Beta-gamma integrals

Question

I know that $B (m,n) =\frac{\Gamma (m) \Gamma (n)}{\Gamma(m+n)}$.

But $B (m,m)$ = $2^{1-2m} B(m, 1/2) $, I didn't get how this equation arrived, I tried applying the formula but cant obtain all the terms.

http://mathworld.wolfram.com/LegendreDuplicationFormula.html and http://math.stackexchange.com/questions/34740/proving-and-deriving-a-gamma-function — Noam Shalev - nospoon, Jul 14 '15 at 05:55

score 2 · Accepted Answer · answered Jul 14 '15 at 05:55

2

Using the duplication formula $\Gamma(x) \Gamma(x+1/2) = 2^{1-2x} \Gamma(1/2) \Gamma(2x)$ found in the Wiki page it can be seen that \begin{align} B(m,m) &= \frac{\Gamma(m) \Gamma(m)}{\Gamma(2m)} = \frac{\Gamma(m) \, 2^{1-2m} \Gamma(1/2)}{\Gamma(m+1/2)} = 2^{1-2m} \, B\left(m , \frac{1}{2}\right). \end{align}

answered Jul 14 '15 at 05:55

Leucippus

26,329

Well done! a +1. – Mark Viola Jul 14 '15 at 06:28

Beta-gamma integrals

1 Answers1