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mathpieuler November 17th, 2017 04:36 AM

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?

mathman November 17th, 2017 12:44 PM

$\displaystyle B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$.

I don't see how to translate you question into this form, but it does seem close.

mathpieuler November 17th, 2017 01:42 PM

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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