Evaluate $\int_{0}^{\frac {\pi}{3}}x\log(2\sin\frac {x}{2})\,dx$ Prove that
$$\int_0^{\pi/3}x\log \left(2 \sin\frac {x}{2}\right)\,dx = \frac {2\zeta(3)}{3}-\frac {\pi^2}{9}\log (2\pi)+\frac {2\pi ^2}{3}\log \left|\frac {\Gamma_2 \left(\frac {5}{6}\right)}{\Gamma_2 \left(\frac {7}{6}\right)}\right|$$

(Editor's note, July 2019: According to Adamchik's paper "On the Barnes Function" (p. 4), the notation $\Gamma_n(z)$ follows Vigneras and Vardi and is the reciprocal of the multiple gamma function $G_n$,
$$\Gamma_n(z) = \frac{1}{G_n(z)}$$ 
with the special case,
$$\Gamma_2(z) = \frac{1}{G_2(z)} = \frac{1}{G(z)}$$ 
where $G(z) = G_2(z)$ is the double Gamma function aka Barnes G function.
 A: I don't know what a double Gamma function is, so I will use the Barnes G function.
Use the fourier expansion of $\log\sin x$ which is $$\log\sin x=-\log2-\sum_{k=1}^\infty\frac{\cos 2kx}{k}$$
Then the rest is pretty straightforward.
\begin{align}
\int_0^{\pi/3}x\log\left(2\sin\frac x2\right)\,dx&=-\int_0^{\pi/3}x\sum_{k=1}^\infty\frac{\cos kx}{k}\,dx\\&=-\sum_{k=1}^\infty\int_0^{\pi/3}\frac xk\cos kx\,dx\\&=-\sum_{k=1}^\infty\left(\frac{\pi}{3k^2}\sin\frac{k\pi}{3}+\frac{1}{k^3}\cos\frac{k\pi}{3}-\frac{1}{k^3}\right)\\&=-\frac{\pi}{3}\operatorname{Cl}_2\left(\frac\pi3\right)-\operatorname{Cl}_3\left(\frac\pi3\right)+\zeta(3)
\end{align}
where $\operatorname{Cl}_n(x)$ are the Clausen function. Let $a_n$ be $(-1)^{n/3}$ if $n$ is a multiple of $3$ and otherwise zero. Then we have
\begin{align}\operatorname{Cl}_3\left(\frac\pi3\right)&=\sum_{n=1}^\infty\frac{1}{n^3}\cos\frac{n\pi}{3}=\sum_{n=1}^\infty\frac{1}{n^3}\left(\frac{(-1)^{n-1}}{2}+\frac{3}{2}a_n\right)\\&=\frac12\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}-\frac{3}{2}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(3n)^3}\\&=\frac{\zeta(3)}{3}\end{align}
and from the relation of the clausen function and the Barnes G function, we have $$\operatorname{Cl}_2\left(\frac\pi3\right)=2\pi\log\left(\frac{G(5/6)}{G(7/6)}\right)+\frac\pi3\log(2\pi)$$
Plugging altogether gives $$\int_0^{\pi/3}x\log\left(2\sin\frac x2\right)\,dx=\frac{2\zeta(3)}{3}-\frac{\pi^2}9\log(2\pi)-\frac{2\pi^2}{3}\log\left(\frac{G(5/6)}{G(7/6)}\right)$$
A: $\def\Li{{\rm Li}}$Not an answer yet, but maybe it could help (I hope...).
From Fourier analysis we have
\begin{equation}
\ln\left(2\sin \frac{x}{2}\right)=-\sum_{n=1}^{+\infty}\frac{\cos(nx)}{n}
\end{equation}
Therefore
\begin{align}
\int_0^{\pi/3}x\ln\left(2\sin \frac{x}{2}\right)\,dx&=-\int_0^{\pi/3}x\sum_{n=1}^{+\infty}\frac{e^{inx}+e^{-inx}}{2n}\,dx\\
&=-\sum_{n=1}^{+\infty}\int_0^{\pi/3}\frac{xe^{inx}+xe^{-inx}}{2n}\,dx\\
&=-\frac{1}{2}\sum_{n=1}^{+\infty}\left(\left.\frac{xe^{inx}}{in^2}\right|_{x=0}^{\pi/3}-\int_0^{\pi/3}\frac{e^{inx}}{in^2}\,dx-\left.\frac{xe^{-inx}}{in^2}\right|_{x=0}^{\pi/3}+\int_0^{\pi/3}\frac{e^{-inx}}{in^2}\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\pi}{3i}\Li_2\left(e^{\frac{\pi i}{3}}\right)+\Li_3\left(e^{\frac{\pi i}{3}}\right)-\Li_3(1)-\frac{\pi}{3i}\Li_2\left(e^{-\frac{\pi i}{3}}\right)+\Li_3\left(e^{-\frac{\pi i}{3}}\right)-\Li_3(1)\right)\\
&=-\frac{1}{2}\left(\frac{\pi}{3i}\left[\Li_2\left(e^{\frac{\pi i}{3}}\right)-\Li_2\left(e^{-\frac{\pi i}{3}}\right)\right]+\Li_3\left(e^{\frac{\pi i}{3}}\right)+\Li_3\left(e^{-\frac{\pi i}{3}}\right)-2\zeta(3)\right)
\end{align}
Interestingly, from Wolfram Alpha I get the following results
\begin{align}
\Li_2\left(e^{\frac{\pi i}{3}}\right)-\Li_2\left(e^{-\frac{\pi i}{3}}\right)&=\frac{i\sqrt{3}}{36}\left[\psi^{(1)}\left(\frac{1}{6}\right)-\psi^{(1)}\left(\frac{5}{6}\right)+\psi^{(1)}\left(\frac{1}{3}\right)-\psi^{(1)}\left(\frac{2}{3}\right)\right]\\[12pt]
\Li_3\left(e^{\frac{\pi i}{3}}\right)+\Li_3\left(e^{-\frac{\pi i}{3}}\right)&=\frac{2\zeta(3)}{3}
\end{align}
Substituting the above results, we get
\begin{align}
\int_0^{\pi/3}x\ln\left(2\sin \frac{x}{2}\right)\,dx
&=\frac{2\zeta(3)}{3}+\frac{\pi}{72\sqrt{3}}\left[\psi^{(1)}\left(\frac{5}{6}\right)-\psi^{(1)}\left(\frac{1}{6}\right)+\psi^{(1)}\left(\frac{2}{3}\right)-\psi^{(1)}\left(\frac{1}{3}\right)\right]
\end{align}
