<s>[INST] How to find $\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$? Integration by parts is of no success. What else to try?
$$\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$$
 [/INST] If $a$ is real-valued and $|a| >1$, $$\sum_{n=1}^{\infty} \frac{\cos (n \theta)}{n} \left(\frac{1}{a} \right)^{n} = - \frac{1}{2} \log \left(1- \frac{2}{a} \cos \theta+ \frac{1}{a^{2}} \right) = \log|a|  - \frac{1}{2} \log \left(1-2 a \cos \theta +a^{2} \right). $$
This identity can be derived from the Maclaurin series $$\sum_{n=1}^{\infty} \frac{z^{n}}{n} = - \log(1-z) \ , \ |z| <1$$ by replacing $z$ with $\displaystyle \frac{e^{i \theta}}{a}$ and equating the real parts on both sides.
Now rearrange the identity, square both sides, and integrate.
$$ \begin{align}  \int_{0}^{\pi} \log^{2}(1-2a \cos \theta+a^{2}) \ d \theta  &= 4 \log^{2}|a|  \int_{0}^{\pi} d \theta -4 \log|a| \int_{0}^{\pi} \sum_{n=1}^{\infty} \frac{\cos (n \theta)}{n} \left(\frac{1}{a} \right)^{n} \ d \theta \\ &+ 4 \int_{0}^{\pi} \sum_{k=1}^{\infty} \frac{\cos(k \theta)}{k} \left(\frac{1}{a} \right)^{k} \sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n} \left(\frac{1}{a} \right)^{n} \ d \theta \\ &= 4 \pi \log^{2}|a| - 4 \log|a| \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{1}{a} \right)^{n} \int_{0}^{\pi} \cos (n \theta) \ d \theta \\ &+ 4 \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{kn} \left(\frac{1}{a} \right)^{k+n} \int_{0}^{\pi} \cos(k \theta) \cos(n \theta) \  d \theta \\ &= 4 \pi \log^{2}|a| - 0 + 4 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \left(\frac{1}{a} \right)^{2n} \int_{0}^{\pi} \cos^{2}(n \theta) \ d \theta  \qquad (1) \\ &= 4 \pi \log^{2}|a| + 2 \pi \sum_{n=1}^{\infty} \frac{1}{n^{2}} \left( \frac{1}{a^{2}}\right)^{n} \\ &= 4 \pi \log^{2}|a| + 2 \pi \ \text{Li}_{2} \left(\frac{1}{a^{2}} \right) \end{align}$$
$(1)$ $\displaystyle \int_{0}^{\pi} \cos(k\theta) \cos(n \theta) \ d \theta = 0$ unless $k=n$
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