Evaluating $\int_0^\pi\arctan\bigl(\frac{\ln\sin x}{x}\bigr)\mathrm{d}x$ I found the following integral as a by product of another one. 
It has a nice closed form.

$$
\int_{0}^{\pi}
\arctan\left(\ln\left(\sin x \right) \over x\right)\,{\rm d}x 
$$

Mathematica and Maple fail to give the answer. Could you find it?
Hint 1: 
The closed form is 

$$
-\pi\arctan \left(2\ln 2  \over \pi\right) 
$$

Hint 2: 
The following integral may help 

$$
\int_{0}^{\pi}{x \over x^{2} + \ln^{2}\left(\alpha\sin x \right)}
\,{\rm d}x
$$ 

(see this post).
 A: $\def\Im{\mathrm{Im}}$
I hope it is not too late to present an alternative solution.
Let
\begin{align}
I(s)=\int^\pi_0\arctan\left(\frac{\ln(s\sin{x})}{x}\right)\mathrm{d}x
\end{align}
Differentiate under the integral sign to get
\begin{align}
I'(s)
&=\frac{1}{s}\int^\pi_0\frac{x}{x^2+\ln^2(s\sin{x})}\mathrm{d}x\\
&=-\frac{1}{s}\Im\int^\pi_0\frac{1}{\ln\left(\frac{se^{i2x}-s}{2i}\right)}\mathrm{d}x\\
&=-\frac{1}{s}\Im\int_{|z|=1}\frac{1}{\ln\left(\frac{sz-s}{2i}\right)}\frac{\mathrm{d}z}{2iz}\\
&=-\frac{1}{s}\Im\frac{\pi}{\ln\left(-\frac{s}{2i}\right)}\\
&=-\frac{1}{s}\Im\frac{\pi}{\ln\left(\frac{s}{2}\right)+\frac{\pi i}{2}}\frac{\ln\left(\frac{s}{2}\right)-\frac{\pi i}{2}}{\ln\left(\frac{s}{2}\right)-\frac{\pi i}{2}}\\
&=\frac{1}{2s}\frac{\pi^2}{\ln^2\left(\frac{s}{2}\right)+\frac{\pi^2}{4}}
\end{align}
where the fourth equality follows from the residue theorem and the fact that the indent around the branch point $z=1$ produces no contribution as $\epsilon\to0$.
Integrating back,
\begin{align}
I(1)
&=I(\infty)+\frac{\pi^2}{2}\int^{s=1}_{s=\infty}\frac{1}{\ln^2\left(\frac{s}{2}\right)+\frac{\pi^2}{4}}\mathrm{d}\ln\left(\frac{s}{2}\right)\\
&=\frac{\pi^2}{2}+\left.\frac{\pi^2}{2}\frac{2}{\pi}\arctan\left(\frac{2\ln\left(\frac{s}{2}\right)}{\pi}\right)\right|^1_\infty\\
&=-\pi\arctan\left(\frac{2\ln{2}}{\pi}\right)
\end{align}
as desired.
A: Consider the integral of more general form $$I(\alpha)=\int_0^\pi \arctan\left(\frac{\ln(\alpha \sin x)}{x}\right)\,dx,\qquad 0<\alpha\leq 1.$$
Then for $\alpha\in(0,1)$
$$I'(\alpha)=\frac{1}{\alpha}\int_0^\pi \frac{x}{x^2+\ln^2(\alpha\sin x)}\,dx,$$
as in the hint. To calculate the last integral we use the following identity, mentioned by Jack D'Aurizio in his comment to this question: $$\frac{b}{a^2+b^2}=\int_0^{+\infty} e^{-ay}\sin by \,dy,\quad a>0.\quad (*)$$
Setting $a=-\ln(\alpha\sin x)>0$ and $b=x$, we get
$$I'(\alpha)=\frac{1}{\alpha}\int_0^\pi \left(\int_0^{+\infty} e^{y\ln(\alpha\sin x)}\sin xy \,dy\right)\,dx=\frac{1}{\alpha}\int_0^{+\infty}\left(\int_0^\pi (\alpha\sin x)^y\sin xy \,dx\right)\,dy.$$
(Changing order of integration is legitimate, since $|e^{y\ln(\alpha\sin x)}\sin xy|\leq e^{y\ln \alpha}$, so integral $\int_0^{+\infty} e^{y\ln(\alpha\sin x)}\sin xy \,dy$ converges uniformly by $x\in[0,\pi]$.) Therefore
