$\alpha_1, \alpha_2 \gt0$ Tenemos
$$ \int_{0}^1 u ^ {\alpha_1-1} (1-u) ^ {\alpha_2-1} \, \mathrm du = \text{B}(\alpha_1,\alpha_2) \tag{1} $$
Donde $\text{B}(a,b)$ es la Función Beta $$\begin{align} \Gamma(\alpha_1) & =\int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \tag{2}\ \Gamma(\alpha_2) & =\int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y \tag{3}\ \end {Alinee el} $$
$$\begin{align} \Gamma(\alpha_1)\Gamma(\alpha_2) & = \int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y \tag{4}\ & =\int\limits_0^\infty\int\limits_0^\infty\ e^{-x-y} x^{\alpha_1-1}y^{\alpha2-1}\,\mathrm{d}x \,\mathrm{d}y \tag{5}\ & =\int{z=0}^\infty\int_{t=0}^1 e^{-z} \Big(zt\Big)^{\alpha_1-1}\Big(z(1-t)\Big)^{\alpha2-1}z\,\mathrm{d}z \,\mathrm{d}t \tag{6}\ & =\int{z=0}^\infty\int_{t=0}^1 e^{-z} z^{\alpha_1+\alpha_2-1}t^{\alpha_1-1}(1-t)^{\alpha2-1}\,\mathrm{d}z \,\mathrm{d}t \tag{7}\ & =\int{0}^\infty e^{-z}z^{\alpha_1+\alpha2-1} \,\mathrm{d}z\int{0}^1t^{\alpha_1-1}(1-t)^{\alpha_2-1}\,\mathrm{d}t \tag{8}\ \Gamma(\alpha_1)\Gamma(\alpha_2) & =\Gamma(\alpha_1+\alpha_2){\rm B}(\alpha_1,\alpha_2) \tag{9}\ \end {Alinee el} $$
Por lo tanto
$$\large{\rm B}(\alpha_1,\alpha2)=\int{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, du =\frac{\Gamma(\alpha_1)\,\Gamma(\alpha_2)}{\Gamma(\alpha_1+\alpha_2)}$$
Sustitución de % de $\text{ Explanation: } (6) $ $x=zt $y $ y=z(1-t)$
