Beta Function: Prove $B(x,y) = B (x, y +1) + B (x+1 , y)$

Question

How can I prove the following $B(x,y) = B (x, y +1) + B (x+1 , y)$? Perhaps I can use Gamma functions to prove it? Does anyone know the proof?

Answer 1

$$B(x,y)= {\Gamma{(x)}\Gamma{(y)} \over {\Gamma(x+y)}}, \text{ and }\Gamma(r+1)=r\Gamma(r), \text{therefore,} $$

$$ B(x+1,y)= {\Gamma{(x+1)}\Gamma{(y)} \over {\Gamma(x+y+1)}} = \left({x\over{x+y}} \right){\Gamma{(x)}\Gamma{(y)} \over {\Gamma(x+y)}} = B(x,y)\left({x\over{x+y}} \right) $$

$$ B(x,y+1)= {\Gamma{(x)}\Gamma{(y+1)} \over {\Gamma(x+y+1)}} = \left({y\over{x+y}} \right){\Gamma{(x)}\Gamma{(y)} \over {\Gamma(x+y)}} = B(x,y)\left({y\over{x+y}} \right) $$

Therefore,

$$ B(x+1,y)+B(x,y+1)=B(x,y)\left({x\over{x+y}} + {y\over{x+y}}\right)=B(x,y) $$

Answer 2

just write down the RHS explicitly

$$B (x, y +1) + B (x+1 , y) = \int_0^1 t^{x-1}(1-t)^{y} + t^{x}(1-t)^{y-1}\,dt = \int_0^1 t^{x-1}(1-t)^{y-1}(1-t+t)\,dt = \int_0^1 t^{x-1}(1-t)^{y-1} = B (x, y) $$

Answer 3

$$\begin{align*}\operatorname{B}(x,y) & =\int\limits_0^1dt\, t^{x-1}(1-t)^{y-1}\\ & =\int\limits_0^1dt\, t^{x-1}(1-t)^{y-1}(1-t+t)\\ & =\int\limits_0^1dt\, t^{x-1}(1-t)^y+\int\limits_0^1dt\, t^x(1-t)^{y-1}\\ & =\operatorname{B}(x,y+1)+\operatorname{B}(x+1,y)\end{align*}$$