Evaluate $\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy$ How to prove
$$\small\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy=
-4 \pi ^2 \left(\frac{\pi }{\sqrt{3}}+\log (2)-\frac{\psi ^{(1)}\left(\frac{1}{6}\right)}{2 \sqrt{3} \pi }\right)$$
Where $\psi^{(1)}$ denotes Trigamma? This identity arises from J. Borwein's review on experimental mathematics (which refer this formula to V. Adamchik) with no related source offered. Any help will be appreciated!

This is easily tackled by a series of substitution, let $I$ be your integral, then 
$$\begin{aligned}I &= \int_0^{2\pi} \int_0^{2\pi} \log(3-2\cos \frac y2 \cos(x+\frac y2)-\cos y) dxdy\\
&= \int_0^{2\pi} \int_0^{2\pi} \log(3-2\cos \frac y2 \cos x-\cos y) dxdy\\
&= 4\pi\int_0^{\pi} \log\left[ \frac{1}{2} \left(-\cos \frac{y}{2}+2 \sec \frac{y}{2}+\sqrt{\cos ^2\frac{y}{2}+4 \sec ^2\frac{y}{2}-5}\right)\right] dy\\
&= 8\pi\int_0^1 \log\left(\frac{2-u^2+\sqrt{u^4-5u^2+4}}{2u}\right) \frac{1}{\sqrt{1-u^2}} du \\
&= 64\pi \int_{\pi/6}^{\pi/2} \log(2\sin t) \frac{2-\cos 2 t}{5-4 \cos 2 t} dt
\end{aligned}$$
where $4\sin^2 t = 2-u^2+\sqrt{u^4-5u^2+4}$, and $\int_0^1 \frac{\log(2u)}{\sqrt{1-u^2}}du = 0$ is used.
The last integral is straightforward by writing
$$\log(2\sin t) = -\sum_{k=1}^\infty \frac{\cos 2kt}{k}\qquad \frac{2-\cos 2 t}{5-4 \cos 2 t} = \frac{1}{2}+\frac{1}{2}\sum_{k=1}^\infty\frac{\cos 2kt}{2^k}$$
Therefore $$I = \sum_{r\geq 0, s>0} \frac{a_{r,s}}{2^r s^2}$$
where $a_{r,s}$ only depends on $r+s \pmod{3}$, the expression of $I$ in terms of polygamma follows immidiately.