$$I'(\alpha)=\frac{1}{\alpha}\int_0^{+\infty}\left(\frac{\alpha}{2}\right)^y\left(\int_0^\pi (2\sin x)^y\sin xy \,dx\right)\,dy.$$
Now we need to deal with
$$J=\int_0^\pi (2\sin x)^y\sin xy \,dx.$$
Changing the variable $x=t+\frac{\pi}{2}$ yields
$$J=\int_{-\pi/2}^{\pi/2}(2\cos x)^y\left(\sin ty\cos\frac{\pi y}{2}+\cos ty\sin\frac{\pi y}{2}\right)\,dt=\sin\frac{\pi y}{2}\int_{-\pi/2}^{\pi/2}(2\cos t)^y\cos ty\,dt.$$
For the last integral we observe that
$$\int_{-\pi/2}^{\pi/2}(2\cos t)^y\cos ty\,dt=\int_{-\pi/2}^{\pi/2}e^{y(\ln(2\cos t)-it)}\,dy=\int_{-\pi/2}^{\pi/2}(1+e^{-2it})^y\,dt=$$
$$=-\frac{1}{2i}\int_{-\pi/2}^{\pi/2}(1+e^{-2it})^y\frac{de^{-2it}}{e^{-2it}}=-\frac{1}{2i}\int_{C^-}\frac{(1+z)^y}{z}\,dz,$$
where $C^-$ is the unit circle (clockwise). Since $f(z)=\frac{(1+z)^y}{z}$ has just one simple pole $z=0$ inside $C^-$ with residue $\mathop{\mathrm{Res}}\limits_{z=0}f(z)=\lim\limits_{z\to 0}(1+z)^y=1$, we get
$$-\frac{1}{2i}\int_{C^-}\frac{(1+z)^y}{z}\,dz=-\frac{1}{2i}(-2\pi i\cdot 1)=\pi$$
and
$$J=\pi\sin\frac{\pi y}{2}.$$
Using $(*)$ one more time, we get
$$I'(\alpha)=\frac{\pi}{\alpha}\int_0^{+\infty}\left(\frac{\alpha}{2}\right)^y\sin\frac{\pi y}{2}\,dy=\frac{\pi^2}{2\alpha(\ln^2\frac{\alpha}{2}+\frac{\pi^2}{4})}.$$
Now we can restore $I(\alpha)$ from its derivative:
$$I(\alpha)=\frac{\pi^2}{2}\int\frac{d\alpha}{\alpha(\ln^2\frac{\alpha}{2}+\frac{\pi^2}{4})}=\frac{\pi^2}{2}\frac{2}{\pi}\arctan\left(\frac{2}{\pi}\ln\frac{\alpha}{2}\right)+c=\pi\arctan\left(\frac{2}{\pi}\ln\frac{\alpha}{2}\right)+c.$$
The next step is to show that $c=0$. For that purpose we observe that
$$\ln(\alpha\sin x)\leq\ln\alpha\quad\Rightarrow\quad -\frac{\pi}{2}<\arctan\left(\frac{\ln(\alpha \sin x)}{x}\right)\leq \arctan\left(\frac{\ln\alpha}{x}\right)\quad\Rightarrow$$
$$-\frac{\pi^2}{2}\leq I(\alpha)\leq \int_0^\pi \arctan\left(\frac{\ln\alpha}{x}\right)\,dx\to -\frac{\pi^2}{2}$$
as $\alpha\to 0+$, so $\lim\limits_{\alpha\to 0+}I(\alpha)=-\frac{\pi^2}{2}$. Also it has to equal to
$$\lim\limits_{\alpha\to 0+}\left(\pi\arctan\left(\frac{2}{\pi}\ln\frac{\alpha}{2}\right)+c\right)=-\frac{\pi^2}{2}+c,\quad\Rightarrow \quad c=0.$$
By now we have established that
$$I(\alpha)=\int_0^\pi \arctan\left(\frac{\ln(\alpha \sin x)}{x}\right)\,dx=\pi\arctan\left(\frac{2}{\pi}\ln\frac{\alpha}{2}\right),\qquad 0<\alpha<1.$$
Letting $\alpha\to1-0$ we get the desired value
$$I(1)=-\pi\arctan\left(\frac{2}{\pi}\ln2\right).$$
(we can change here limit and integral, since our integral is just proper).
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\pi}\arctan\pars{\ln\pars{\sin\pars{x}} \over x}\,\dd x
     =-\pi\,\arctan\pars{2\ln\pars{2} \over \pi}}$

