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.

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
$$I(\theta)=\pi\theta^2-2\pi^2\theta+\frac{2\pi^3}{3}$$
