Analytic Continuation of Fractional Derivatives

Question

I've been doing some work with fractional calculus, and I have been running into a problem. For a given meromorphic function $f: \mathbb{C}\to \mathbb{C}$, if a formula of the form $$(D^{\alpha}f)(0)=g(\alpha)$$ has been found where $g$ is as well a meromorphic function has been found, but the proof only works for an open interval $\alpha \in (0,1)$, can a sort of analytic continuation be used to show that this formula holds for all $\alpha$? If not, what conditions would be needed to be able to take such a continuation?

Answer 1

In the specific case of the Caputo derivative defined by

$$D^\alpha f(x)=\frac1{\Gamma(\lceil\alpha\rceil-\alpha)}\frac{\mathrm d^{\lceil\alpha\rceil}}{\mathrm d^{\lceil\alpha\rceil}x}\int_a^xf(t)(x-t)^{\lceil\alpha\rceil-\alpha-1}~\mathrm dt$$

for non-integer $\alpha>0$, we have the identity

$$D^\alpha f(x)=\frac1{\Gamma(n-\alpha)}\frac{\mathrm d^n}{\mathrm d^nx}\int_a^xf(t)(x-t)^{n-\alpha-1}~\mathrm dt$$

for any natural $n>\alpha$. As this last statement holds for all $\alpha$ actually, and is somewhat easily seen to be analytic in $\alpha$ by the same argument used to show it for $\alpha\in(0,1)$, we have an analytic fractional derivative for all $\alpha<n$ for arbitrarily large $n$, and hence analytic everywhere.

The above is usually used to generalize this to complex $\alpha$ by setting $n>\Re(\alpha)$, and the same argument will work there as well.