Integrals of $\log^2(\sin x)$ and $\log^3(\sin x)$ I have to evaluate the integral $$\int_{0}^{\frac{\pi}{2}}\log(\sin(x))^2\,dx$$ What about $\int_{0}^{\frac{\pi}{2}}\log(\sin(x))^3dx$ and the other powers?
 A: It is well-known that
$$\log(\sin\theta) = \log(2)+\sum_{k\geq 1}\frac{\cos(2k\theta)}{k}\tag{1}$$
hence by Parseval's theorem
$$ \int_{0}^{\pi/2}\log^2(\sin\theta)\,d\theta = \frac{\pi}{2}\log^2(2)+\frac{\pi}{4}\zeta(2).\tag{2}$$

In order to compute $\int_{0}^{\pi/2}\log^3(\sin\theta)\,d\theta$ through the same approach (orthogonality), we have to compute
$$ \int_{0}^{\pi/2}\cos(2ax)\cos(2bx)\cos(2cx)\,dx $$
and that leads to a more complicated series. However,
$$ \int_{0}^{\pi/2}\log^3(\sin\theta)\,d\theta = \int_{0}^{1}\frac{\log^3(u)}{\sqrt{1-u^2}}\,du = \left.\frac{d^3}{d\alpha^3}\int_{0}^{1}\frac{x^\alpha}{\sqrt{1-x^2}}\,dx\right|_{\alpha=0} $$
and the last integral can be computed through Euler's Beta function. So we have:
$$ I_k = \int_{0}^{\pi/2}\log^k(\sin\theta)\,d\theta = \left.\frac{d^k}{d\alpha^k}\frac{\sqrt{\pi}\,\Gamma\left(\frac{1}{2}+\frac{\alpha}{2}\right)}{2\,\Gamma\left(1+\frac{\alpha}{2}\right)}\right|_{\alpha=0}\tag{3} $$
and if $k=3$:
$$ \int_{0}^{\pi/2}\log^3(\sin\theta)\,d\theta = \color{red}{-\frac{\pi}{8}  \left(\pi ^2 \log 2+4\log^3 2+6\, \zeta(3)\right)}.\tag{4} $$
