Convolution integral of the $t^{\alpha}$ and $t^{\beta}$, where $\alpha>-1$ and $\beta>-1$

Question

I would like to evaluate integral $$ \int\limits_{0}^{t} (t-\tau)^{\alpha}\tau^{\beta}d\tau \ \ \ \ \ \ \ \ (\star) $$ where $t>0$, $\alpha>-1$, $\beta>-1$.

I did it in the following way. By binomial formula I have: $$ (t-\tau)^{\alpha} = \sum\limits_{k = 0}^{\infty} (-1)^k\binom{\alpha}{k} t^{\alpha - k}\tau^k $$ I put it into $(\star)$ and obtain after some mainipulations: $$ \int\limits_{0}^{t} (t-\tau)^{\alpha}\tau^{\beta}d\tau = \text{constant} \cdot t^{\alpha + \beta + 1} $$ where $$ \text{constant} = \sum\limits_{k=0}^{\infty} \frac{(-1)^k\binom{\alpha}{k}}{\beta + k + 1} $$ On the other hand, when I compute $(\star)$ using some software for symbolic calculations I obtained: $$ \int\limits_{0}^{t} (t-\tau)^{\alpha}\tau^{\beta}d\tau = \frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}t^{\alpha + \beta + 1}\ \ \ \ \ \ (\star\star) $$ Ths means that: $$ \sum\limits_{k=0}^{\infty} \frac{(-1)^k\binom{\alpha}{k}}{\beta + k + 1} = \frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)} $$

My question: How to prove it?

I have also two additional questions:

1) Could I use integration term by term binomial series?

2) Is it any simpler method to obtain $(\star\star)$ without using software for symbolic calculations?

Answer 1

1. Yes, termwise integration is valid here, the argument is monotone convergence theorem (the terms of this binomial series eventually stop changing signs). In addition, it's not hard to see (using Stirling's asymptotics) that the terms of the resulting series are $\mathcal{O}(k^{-2-\alpha})$, thus it converges (as expected).
2. I don't think so. The series is inherently linked with the integral, so each possible proof of this must be a step away from a proof of the expression of $\mathrm{B}$-function in terms of $\Gamma$-function.