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November 17th, 2017, 05:36 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/(1t^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, 01:44 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622 
$\displaystyle B(x,y)=\int_0^1 t^{x1}(1t)^{y1}dt$. I don't see how to translate you question into this form, but it does seem close. 
November 17th, 2017, 02:42 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) * (1x)^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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betafunction, gamma, gammafunction 
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