Is it true
$$ \frac{1}{\Gamma(1-\nu)}\frac{1}{\Gamma(\nu)} \int\_{0}^{x}(x-y)^{-\nu}dy\int\_0^y (y-t)^{\nu-1}f(t)dt = \int\_0^x f(u)du$$
for any continuous function $f(x)$ such that $f(0)=0$ and $0<\nu<1$?



How to validate the exponentiality of fractional calculus?
Without loss of generality we may assume $x>0$. Then also $0<t<x$ and I can use the identity
$$\int\_t^x(y-t)^{\nu-1}(x-y)^{-\nu}\,dy=\frac{\pi}{\sin\pi \nu}=\Gamma(1-\nu)\Gamma(\nu),$$
valid for $0<\nu<1$, $0<x<t$, to conclude that
$$\frac{1}{\Gamma(1-\nu)}\frac{1}{\Gamma(\nu)} \int\_{0}^{x}(x-y)^{-\nu}\,dy\int\_0^y (y-t)^{\nu-1}f(t)\,dt = \int\_0^x f(t)dt.$$
The assumption $f(0)=0$ is not needed.


