Prove about a integral with the special integral Beta $\mathbb{B}(\alpha,\beta)$.

Question

How can I prove this identity: \begin{eqnarray*} \int_{0}^{1}y^{a}\frac{y^{\alpha-1}(1-y)^{\beta-1}}{\mathbb{B}(\alpha,\beta)}dy=\frac{\Gamma(\alpha+a)\Gamma(\alpha+\beta)}{\Gamma({\alpha})\Gamma({\alpha+\beta+a})} \end{eqnarray*} for all $a$ such that $a+\alpha>0$ and $\alpha,\beta>0$.

My attempt: This problem it comes from the context of a probability and distribution theory problem that I am trying to solve. Where the random variable has a beta distribution. I know, that $$\mathbb{B}(\alpha,\beta)=\int_{0}^{1}y^{\alpha-1}(1-y)^{\beta-1}dy=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$ and can proved that $$\int_{0}^{1}y\left[\frac{y^{\alpha-1}(1-y)^{\beta-1}}{\mathbb{B}(\alpha,\beta)}\right]dy=\frac{\alpha}{\alpha+\beta}$$ but $\alpha \in \mathbb{R}$ so I can't use induction on $\alpha$. How can I prove that identity?

Answer 1

The integral can be rewritten as

$$\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int_0^1y^{2a-1}(1-y)^{b-1}dy=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\frac{\Gamma(2a)\Gamma(b)}{\Gamma(2a+b)}=\frac{\Gamma(a+b)\Gamma(2a)}{\Gamma(a)\Gamma(2a+b)}$$

Answer 2

If we were to move the Beta function outside, we would expect to obtain

$$ \int_0^1y^\alpha{y^{\alpha-1}(1-y)^{\beta-1}\over B(\alpha,\beta)}\mathrm dy={1\over B(\alpha,\beta)}\int_0^1y^{2\alpha-1}(1-y)^{\beta-1}\mathrm dy={B(2\alpha,\beta)\over B(\alpha,\beta)} $$

Now, apply the Gamma function representation for the Beta function, so that

$$ \int_0^1y^\alpha{y^{\alpha-1}(1-y)^{\beta-1}\over B(\alpha,\beta)}\mathrm dy={\Gamma(2\alpha)\Gamma(\alpha+\beta)\over\Gamma(2\alpha+\beta)\Gamma(\alpha)} $$