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Evaluating $\int_0^\pi\arctan\bigl(\frac{\ln\sin x}{x}\bigr)\mathrm{d}x$

In the first two solutions of the above post,why is the parameter value $\alpha\in(0,1)$ or $\mu\in(0,1)$ . The solution does seem to work for $\alpha\geqslant 1$ or $\mu\geqslant 1$ also.

Bumblebee
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shanrrg
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1 Answers1

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In the linked post, the first answer by @M.N.C.E. gave the answer.

Using your notation and his/her result $$I'(\alpha)=\frac {\pi^2}{2\alpha} \frac{1}{\log ^2\left(\frac{\alpha}{2}\right)+\frac{\pi ^2}{4}}$$ and $$\int_0^\pi \tan ^{-1}\left(\frac{\log (\alpha \sin (x))}{x}\right)\,dx=\pi \tan ^{-1}\left(\frac{2 }{\pi }\log \left(\frac{\alpha }{2}\right)\right)$$ which works $\forall \alpha >0$.

Claude Leibovici
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  • But i am not able to understand why @CuriousGuest & Felix Marin both assume the parameter to be in the range (0,1)? – shanrrg Apr 02 '23 at 06:36
  • @shanrrg I’ve looked at this problem for about a week now, and I can say with some certainty that the choice of the parameter is largely one of convenience rather than anything else. I think in one case, a Beta Function integral is used to prove the result, which hinges on the logarithm term being negative. This is only valid in that range. If the parameter is bigger than the one the logarithmic term changes sign and you have ti be mindful of that sign change throughout the integration. – Jessie Christian Jan 24 '25 at 03:16
  • However, the top solution listed, is valid for any value of the parameter that is positive. It can shown rigorously and it can also be shown without using Feynman’s trick at all. I think in general when you use Feynman’s trick, you really only care about the value of the parameter that gives you the integral you want, and one other value to determine the missing constant. – Jessie Christian Jan 24 '25 at 03:21