# Evaluating $\int_0^{\pi/2} \frac{t \ln (1-\sin{t})}{\sin t} dt$

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 $$-\pi^3/8$$.

• 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? – Ethan Bolker Nov 16 '19 at 19:31
• You may get some inspirations here – A.Γ. Nov 16 '19 at 20:16
• 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. – Nap D. Lover Nov 16 '19 at 20:30

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