Calculate $ \int_0^{2\pi} \ln(2-2\cos(t)) \ln(2-2\cos(t+\theta)) dt$ I'm trying to evaluate 
$$ \int_0^{2\pi} \ln(2-2\cos(t)) \ln(2-2\cos(t+\theta)) dt$$
but I'm not sure the best way to proceed.  I've been trying to factor the inner terms in to rational functions of $e^{it}$ and use logarithm identities to split the integral in to the sum of simpler complex integrals, but to limited success.
 A: This integral has an expression in terms of special functions.
Using the fact that
$$-\sum_{n=1}^{\infty}\frac{\cos(nx)}{n}=\ln(2|\sin(x/2)|)$$
the above integral can be directly rewritten as 
$$I=4 \int_{0}^{2\pi}\ln\Big(2\Big|\sin\frac{t}{2}\Big|\Big)\ln\Big(2\Big|\sin\frac{t+\theta}{2}\Big|\Big)dt=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{4}{mn}\int_{0}^{2\pi}dt~\cos(nt)\cos(mt+m\theta)$$
However it is true that
$$\int_{0}^{2\pi}dt~\cos(nt)\cos(mt+m\theta)=\pi\cos m\theta(\delta_{m+n,0}+\delta_{m-n,0})$$
and thus the desired quantity reduces to
$$I(\theta)=4\pi\sum_{n=1}^{\infty}\frac{\cos n\theta}{n^2}=2\pi\Big[\text{Li}_2(e^{i\theta})+\text{Li}_2(e^{-i\theta})\Big]$$
It seems that from this very simple form, a simpler one is unattainable, however, finding particular values of the integral in terms of known constants is not impossible. For example
$$I(0)=4\pi \zeta(2)=\frac{2\pi^3}{3}\\I(\pi)=-2\pi\zeta(2)=-\frac{\pi^3}{3}$$
EDIT:
It is actually not difficult to obtain a closed form for this sum as in here (thanks @Jay Lemmon for catching this).
Here's a way to show it different from the one cited on the link: 
We know that
$$\sum_{n=1}^{\infty}\frac{\sin nx}{n}=\frac{\pi-x}{2}~~, x\in(0,2\pi)$$
Integrating this equation in $[0,\theta]$ we obtain
$$-\sum_{n=1}^{\infty}\frac{\cos n\theta}{n^2}+\frac{\pi^2}{6}=\frac{\pi \theta}{2}-\frac{\theta^2}{4}$$
which shows that the integral in question has a very simple form
$$I(\theta)=\pi\theta^2-2\pi^2\theta+\frac{2\pi^3}{3}$$
