Prove that $\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, {\rm d}u =\frac{\Gamma(\alpha_1)\Gamma(\alpha_2)} {\Gamma(\alpha_1+\alpha_2)}$ I have the following equality in a textbook of mine
$$\frac{y^{\alpha_1+\alpha_2-1} e^{-y/\beta}}{\Gamma(\alpha_1+\alpha_2) \beta^{\alpha_1+\alpha_2}} \cdot \frac{\Gamma(\alpha_1+\alpha_2)}{\Gamma(\alpha_1)\Gamma(\alpha_2)} \int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du = \frac{y^{\alpha_1+\alpha_2-1} e^{-y/\beta}}{\Gamma(\alpha_1+\alpha_2) \beta^{\alpha_1+\alpha_2}}$$
and I see that for the equality to be true we must have 

$$\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du =\frac{\Gamma(\alpha_1)\Gamma(\alpha_2)} {\Gamma(\alpha_1+\alpha_2)}$$

However can someone give me an explanation why this is the case ?
 A: For $\alpha_1, \alpha_2 \gt0$ We have 
$$\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du = \text{B}(\alpha_1,\alpha_2) \tag{1}
$$
Where $\text{B}(a,b)$ is Beta Function
$$
\begin{align}
\Gamma(\alpha_1)
&
=\int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x
\tag{2}\\
\Gamma(\alpha_2)
&
=\int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y 
\tag{3}\\
\end{align}
$$
$$
\begin{align}
\Gamma(\alpha_1)\Gamma(\alpha_2) 
&
= \int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y
\tag{4}\\
&
=\int\limits_0^\infty\int\limits_0^\infty\ e^{-x-y} x^{\alpha_1-1}y^{\alpha_2-1}\,\mathrm{d}x  \,\mathrm{d}y
\tag{5}\\
&
=\int_{z=0}^\infty\int_{t=0}^1 e^{-z} \Big(zt\Big)^{\alpha_1-1}\Big(z(1-t)\Big)^{\alpha_2-1}z\,\mathrm{d}z  \,\mathrm{d}t
\tag{6}\\
&
=\int_{z=0}^\infty\int_{t=0}^1 e^{-z} z^{\alpha_1+\alpha_2-1}t^{\alpha_1-1}(1-t)^{\alpha_2-1}\,\mathrm{d}z  \,\mathrm{d}t
\tag{7}\\
&
=\int_{0}^\infty  e^{-z}z^{\alpha_1+\alpha_2-1} \,\mathrm{d}z\int_{0}^1t^{\alpha_1-1}(1-t)^{\alpha_2-1}\,\mathrm{d}t
\tag{8}\\
\Gamma(\alpha_1)\Gamma(\alpha_2) 
&
=\Gamma(\alpha_1+\alpha_2){\rm B}(\alpha_1,\alpha_2)
\tag{9}\\
\end{align}
$$
Hence

$$\large{\rm B}(\alpha_1,\alpha_2)=\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, du =\frac{\Gamma(\alpha_1)\,\Gamma(\alpha_2)}{\Gamma(\alpha_1+\alpha_2)}$$


$\text{ Explanation: } (6) $
Substituting $x=zt $ and  $ y=z(1-t)$
