Evaluate $\int_0^\pi \log(1-2a\cos x+a^2)\;dx$

From the identity $$1-a^{2n}=(1-a^2)\prod_{i=1}^{n-1}(1-2a\cos \dfrac{i\pi}{n}+a^2),\quad a\in \Bbb R$$ prove that $$\int_0^\pi \log(1-2a\cos x+a^2)\;dx=\begin{align}\begin{cases}0\quad &|a|<1\\ 2\pi\log|a|\quad &|a|>1.\end{cases}\end{align}$$

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The identity follows from finding the roots of $\dfrac{1-a^{2n}}{1-a^2}$, whose proof is similar to this problem.

I: Elementary real analysis
$$1-a^{2n}=(1-a^2)\prod_{i=1}^{n-1}(1-2a\cos \dfrac{i\pi}{n}+a^2)\\
\dfrac{1-a^{2n}}{1-a^2}=\prod_{i=1}^{n-1}(1-2a\cos \dfrac{i\pi}{n}+a^2)\\
1+a^2+a^4+\cdots+a^{2n-2}=\prod_{i=1}^{n-1}(1-2a\cos \dfrac{i\pi}{n}+a^2)\\
\dfrac{\pi}{n}\log(1+a^2+a^4+\cdots+a^{2n-2})=\dfrac{\pi}{n}[\log(1-2a\cos \dfrac{\pi}{n}+a^2)+\log(1-2a\cos \dfrac{2\pi}{n}+a^2)+\cdots\\+\log(1-2a\cos \dfrac{(n-1)\pi}{n}+a^2)]$$Recall the motivation of the definition of Riemann sum: we want to approximate an area bounded by a curve by calculating areas of rectangles. Riemann sum is the sum of areas of rectangles. We partition $[0,\pi]$ into $n$ subintervals and find the sum of areas of rectangles each with width $\dfrac{\pi}{n}$ and lengths $\log(1-2a\cos \dfrac{i\pi}{n}+a^2)$ for $i=1,2,\cdots,n$. Therefore, we have $$\begin{align}\int_0^\pi \log(1-2a\cos x+a^2)\;dx&=\lim_{n\to \infty}\dfrac{\pi}{n}\sum_{i=1}^n \log(1-2a\cos \dfrac{i\pi}{n}+a^2)\\
&=\lim_{n\to \infty}\dfrac{\pi}{n}\log(1+a^2+a^4+\cdots+a^{2n-2}).\end{align}$$ When $|a|<1$, $\log(1+a^2+a^4+\cdots+a^{2n-2})=0$. It follows that the integral is $0$ when $|a|<1$. As for $|a|>1$, the trick is to make use of the case $|a|<1$. Note that $a>1$ implies that $\dfrac{1}{a}<1$: $$\lim_{n\to \infty} \dfrac{\pi}{n}\log\bigg(a^{2n-2}(\dfrac{1}{a^{2n-2}}+\dfrac{1}{a^{2n-4}}+\cdots+1)\bigg)=\lim_{n\to \infty} \dfrac{2n-2}{n}\pi \log a=2\pi\log a.$$
II: Complex analysis

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other proofs
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