The Riemann Liouville integral is defined as: $\frac{1}{\Gamma\left(\nu\right)}\int\limits _{h}^{t}\left(t-\xi\right)^{\nu-1}f\left(\xi\right)d\xi$
It is supposed it does exist for all $\nu>0$ and $t>0$ if f continous in $\left(0,\infty\right)$ and is Riemann integrable in any finite subinterval of $\left[0,\infty\right]$. However I didn't find a prove of that. Can anyone sketch a prove or advise on where to find it?
Thanks a lot, Karan
I think I got this right.
For $0<\epsilon<t$ we can split the integral
$\int\limits _{h}^{t}\left(t-\xi\right)^{\nu-1}f\left(\xi\right)d\xi=\int\limits _{h}^{\epsilon}\left(t-\xi\right)^{\nu-1}f\left(\xi\right)d\xi+\int\limits _{\epsilon}^{t}\left(t-\xi\right)^{\nu-1}f\left(\xi\right)d\xi$
The first integral is bounded because:
$\int\limits _{h}^{\epsilon}\left|\left(t-\xi\right)^{\nu-1}f\left(\xi\right)\right|d\xi\leq K_{1}\int\limits_{h}^{\epsilon}\left|f\left(\xi\right)\right|d\xi$ ,
where $K_{1}=\underset{\xi\in\left[h,\epsilon\right]}{\sup}\left(t-\xi\right)^{\nu-1}$ that exists because $\left(t-\xi\right)^{\nu-1}$ is bounded in [$h$,$\epsilon$] and $f$ is Riemman integrable on the same interval.
The second integral is also bounded because:
$\int\limits _{\epsilon}^{t}\left|\left(t-\xi\right)^{\nu-1}f\left(\xi\right)\right|d\xi\leq K_{2}\int\limits_{h}^{\epsilon}\left|\left(t-\xi\right)^{\nu-1}\right|d\xi$
where $K_{2}=\underset{\xi\in\left[\epsilon,t\right]}{\sup}f\left(\xi\right)$ that exists because $f\left(\xi\right)$ is bounded in $\left[\epsilon,t\right]$, and $\left(t-\xi\right)^{\nu-1}$ is Riemman integrable on the same interval. Of course the integral is improper if $0<\nu<1$.