1
$\begingroup$

I'm trying to prove: $$ \int_0^{\pi/2} (\cos \theta)^{2m-1}(\sin \theta)^{2n-1}d\theta =\dfrac {\Gamma (m)\Gamma (n)}{2\Gamma (m+n)} $$ Could you help me do that?

$\endgroup$
  • $\begingroup$ what do you know on the subject ? what did you try? $\endgroup$ – G Cab Nov 17 '16 at 18:57
  • $\begingroup$ I tried to convert polar coordinates to catesian and i got some result but I couldn't prove that exactly! $\endgroup$ – H.H Nov 17 '16 at 19:02
2
$\begingroup$

It's Beta function. Make a change of variable. Take $sin\theta = x$, then, $cos\theta = \sqrt{1-x^{2}}$. And the derivative $d\theta = \frac{1}{\sqrt{1-x^{2}}}$, and $\frac{\pi}{2}$ will become one. Than you will have form of $\int_{0}^{1}(1-x^{2})^{m-1}x^{2n-1}dx$. And make one more change of variable. $x^2 := t$. Than you will have $dx = \frac{1}{2}t^{-\frac{1}{2}}dt$. And you will have $$\int_{0}^{1}\frac{1}{2}t^{-\frac{1}{2}} (1-t)^{m-1}t^{n-\frac{1}{2}}dt=\frac{1}{2} \int_{0}^{1}(1-t)^{m-1}t^{n-1}dt= \frac{1}{2}B(m,n)= \frac{\Gamma(n)\Gamma(m)}{2\Gamma(n+m)}$$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you so much but I don't want to use beta function. Just trying to prove that integral itself! $\endgroup$ – H.H Nov 17 '16 at 19:33
  • $\begingroup$ In another way: $$ \eqalign{ & {\rm B}(x,y) = 2\int_{\theta \, = \,0}^{\pi /2} {\cos (\theta )^{\,2\,x - 1} \sin (\theta )^{\,2\,y - 1} d\theta } = \cr & = 2\int_{\theta \, = \,0}^{\pi /2} {\cos (\theta )^{\,2\,x - 2} \sin (\theta )^{\,2\,y - 2} \cos (\theta )\sin (\theta )d\theta } = \cr & = \int_{\theta \, = \,0}^{\pi /2} {\cos (\theta )^{\,2\,x - 2} \sin (\theta )^{\,2\,y - 2} d\left( {\sin (\theta )^{\,2\,} } \right)} = \cr & = \int_{t\, = \,0}^{\,1} {\left( {1 - t} \right)^{\,x - 1} t^{\,\,y - 1} dt} \cr} $$ $\endgroup$ – G Cab Nov 17 '16 at 23:18
  • $\begingroup$ @H.H: so you want to connect the definition of Beta as ratio of Gamma with the integral definition ? Then see e.g. this link $\endgroup$ – G Cab Nov 17 '16 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.