TITLE: Evaluating $\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx$
QUESTION [11 upvotes]: One of the ways to compute the integral 
$$\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx=\frac{\pi}{4}\left(\operatorname{Li_3}(e^{-2})+2\operatorname{Li_2}(e^{-2})-2\log(2)-\zeta(3)\right)$$
is to make use of the series of $\log(\sin(x))$, but the result I got after doing that wasn't that friendly.
Is it possible to find a neat way of evaluating the integral?

REPLY [7 votes]: We start with: $\displaystyle \int_0^1 \frac{t^2}{x^2+t^2}\,dt = 1-x\tan^{-1}\frac{1}{x}$
Then, 
$\displaystyle \begin{align} \int_0^{\infty} \log (2\sin x)\left(1-x\tan^{-1}\frac{1}{x}\right)\,dx &= \int_0^{\infty}\int_0^1 \frac{t^2\log (2\sin x)}{t^2+x^2}\,dt\,dx\\&= -\sum\limits_{n=1}^{\infty} \int_0^1 t^2\int_0^{\infty} \frac{1}{n}\frac{\cos 2nx}{t^2+x^2}\,dx\,dt\tag{1}\\&= -\frac{\pi}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n}\int_0^1 t^2\frac{e^{-2nt}}{t}\,dt\\&= -\frac{\pi}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n}\left(\frac{1}{4n^2} - \frac{e^{-2n}}{4n^2} - \frac{e^{-2n}}{2n}\right)\\&= -\frac{\pi}{2}\left(\frac{1}{4}\operatorname{Li}_3(1)-\frac{1}{4}\operatorname{Li}_3(e^{-2})-\frac{1}{2}\operatorname{Li}_2(e^{-2})\right)\end{align}$
where, we used: $\displaystyle \int_0^{\infty} \frac{\cos ax}{b^2+x^2}\,dx  = \frac{\pi e^{-ab}}{2b}$ in $(1)$.
Combining with the fact that: $\displaystyle \int_0^{\infty} \left(1-x\tan^{-1}\frac{1}{x}\right)\,dx = \frac{\pi}{4}$ we get the desired result.