Prove an equation about fractional integral I'm doing this exercise from Real Analysis of Folland and got stuck on this problem. Can anyone help me solve this? I really appreciate.


If $f$ is continuous on $[0, \infty)$, for $\alpha \gt 0$ and $x \ge 0$, let $$I_{\alpha}f(x) = \frac{1}{\Gamma(\alpha)}\int_{0}^{x}(x-t)^{\alpha -1}f(t)dt$$$I_{\alpha}f$ is called the $\alpha th$ fractional integral of $f$, here $\Gamma(x) = \int_{0}^{\infty}t^{x-1}e^{-t}dt$. Prove that $I_{\alpha + \beta}f = I_{\alpha}(I_{\beta}f)$ for all $\alpha, \beta \gt 0$.


I know the related Gamma equation: $\Gamma(x)\Gamma(y)/\Gamma(x+y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$, but I still can't solve this.
Thanks so much.

This is just direct computation
\begin{align*}
I_\alpha(I_\beta f) &= \frac{1}{\Gamma(\alpha)}\int_0^x(x-t)^{\alpha-1}I_\beta f dt\\
&=\frac{1}{\Gamma(\alpha)}\int_0^x(x-t)^{\alpha-1}\frac{1}{\Gamma(\beta)}\int_0^t (t-s)^{\beta-1}f(s)dsdt\\
&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_0^x\int_0^t (x-t)^{\alpha-1}(t-s)^{\beta-1}f(s) dsdt\\
&= \frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_0^x\int_s^x(x-t)^{\alpha-1}(t-s)^{\beta-1}f(s)dtds\\
&= \frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_0^x\int_0^1 (x-s)^{\alpha-1}(1-r)^{\alpha-1}(x-s)^{\beta-1}r^{\beta-1}f(s)(x-s)drds\\
&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_0^x(x-s)^{\alpha+\beta-1}f(s)\int_0^1(1-r)^{\alpha-1}r^{\beta-1}drds\\
&=\frac{1}{\Gamma(\alpha+\beta)}\int_0^x(x-s)^{\alpha+\beta-1}f(s)ds\\
&=I_{\alpha+\beta}f(x),
\end{align*}
where the fourth equality follows from the Fubini and the fifth equality uses substitution $r= (t-s)/(x-s)$.
