equation for the beta function

Question

Using only the definition

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

for the Beta function, proof the term: $(x + y)B(x + 1, y) = xB(x, y) \space\space \forall x, y > 0$ .

Thanks in advance! I've tried rewriting the expression and getting the (x + y) of the first term into the integral, but that didn't lead anywhere so far.

Answer 1

Show that from the definition of a Gamma function:

$$ \Gamma(x) = \int_0^\infty t^{x-1}e^{-t}dt\,. $$

that $\Gamma(x) \Gamma(y) = \Gamma(x+y) B(x,y)$. Then recall that $\Gamma(x+1) = x\Gamma(x)$ and the result will follow.

Answer 2

First note that

$$\begin{align} B(x+1,y)+B(x,y+1)&=\int_0^1 \left(t^x(1-t)^{y-1}+t^{x-1}(1-t)^y\right) dt\\\\ &=\int_0^1\left( t^{x-1}(1-t)^{y-1}\left[t+(1-t)\right] \right)dt\\\\ &=B(x,y) \end{align}$$

Then,

$$\begin{align} yB(x+1,y)&= y\int_0^1 t^{x} (1-t)^{y-1}dt\\\\ &=-\int_0^1 t^x \frac{d(1-t)^{y}}{dt} dt \,\,\,\,\ \dots \text{using} \,\,\frac{d(1-t)^{y}}{dt}=-y\,(1-t)^{y-1}\\\\ &=\int_0^1 \frac{dt^{x}}{dt}(1-t)^{y}dt\,\,\,\,\ \dots \text{using integration by parts with}\,\, u=t^x\,\,\text{and}\,\,v=(1-t)^y\\\\ &=x\int_0^1 t^{x-1}(1-t)^{y}dt \\\\ &=xB(x,y+1)\\\\ &=x\left(B(x,y)-B(x+1,y)\right) \end{align}$$

whereupon we get the desired result!