Defined integral of $x^{\alpha} (1-x)^{\beta-1}$ How can one prove the following equality?
$$\int_{0}^{1} x^{\alpha} (1-x)^{\beta-1} \,dx = \frac{\Gamma(\alpha+1) \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)}$$
 A: We show that using Convolution 
$$\beta(x+1,y+1)=\int^{1}_{0}t^{x}\, (1-t)^{y}\,dt= \frac{\Gamma(x+1)\Gamma
{(y+1)}}{\Gamma{(x+y+2)}}$$ 
$$proof$$
Let us choose some functions $f(t) = t^{x} \,\, , \, g(t) = t^y$
Hence we get 
$$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds $$ 
So by definition we have 
$$\mathcal{L}\left(t^x*t^y\right)= \mathcal{L}(t^x) \mathcal{L}(t^y ) $$
We can now use the laplace of the power 
$$\mathcal{L}\left(t^x*t^y\right)= \frac{x!\cdot y!}{s^{x+y+2}}$$
Notice that we need to find the inverse of Laplace $\mathcal{L}^{-1}$
$$\mathcal{L}^{-1}\left(\mathcal{L}(t^x*t^y)\right)=\mathcal{L}^{-
1}\left( \frac{x!\cdot y!}{s^{x+y+2}}\right)=t^{x+y+1}\frac{x!\cdot y!}
{(x+y+1)!}$$
So we have the following 
$$(t^x*t^y) =t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$$
By definition we have 
$$t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!} = \int^{t}_0 s^{x}(t-s)^{y}\,ds
$$
This looks good ,  put $t=1$ we get  
$$\frac{x!\cdot y!}{(x+y+1)!} = \int^{1}_0 s^{x}(1-s)^{y}\,ds$$
By using that $n! = \Gamma{(n+1)}$
We arrive happily to our formula  
$$ \int^{1}_0 s^{x}(1-s)^{y}\,ds= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma
{(x+y+2)}}$$
which can be written as 
$$ \int^{1}_0 s^{x-1}(1-s)^{y-1}\,ds= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma
{(x+y+1)}}$$
A: Let us prove the symmetrical form
$$ f(\alpha,\beta)=\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}\,dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\tag{1} $$
for any positive $\alpha,\beta$. $f(\cdot,\beta)$ and $f(\alpha,\cdot)$ are continuous functions and moments, in particular they are log-convex functions by the Cauchy-Schwarz inequality. Clearly $f(\alpha,\beta)=f(\beta,\alpha)$ (due to the substitution $x\mapsto 1-x$) and $f(\alpha,1)=\frac{1}{\alpha}$.
By integration by parts
$$ f(\alpha+1,\beta)= \frac{\alpha}{\beta}\cdot f(\alpha,\beta+1) \tag{2}$$
hence, by induction, $(1)$ holds for any $\alpha\in\mathbb{N}^*$ or $\beta\in\mathbb{N}^*$. By log-convexity and the Bohr-Mollerup theorem it follows that $(1)$ holds for any positive $\alpha,\beta$. With a little extra effort, it is not difficult to show that the same holds for any $\alpha,\beta$ with positive real part.
A: 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.
