# How to show that $\int_0^{\pi/2}\int_0^{\pi/2}\left(\frac{\sin\phi}{\sin\theta}\right)^{1/2}\,d\theta\,d\phi=\pi$?

$$\int_0^{\pi/2}\int_0^{\pi/2}\left(\frac{\sin\phi}{\sin\theta}\right)^{1/2}\,d\theta\,d\phi=\pi$$ 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.

• More generally, for every $|a|<1$, $$\int_0^{\pi/2}\int_0^{\pi/2}\left(\frac{\sin\phi}{\sin\theta}\right)^{a}\,d\theta\,d\phi=\frac{\pi}{2a} \tan\left(\frac{\pi a}2\right)$$ – Did Nov 28 '16 at 21:52
• I am looking for a simpler solution @Did – FreeMind Nov 28 '16 at 22:02
• "Simpler" than what? And how? – Did Nov 28 '16 at 22:05
• @Did Simpler answer. Another way than using beta function. I know there should be a simpler answer. – FreeMind Nov 28 '16 at 22:06
• @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. – FreeMind Nov 28 '16 at 22:10

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.