Suppose that $f(x) \in C^1$ for a $x \in [a, x]$. Then a regularization of Riemann-Liouville fractional derivative is defined as:
$ \frac{1}{\Gamma(1-b)} \frac{d}{dx} \int_{a}^{x}\left( \mathrm{f}\left( t\right) -\mathrm{f}\left( a\right) \right) \,\left( \frac{d}{d\,t}\,\left( -\frac{{\left( x-t\right) }^{1-b}}{1-b}\right) \right) dt $
assuming $0<b \leq 1$.
Is this definition equivalent to the Caputo definition? $ \frac{1}{\Gamma(1-b)} \int_{a}^{x}\frac{ \mathrm{f^\prime}\left( t\right) }{{\left( x-t\right) }^{b}}dt $
(As suggested ) here is the calculation: The first step is to integrate by parts:
$ I= \frac{\int_{a}^{x}\left( \frac{d}{d\,t}\,\left( \mathrm{f}\left( t\right) -\mathrm{f}\left( a\right) \right) \right) \,{\left( x-t\right) }^{1-b}dt}{1-b} - \left. \frac{\left( \mathrm{f}\left( t\right) -\mathrm{f}\left( a\right) \right) \,{\left( x-t\right) }^{1-b}}{1-b}\right|_{t=x} +\left. \frac{\left( \mathrm{f}\left( t\right) -\mathrm{f}\left( a\right) \right) \,{\left( x-t\right) }^{1-b}}{1-b}\right|_{t=a} $.
Since $0<b \leq 1$ the second term vanishes. The third term also vanishes. Then we are left with:
$ I = \frac{\int_{a}^{x}\left( \frac{d}{d\,t}\,\mathrm{f}\left( t\right) \right) \,{\left( x-t\right) }^{1-b}dt}{1-b} $
Finally, differentiating gives
$ \frac{d}{d\,x}\,I = \int_{a}^{x}\frac{ \mathrm{f^{\prime}}\left( t\right) }{{\left( x-t\right) }^{b}}dt $.
So applying also the factor yields Caputo's definition.