$\int_{0}^{\pi}\ln^2\left ( a^2-2a\cos x+1 \right )dx$

[Tolaso J Kos]

Evaluate the integral:

$$\int_{0}^{\pi}\ln^2\left ( a^2-2a\cos x+1 \right )\, {\rm d}x$$

[r9m]

We assume $a > 0$ (is a positive real number).

We'll use $\displaystyle \int_0^{2\pi} f\left(a+re^{\pm ix}\right)\,dx = 2\pi f(a)$ for $f$ analytic in $|z-a| < R$ where, $ 0< r < R$ to simplify the integral.

\begin{align*}I(a)&=\int_0^{\pi} \log^2 (1-2a\cos x + a^2)\,dx \\&= \frac{1}{2}\int_0^{2\pi} \log^2 \left((a-e^{ix})(a-e^{-ix})\right)\,dx\\&= \frac{1}{2}\int_0^{2\pi} \log^2 (a-e^{ix})\,dx+\frac{1}{2}\int_0^{2\pi} \log^2 (a-e^{-ix})\,dx+\int_0^{2\pi} \log \left(a-e^{-ix}\right)\log \left(a-e^{ix}\right)\,dx\\&= 2\pi\log^2 a + \int_0^{2\pi} \log \left(a-e^{-ix}\right)\log \left(a-e^{ix}\right)\,dx\end{align*}

Case-I: $a>1$

\begin{align*}I(a)&= 2\pi \log^2 a + \int_0^{2\pi} \log^2 a + \log a \log \left(1-a^{-1}e^{ix}\right)+\log a \log \left(1-a^{-1}e^{-ix}\right)+\log \left(1-a^{-1}e^{ix}\right) \log \left(1-a^{-1}e^{-ix}\right)\,dx\\&= 4\pi\log^2 a + \int_0^{2\pi} \log \left(1-a^{-1}e^{ix}\right) \log \left(1-a^{-1}e^{-ix}\right)\,dx\\&= 4\pi\log^2 a + \sum\limits_{n,m=1}^{\infty} \int_0^{2\pi} \frac{a^{-(m+n)}}{mn}e^{i(m-n)x}\,dx\\&= 4\pi\log^2 a + 2\pi \sum\limits_{m=1}^{\infty} \frac{a^{-2m}}{m^2} = 4\pi\log^2 a +2\pi \operatorname{Li}_2 (a^{-2})\end{align*}

Case-II: $a<1$

\begin{align*}I(a) &= 4\pi\log^2 a + \frac{1}{2}\log a\int_0^{2\pi} \log (1-2a^{-1}\cos x+a^{-2})\,dx+ I(a^{-1})\\&= 8\pi \log^2 a + 2\pi \operatorname{Li}_2 (a^2)\end{align*}

[Tolaso J Kos]

We'll use $\displaystyle \int_0^{2\pi} f\left(a+re^{\pm ix}\right)\,dx = 2\pi f(a)$ for $f$ analytic in $|z-a| < R$ where, $ 0< r < R$

Excellent. This is known as Gauss Mean Value Theorem for analytic functions. Nice use of it.