With $\ds{{\large\tt 0 < \mu < 1}}$:
  \begin{align}
\mbox{Lets define}\quad{\cal F}\pars{\mu}&\equiv
\int_{0}^{\pi}\arctan\pars{\ln\pars{\mu\sin\pars{x}} \over x}\,\dd x\quad
\mbox{such that}
\\[3mm]{\cal F}'\pars{\mu}&=\int_{0}^{\pi}
{1 \over \bracks{\ln\pars{\mu\sin\pars{x}}/x}^{2} + 1}
\,{1 \over x}\,{1 \over \mu\sin\pars{x}}\,\sin\pars{x}\,\dd x
\\[3mm]&={1 \over \mu}\int_{0}^{\pi}{x \over \ln^{2}\pars{\mu\sin\pars{x}} + x^{2}}
\,\dd x
=-\,{1 \over \mu}\,\Im\int_{0}^{\pi}{\dd x \over \ln\pars{\mu\sin\pars{x}} + x\ic}
\end{align}

$$
\mbox{We are interested in}\quad{\cal F}\pars{1^{-}}:\ {\large ?}.
\quad\mbox{Note that}\quad{\cal F}\pars{0^{+}} = -\,{\pi^{2} \over 2}\tag{1}
$$

\begin{align}
{\cal F}'\pars{\mu}&=
-\,{1 \over \mu}\,\Im
\int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}}
{1 \over \ln\pars{\mu\bracks{z^{2} - 1}/\bracks{2\ic z}} + \ln\pars{z}}\,{\dd z \over \ic z}
\\[3mm]&={1 \over \mu}\,\Re
\int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}}
{1 \over \ln\pars{\mu\bracks{1 - z^{2}}\ic/2}}\,{\dd z \over z}
\end{align}

