Proof that Beta-function $B(m,n)$ = $\frac{n-1}{m}B(m+1,n-1)$?

Question

When m and n are positive integers.

It probably has to do with the incomplete Beta-function $B_{sin^2(x)}(m,n)$.

Answer 1

By definition, we have $$ B(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx, \qquad n,\,m=1,2,\cdots. $$ If we integrate by parts, we get

$$ \begin{align} B(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx &=\left. \frac{x^m}m(1-x)^{n-1}\right|_0^1+\frac{n-1}m\int_0^1x^m(1-x)^{n-2}dx\\\\ &=0+\frac{n-1}mB(m+1,n-1)\\\\ &=\frac{n-1}mB(m+1,n-1)\qquad m\geq1,\,n\geq2, \end{align} $$

as announced.

Answer 2

This is easily demonstrated using the integral definition of the Beta-function and then integrating by parts.