Integral ${\large\int}_0^{\pi/2}\frac{x\,\log\tan x}{\sin x}\,dx$

Question

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

Answer 1

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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,

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\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}

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$\ds{\mrm{f}\pars{a}}$ 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}

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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}