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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

jean-francois Niglio
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    If you change the argument of Gamma-function in your last expression (writing $\Gamma(\alpha)$ instead of $\Gamma(\beta)$ (typo?) and given that $\int_{0}^{1}(1-\theta)^{\alpha-1}\theta^{\beta-1}d\theta=B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ - you got the correct result – Svyatoslav May 18 '21 at 11:07
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    Hi good morning thanks. I just realized that this morning when I got up. Thanks for your message though. Appreciate it. Sorry about the typo. – jean-francois Niglio May 19 '21 at 02:33

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