Integrate $\int\limits_{0}^{1} \frac{\log(x^{2}-2x \cos a+1)}{x} dx$

Assume $0<a<2 \pi$. In this case $x^2-2 \, x \, \cos a+1 > 0$ for $0\le x \le 1$, so integral converges.

Denote $I(a)$ the value of the integral. Consider $\partial_a I(a)$. You can exchange the order of integration and differentiation, getting

$$ \partial_a I(a) = \int_0^1 \mathrm{d} x \frac{2 \sin a}{x^2 + 1 - 2 \, x \, \cos a} = \pi - a $$

The integral above is computed after a substitution $x = \cos a + \sin a \cdot \tan \frac{t}{2}$ and carefully considering integration region.

From here, knowing that for $I(\frac{\pi}{2}) = \frac{\pi^2}{24}$, we get

$$I(a) = \frac{\pi^2}{24} + \int_{\frac{\pi}{2}}^a \mathrm{d} \alpha ( \pi - \alpha) = \pi a - \frac{a^2}{2} - \frac{\pi^2}{3}.$$

The formula also extends to $a=0$ because the $I(0)$ converges.

The answer is extended to real $a$ by periodicity.

Rewriting the integral as $$ \int\limits_{0}^{1} \frac{\log|x-e^{i a}|^2}{x} \ dx= \int\limits_{0}^{1} \frac{\log|1-xe^{-i a}|^2}{x} \ dx= $$ $$ \int\limits_{0}^{1} \frac{\log(1-xe^{i a })}{x} \ dx+ \int\limits_{0}^{1} \frac{\log(1-xe^{-i a })}{x} \ dx, $$ expanding and integrating termwise leads to an answer in the form of series.