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In a problem in scattering theory, this integral arises: $$\displaystyle{\int\limits_0^{\pi/2} \frac{t \ln (1-\sin{t})}{\sin t} dt}$$ I have tried a number of approaches to evaluating the integral, which I suspect has a closed form solution. The reason is that I generated a numerical value for the integral, $-3.87578458503\ldots$ and after a bit of numerical exploration I found this to agree with $-\pi^3/8$.

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  • $\begingroup$ Interesting. Have you tried Wolfram Alpha? Do you have any reason from physics to suspect a power of $\pi$? Are there related scattering theory problems that lead to similar integrals? Doe the appearance of $\pi$ matter, or is all you need the numerical value? $\endgroup$ – Ethan Bolker Nov 16 '19 at 19:31
  • $\begingroup$ You may get some inspirations here $\endgroup$ – A.Γ. Nov 16 '19 at 20:16
  • $\begingroup$ Any chance you can (briefly) describe or reference the problem in scattering theory that leads to this integral? Just curious to see it arise in context. $\endgroup$ – Nap D. Lover Nov 16 '19 at 20:30
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First set $\sin t=u$ then $u=\frac{2x}{1+x^2}$ and use the fact that $\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\tan^{-1}(x)$ we get

$$\mathcal{I}=\int_0^{\pi/2}\frac{t\ln(1-\sin t)}{\sin t}dt=\int_0^1\frac{\sin^{-1}(u)\ln(1-u)}{u\sqrt{1-u^2}}du\\=-2\int_0^1\frac{\tan^{-1}(x)}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=-2\left(\frac{\pi^3}{16}\right)=-\frac{\pi^3}{8}$$

where the last integral is proved here

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