Dayton Burnett

2022-01-23

Verifing ${\int }_{0}^{\pi }x\mathrm{ln}\left(\mathrm{sin}x\right)dx=-\mathrm{ln}\left(2\right)\frac{{\pi }^{2}}{2}$

pacetfv

Expert

Let $I={\int }_{0}^{\pi }x\mathrm{ln}\left(\mathrm{sin}x\right)dx$. Then, changing variables $x\to \pi -x$:
$I={\int }_{0}^{\pi }\left(\pi -x\right):$
Therefore:
$I=\frac{\pi }{2}{\int }_{0}^{\pi }\mathrm{ln}\left(\mathrm{sin}x\right){dx}^{\text{symmetry}}=\pi {\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(\mathrm{sin}x\right)dx$
The latter integral had been solved elsewhere.

Frauental91

Expert

This answer is meant to offer an alternative from the last integral of Sashas

RizerMix

Expert

For what is worth, the general case of what Sasha did is If $I={\int }_{0}^{\pi }xf\left(\mathrm{sin}x\right)dx$ then $I=\frac{\pi }{2}{\int }_{0}^{\pi }f\left(\mathrm{sin}x\right)dx$ and the argument is completely analogous.