Proving recurrence relation for the Beta function

Question

How can I prove the following: $$B(x+1,y)=B(x,y) \cdot \frac{x}{x+y}$$ Does anyone know the proof? I know the Beta function is related to the Gamma function, but I would like to prove it with partial integration. When I used partial integration, I got stuck.

Answer 1

Starting with the right hand of the equation: $$B(x,y)\frac{x}{x+y}=B(x,y)\frac{\frac{\Gamma(x+1)}{\Gamma(x)}}{\frac{\Gamma(x+y+1)}{\Gamma(x+y)}}=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\frac{\frac{\Gamma(x+1)}{\Gamma(x)}}{\frac{\Gamma(x+y+1)}{\Gamma(x+y)}}=\frac{\Gamma(x+1)\Gamma(y)}{\Gamma(x+y+1)}=B(x+1,y)$$ Here only the definition of the beta function $$B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$ and the relation between the gamma functions $$\Gamma(x+1)=\Gamma(x)x$$ were applied.

Answer 2

First euler beta function:

$B(z,w) = \int_{0}^{1}t^{z-1}(1-t)^{w-1}dt$

where $z, w \in \{z \space| \space Rez > 0\}$, substitute t with $t=\frac{s}{1+s}$

$B(z, w) =\int_0^{\infty}\frac{t^{z-1}\space \space dt}{(1+t)^{z+w}}$

Use the partial integration with $zB(z, w)=z\int_0^{\infty}\frac{t^{z-1}\space \space dt}{(1+t)^{z+w}}=\int_0^{\infty}\frac{dt^z}{(1+t)^{z+w}}$

Then $zB(z, w)=(z+w)B(z+1,w)$ is obtained