We 'close' the contourn with the line segment $\ds{\braces{\pars{x,0}\ \mid\ x \in \pars{-1,1}}}$. The segment is indented, with arcs of radius $\ds{\epsilon}$
such that $\ds{0 < \epsilon < 1}$, around $\ds{z = -1}$, $\ds{z = 0}$ and
$\ds{z = 1}$. It turns out that the contributions from the 'indented points' at $\ds{z = \pm 1}$ vanishes out in the limit $\ds{\epsilon \to 0^{+}}$. We are left with a principal value along $\ds{\pars{-1,1}}$ and the contribution from the 'indented point' at $\ds{z = 0}$. The above mentioned principal value vanishes out $\ds{\pars{~\mbox{its integrand is odd in}\ \pars{-1,1}~}}$ such that the whole contribution to $\ds{{\cal F}'\pars{\mu}}$, in the limit
$\ds{\epsilon \to 0^{+}}$, arises 'curiously and amusing' just from the 'indented point' at $\ds{z = 0}$.
It's shown as follows:
\begin{align}
{\cal F}'\pars{\mu}&=\left.-\,{1 \over \mu}\,\Re\int_{\pi/2}^{0}
{1 \over \ln\pars{\mu\bracks{1 - z^{2}}\ic/2}}\,{\dd z \over z}
\right\vert_{\,z\ =\ -1\ +\ \epsilon\expo{\ic\theta}}
\\[3mm]&\phantom{=}-\,{1 \over \mu}\,\Re\int_{-1 + \epsilon}^{\epsilon}
{1 \over \ln\pars{\mu\bracks{1 - x^{2}}\ic/2}}\,{\dd x \over x}
\left.-\,{1 \over \mu}\,\Re\int_{\pi}^{0}
{1 \over \ln\pars{\mu\bracks{1 - z^{2}}\ic/2}}\,{\dd z \over z}
\right\vert_{\,z\ =\ \epsilon\expo{\ic\theta}}
\\[3mm]&\phantom{=}-\,{1 \over \mu}\,\Re\int_{\epsilon}^{1 - \epsilon}
{1 \over \ln\pars{\mu\bracks{1 - x^{2}}\ic/2}}\,{\dd x \over x}
\\[3mm]&\phantom{=}\left.-\,{1 \over \mu}\,\Re\int_{\pi}^{\pi/2}
{1 \over \ln\pars{\mu\bracks{1 - z^{2}}\ic/2}}\,{\dd z \over z}
\right\vert_{\,z\ =\ 1\ +\ \epsilon\expo{\ic\theta}}
\end{align}

\begin{align}
{\cal F}'\pars{\mu}&=
-\,{1 \over \mu}\,\Re\pp\
\overbrace{\int_{-1}^{1}
{1 \over \ln\pars{\mu\bracks{1 - x^{2}}\ic/2}}\,{\dd x \over x}}^{\ds{=\ 0}}\
-\,{1 \over \mu}\,\Re\int_{\pi}^{0}{\ic\,\dd\theta \over \ln\pars{\mu\ic/2}}
\\[3mm]&=-\,{\pi \over \mu}\,\Im\bracks{1 \over \ln\pars{\mu\ic/2}}
\end{align}

By using the boundary condition $\pars{1}$:
\begin{align}
{\cal F}\pars{1^{-}}&
=\int_{0}^{\pi}\arctan\pars{\ln\pars{\sin\pars{x}} \over x}\,\dd x
=-\,{\pi^{2} \over 2}
-\pi\,\Im\int_{0^{+}}^{1^{-}}{1 \over \mu}\,{\dd\mu \over \ln\pars{\mu\ic/2}}
\\[3mm]&=-\,{\pi^{2} \over 2}
-\pi\,\Im\int_{0^{+}}^{1^{-}}{\dd\mu/\mu \over \ln\pars{\mu/2} + \pi\ic/2}
=-\,{\pi^{2} \over 2}
-\pi\,\Im\int_{-\infty}^{-\ln\pars{2^{+}}}{\dd t \over t + \pi\ic/2}
\\[3mm]&=-\,{\pi^{2} \over 2}
+\pi\int_{-\infty}^{-\ln\pars{2^{+}}}{\pi\,\dd t/2 \over t^{2} + \pars{\pi/2}^{2}}
=-\,{\pi^{2} \over 2}
+\left.\pi\arctan\pars{2t \over \pi}\right\vert_{\,-\infty}^{\,-\ln\pars{2^{+}}}
\\[3mm]&=-\,{\pi^{2} \over 2} + \pi\bracks{%
\arctan\pars{-\,{2\ln\pars{2} \over \pi}} + {\pi \over 2}}
\end{align}

$$\color{#66f}{\large%
\int_{0}^{\pi}\arctan\pars{\ln\pars{\sin\pars{x}} \over x}\,\dd x
=-\pi\,\arctan\pars{2\ln\pars{2} \over \pi}} \approx {\tt -1.3055}
$$

