-\,{5 \over 64}\,\pi^{3} + 2\,\Im\Li{3}\pars{1 + \ic \over 2} -
{1 \over 16}\,\pi\ln\pars{2}^{2}}
\\[1cm]
\ds{\mrm{f}\pars{\ic}} & \ds{=} & \ds{%
\Im\int_{1}^{\ol{r}}\,\,
{\ln^{2}\pars{y} \over 1 - 2y}\,\dd y -
\Re\int_{0}^{\ol{r}}\,\,{\ln^{2}\pars{y} \over 1 - 2\ic y}\,\dd y\,\qquad\qquad
\fbox{$\ds{\ r \equiv \half + \half\,\ic\ }$}}
\\[5mm] & \ds{=} & \ds{%
-\,{5 \over 64}\,\pi^{3} + 2\,\Im\Li{3}\pars{1 + \ic \over 2} -
G\ln\pars{2} - {1 \over 16}\,\pi\ln\pars{2}^{2}}
\end{array}\right.
\end{equation}


With these 'partial' results we arrive to the $\ds{\ul{final\ one}}$
  $\ds{\pars{~\mbox{see expression}\ \pars{2}~}}$:

\begin{equation}
\begin{array}{|rcl|}\hline\mbox{}
\\
\ds{\quad\color{#f00}{I_{1}}} & \ds{\equiv} &
\ds{\color{#f00}{\int_0^{\pi/2}{x\ \ln\pars{\tan\pars{x}} \over \sin\pars{x}}\,\dd x} =
{3 \over 16}\,\pi^{3} - 8\,\Im\Li{3}\pars{1 + \ic \over 2} - 2G\ln\pars{2} +
{1 \over 4}\pi\ln^{2}\pars{2}\quad}
\\[3mm]
& \ds{\approx} &
\ds{0.36060659738847968961683946939114446475244886455702}
\\[5mm]
&& G\ \mbox{is the}\ Catalan\ Constant.
\\ \mbox{}\\ \hline
\end{array}
\end{equation}