- **Complex Analysis**
(*http://mymathforum.com/complex-analysis/*)

- - **Gamma & betafunction**
(*http://mymathforum.com/complex-analysis/342839-gamma-betafunction.html*)

Gamma & betafunctionDear 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? |

$\displaystyle B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$. I don't see how to translate you question into this form, but it does seem close. |

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).? |

All times are GMT -8. The time now is 11:13 PM. |

Copyright © 2019 My Math Forum. All rights reserved.