2022-01-26

Find: ${\int }_{0}^{\frac{\pi }{4}}x\mathrm{ln}\left(\mathrm{sin}x\right)dx$

Dominique Green

Expert

$\mathrm{ln}\left(\mathrm{sin}x\right)=-\mathrm{ln}2-\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(2nx\right)}{n}$
We can use the Fourier series and integrate by parts to obtain
$I\equiv {\int }_{0}^{\frac{\pi }{4}}x\left[-\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)\right]dx=\frac{{\pi }^{2}}{32}\mathrm{ln}\left(2\right)+{\frac{14}{\sum }}_{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}\left[\frac{\pi }{2}\mathrm{sin}\left(\frac{\pi }{2}n\right)-\frac{1}{n}\left(1-\mathrm{cos}\left(\frac{\pi }{2}n\right)\right)\right]$
$\mathrm{sin}\left(\frac{\pi }{2}n\right)$ is non-zero and alternating for odd n, while $\mathrm{cos}\left(\frac{\pi }{2}n\right)$ is non-zero and alternating for even n. Therefore,
$I=\frac{{\pi }^{2}}{32}\mathrm{ln}\left(2\right)+\frac{\pi }{8}\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{{\left(2k+1\right)}^{2}}-\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{3}}-\frac{1}{32}\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k-1}}{{k}^{3}}$
The first series is Catalan's constant G, the second one is $\zeta \left(3\right)$ and the third one is $\eta \left(3\right)=\frac{34}{\zeta }\left(3\right)$ (with the Riemann zeta function $\zeta$ and the Dirichlet eta function $\eta$), so we obtain
$I=\frac{{\pi }^{2}}{32}\mathrm{ln}\left(2\right)+\frac{\pi }{8}G-\frac{35}{128}\zeta \left(3\right)$