TITLE: Evaluate $\int_{0}^{\pi} \log \left(m^2-2m\cos x+1\right)\: dx$
QUESTION [6 upvotes]: Evaluate $$\int_{0}^{\pi} \log \left(m^2-2m\cos x+1\right)\: dx$$
My Try: I let $$f(m)=\int_{0}^{\pi} \log \left(m^2-2m\cos x+1\right)\: dx$$
Differentiating w.r.t $m$ both sides we get
$$f'(m)=2\int_{0}^{\pi}\frac{(m- \cos x)dx}{(m-\cos x)^2+\sin^2x}=2\int_{0}^{\pi}\frac{(m+ \cos x)dx}{(m+\cos x)^2+\sin^2x}$$
we get
$$f'(m)=2\int_{0}^{\pi}\frac{m \csc^2 x+\csc x \cot x}{(m\csc x+\cot x)^2+1}$$
How to proceed now?

REPLY [6 votes]: For $|m|<1$, we have
$$ \log(1+m^2-2m\cos x) = \log(1-me^{ix})(1-me^{-ix})
= \log{(1-me^{ix})}+\log(1-me^{-ix}) \\
= -\sum_{k=1}^{\infty} \frac{1}{k}m^k (e^{ikx}+e^{-ikx})
= -2\sum_{k=1}^{\infty} \frac{m^k}{k} \cos{kx}.$$
Thus:
$$ \int_0^\pi\log(1+m^2-2m\cos x) dx=-2\sum_{k=1}^{\infty} \frac{m^k}{k}\int_0^\pi \cos{kx} dx=0.
$$
For $|m|>1$, we can write:
$$
\log(1+m^2-2m\cos x)=\log(1+m^{-2}-2m^{-1}\cos x)+\log m^2,
$$
so that after integration one obtains $\pi \log m^2$.
Thus, finally
$$\int_0^\pi\log(1+m^2-2m\cos x) dx=\begin{cases}
0,& |m|<1,\\
\pi\log m^2,& |m|>1.\\
\end{cases}
$$