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$$\int_0^{\pi/2}\int_0^{\pi/2}\left(\frac{\sin\phi}{\sin\theta}\right)^{1/2}\,d\theta\,d\phi=\pi$$ Indeed, I tried to solve this integral by complexifying (using Euler's formula) the $\sin\theta$ and $\sin\phi$.But it didn't work because I faced the exponent which would make things difficult to tackle such integral.

I would appreciate any suggestions for solving this integral.

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    $\begingroup$ More generally, for every $|a|<1$, $$\int_0^{\pi/2}\int_0^{\pi/2}\left(\frac{\sin\phi}{\sin\theta}\right)^{a}\,d\theta\,d\phi=\frac{\pi}{2a} \tan\left(\frac{\pi a}2\right) $$ $\endgroup$ – Did Nov 28 '16 at 21:52
  • $\begingroup$ I am looking for a simpler solution @Did $\endgroup$ – FreeMind Nov 28 '16 at 22:02
  • $\begingroup$ "Simpler" than what? And how? $\endgroup$ – Did Nov 28 '16 at 22:05
  • $\begingroup$ @Did Simpler answer. Another way than using beta function. I know there should be a simpler answer. $\endgroup$ – FreeMind Nov 28 '16 at 22:06
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    $\begingroup$ @Did No, I didn't know the solution. Actually I made several attempts but I failed. The one who gave me the question told me that it can be solved without the knowledge of special functions like Beta. $\endgroup$ – FreeMind Nov 28 '16 at 22:10
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One may use a classic integral representation of the Euler beta function $$ \int_0^{\pi/2}\sin^a(x)\:dx=\frac{\Gamma\left(\frac12\right) \Gamma\left(\frac12+\frac{a}2\right)}{2\,\Gamma\left(1+\frac{a}2\right)} $$ giving $$ \int_0^{\pi/2}\sqrt{\sin(\phi)}\:d\phi=\frac{\Gamma\left(\frac12\right) \Gamma\left(\frac34\right)}{2\,\Gamma\left(\frac54\right)},\quad\int_0^{\pi/2}\frac1{\sqrt{\sin(\theta)}}\:d\theta=\frac{\Gamma\left(\frac12\right) \Gamma\left(\frac14\right)}{2\,\Gamma\left(\frac34\right)} $$ and $$ \int_0^{\pi/2}\sqrt{\frac{\sin(\phi)}{\sin(\theta)}}\:d\phi\:d\theta=\Gamma\left(\frac12\right)\cdot\Gamma\left(\frac12\right)=\pi $$ as announced.

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  • $\begingroup$ Thanks for your response. I am also interested in other approaches if available. $\endgroup$ – FreeMind Nov 28 '16 at 21:36

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