My Math Forum Gamma & betafunction

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 November 17th, 2017, 04:36 AM #1 Newbie   Joined: Feb 2011 Posts: 24 Thanks: 0 Gamma & betafunction Dear all, I am stuck in the follow equation: Show that the integral: 0 to 1 of 1/(1-t^2)^2/3 * dt = 2^(-4/3) Betafunction (1/3, 1/3) I tried to use a substitution t = sin z, but that didn't make it easer. Can somebody please help me?
 November 17th, 2017, 12:44 PM #2 Global Moderator   Joined: May 2007 Posts: 6,822 Thanks: 723 $\displaystyle B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$. I don't see how to translate you question into this form, but it does seem close.
 November 17th, 2017, 01:42 PM #3 Newbie   Joined: Feb 2011 Posts: 24 Thanks: 0 I found the following substitution Let t^2 =x, then t= sqrt(x), so dt = 0.5*t^-0,5 *dt Plugging this in: we get: 0.5 * [Integral from 0 to 1 of (x^-0.5) * (1-x)^-2/3 * dx] But this is 0.5* B(0.5, 1/3). How to get to 2^(-4/3) * B(1/3, 1/3).?

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