# Beta function?

• Feb 10th 2009, 12:55 PM
fardeen_gen
Beta function?
Prove that:
$n\:\int_{0}^{1} x^m \:(1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}$
Is this correct?

L.H.S
= n.B(m, n -1) , where B(x, y) is Beta function
= n. m!(n-1)!/((m + n - 1 + 1)!)
= n(n-1)!m!/((m+n)!)
= R.H.S
• Feb 10th 2009, 03:22 PM
NonCommAlg
Quote:

Originally Posted by fardeen_gen
Prove that:
$n\:\int_{0}^{1} x^m \:(1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}$
Is this correct?

L.H.S
= n.B(m, n -1) , where B(x, y) is Beta function
= n. m!(n-1)!/((m + n - 1 + 1)!)
= n(n-1)!m!/((m+n)!)
= R.H.S

that's not a proof! you're using what you're supposed to prove! do it by induction: fix an $m$ and do the induction over $n.$ nothing to prove for n = 1. so suppose the claim is true for $n \geq 1.$

let $I(m,n)=\int_0^1 x^m (1-x)^{n-1} \ dx.$ since $x^m(1-x)^{n-1}=x^{m+1}(1-x)^{n-1} + x^m(1-x)^n,$ we have: $I(m,n)=I(m+1,n)+I(m,n+1). \ \ \ \ \ \ (1)$

on the other hand by integration by parts: $I(m+1,n)=\frac{m+1}{n}I(m,n+1),$ which with (1) gives us: $I(m,n+1)=\frac{n}{m+n+1}I(m,n)=\frac{1}{m+n+1} \cdot \frac{n!m!}{(m+n)!}=\frac{1}{n+1} \cdot \frac{(n+1)!m!}{(m+n+1)!}. \ \Box$