How to evaluate $\int_0^\pi(\ln(\sin(x)))^2 \, dx$

Question

To find: $$\int_0^\pi(\ln(\sin(x)))^2 \, dx$$ Tried changing limits and adding, also tried using complex numbers but failed.

Answer 1

write $\sin(x)^a=e^{a\log(\sin(x))}$

Then $$ I=\partial_a^2 \left. \int_0^\pi dx\sin^a(x) \right|_{a=0} $$

This integral can be evaluated in terms of Euler's Beta function and yields after simplification

$$ I=\sqrt{\pi }\partial_a^2 \left. \left(\frac{ \Gamma \left(\frac{a+1}{2}\right)}{\Gamma \left(\frac{a}{2}+1\right)}\right) \right|_{a=0} $$

calculating the derivatives and taking the limit as $a\rightarrow 0$ is a little bit exhausting, but finally yields an result (best use a CAS),

$$ I=\frac{\pi ^3}{12}+\pi \log ^2(2) $$

Essentially the limit follows from the fact that: $$ \psi_0\left(\frac{1}{2}\right)=-\gamma-2\log(2),\quad \psi_1\left(\frac{1}{2}\right)=\frac{\pi^2}{2},\quad \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} $$

and

$$ \partial_x \Gamma(x)=\Gamma(x) \psi_0(x), \quad \partial_x \psi_0(x)=\psi_1(x) $$

where $ \psi_{0/1}(x)$ is the Di/Trigamma function and $\gamma$ is the Euler-Mascheroni constant.

Answer 2

Maybe a faster way to recover tired's result is to apply Parseval's identity to the Fourier series of $\log\sin x$.