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!
 A: 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.
