TITLE: Defined integral of $x^{\alpha} (1-x)^{\beta-1}$
QUESTION [5 upvotes]: How can one prove the following equality?
$$\int_{0}^{1} x^{\alpha} (1-x)^{\beta-1} \,dx = \frac{\Gamma(\alpha+1) \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)}$$

REPLY [2 votes]: Function gamma : $\Gamma(z)=\int_0^{\infty}t^{z-1}e^{-t}dt$
Function beta : $B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt$
Your expression is $B(\alpha+1,\beta)=\frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(\alpha+\beta+1)}$

$$\Gamma(\alpha)\Gamma(\beta)=\int_0^{\infty}t^{\alpha-1}e^{-t}dt\int_0^{\infty}u^{\beta-1}e^{-u}du=\int_0^{\infty}\int_0^{\infty}t^{\alpha-1}u^{\beta-1}e^{-(t+u)}\,dt\,du$$
Change of variables $t=xy, u=x(1-y)$ 
With $(x,y)\in\ ]0,\infty[\times]0,1[$ for $(t,u)\in\ ]0,\infty[\times]0,\infty[$ and jacobian $|J|=x$.
(Rem: $x=t+u$ and $0\le y=t/(t+u)\le t/t\le 1$ since $u\ge 0$)
$$\Gamma(\alpha)\Gamma(\beta)=\int_0^{\infty}\int_0^1x^{\alpha-1}y^{\alpha-1}x^{\beta-1}(1-y)^{\beta-1}e^{-x}\,x\,dx\,dy=$$
$$\int_0^{\infty}\int_0^1[x^{\alpha-1}x^{\beta-1}e^{-x}\,x][y^{\alpha-1}(1-y)^{\beta-1}]\,dx\,dy=\quad$$
$$\int_0^{\infty}x^{\alpha+\beta-1}e^{-x}\,dx\int_0^1y^{\alpha-1}(1-y)^{\beta-1}\,dy=\Gamma(\alpha+\beta)B(\alpha,\beta)$$
There are Fubini invocations here and there but since everything is in separated variables, it works everywhere $\Gamma$ and $B$ are defined.