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For $n\in \Bbb{N}$, $$\int_0^{\pi/2} (\log \sin x)^n\text{ d}x=\frac{1}{2^{n+1}}B^{(n)}\left(\frac{1}{2},\frac{1}{2}\right)$$

Can we extend that result a bit further, to $n\in \Bbb{Q},n\gt 1?$

  • I don't know what $B^{(n)}$ is, but does the formula make sense for all $n\in Q$? Did you plug in some values to check if it still holds? – Bananach Feb 06 '16 at 07:17
  • $B^(n)$ is the derivative of the Beta function –  Feb 06 '16 at 07:21
  • It seems to me that $$\int_0^{\pi/2} (\log \sin x)^n\text{ d}x=\frac{1}{2^{n+1}}B^{(n)}\left(\frac{1}{2},\frac{1}{2}\right)$$ – Claude Leibovici Feb 06 '16 at 07:30
  • @ClaudeLeibovici Your totally right, mistake of writing the problem. –  Feb 06 '16 at 07:32

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Use the change of variables $t= \sin(x) $ to get $$ I = \int_0^1 \frac{\ln^n t} {\sqrt{1- t^2}} dt$$ Then make another change of variables $t^2=u$ and simplify . Then see my answer to finish the problem.

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