I am currently reading a book on Fractional Calculus. Using the left Riemann-Liouville integral definition
$$ {I}{_t^\alpha}f(t) = \frac{1}{\Gamma(\alpha)}\int_{a}^{t}(t-\tau)^{\alpha-1}f(\tau)d\tau ~\mathcal{Re}(\alpha) > 0$$
I am trying to prove that if $f(t) = (t-a)^{\beta-1}$ then we have
$$ {I}{_t^\alpha}f(t) = \frac{1}{\Gamma(\alpha)}\int_{a}^{t}(t-\tau)^{\alpha-1}(\tau-a)^{\beta-1}d\tau = \frac{\Gamma(\beta)}{\Gamma(\beta+\alpha)}(t-a)^{\beta+\alpha-1}$$
So far I have tried to use the substitution
$$ \theta= \frac{\tau-a}{t-a}$$ which led me to the follwing result
$$ \frac{(t-a)^{\alpha+\beta-1}}{\Gamma(\beta)}\int_{0}^{1}(1-\theta)^{\alpha-1}\theta^{\beta-1}d\theta $$
Either that's the right thing to do but I am stuck here or the substitution is wrong, to begin with. Does anyone have a solution?
Many thanks