I want to calculate the following integral :

$\displaystyle{\int^{\frac{\pi}{2}}_{0} \tan^{-1}\left(\frac{1}{2 \sin(x)}\right)} \text{ d}x$

But I don't how; I tried by subsituting $u = \frac{1}{2 \sin(x)}$ and $u = \tan^{-1}\left(\frac{1}{2 \sin(x)}\right)$ but it doesn't lead me anywhere.

Thanks for your help.


Well, the integral over $(0,2\pi)$ is clearly zero by symmetry, hence the given problem is equivalent to finding

$$ -\int_{0}^{\pi/2}\arctan\left(\frac{1}{2\sin x}\right)\,dx=-\int_{0}^{1}\frac{\arctan\frac{1}{2x}}{\sqrt{1-x^2}}\,dx=-\frac{\pi^2}{4}+\int_{0}^{1}\frac{\arctan(2x)}{\sqrt{1-x^2}}\,dx $$ or $$ -\frac{\pi^2}{4}+\int_{0}^{2}\int_{0}^{1}\frac{x}{(1+a^2 x^2)\sqrt{1-x^2}}\,dx\,da=-\frac{\pi^2}{4}+\int_{0}^{2}\frac{\text{arcsinh}(a)}{a\sqrt{1+a^2}}\,da$$ or $$ -\frac{\pi^2}{4}+\int_{0}^{\log(2+\sqrt{5})}\frac{u}{\sinh u}\,du =-\frac{\pi^2}{4}+\int_{1}^{2+\sqrt{5}}\frac{2\log v}{v^2-1}\,dv$$ where the last integral depends on the dilogarithm $\text{Li}_2$ evaluated at $\pm(\sqrt{5}-2)$:

$$ \int_{0}^{\pi/2}\arctan\left(\frac{1}{2\sin x}\right)\,dx= \text{arcsinh}\left(\tfrac{1}{2}\right)\text{arcsinh}(2)-\text{Li}_2(\sqrt{5}-2)+\text{Li}_2(2-\sqrt{5})$$

  • $\begingroup$ Thank you a lot ! 😉 $\endgroup$ – Skyost Nov 2 '18 at 11:43

Partial answer: Consider $f(y) = \int_{\pi/2}^{2\pi} \arctan\frac1{y\sin x}\,dx$. You are interested ultimately in $f(2)$. It appears that $f'(y)$ can be integrated/evaluated (pass the derivative through the integral, simplify, and do $u=\cos x$) easily enough. Can you then integrate this result to get back $f(y)$?

  • $\begingroup$ Simpler to say it than actually doing it, isn't it? $\endgroup$ – Jack D'Aurizio Nov 2 '18 at 3:21

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