If my understanding of complex analysis is correct then the arbitrary order generalization of Cauchy's formula for repeated integration $$(J^\alpha f) ( x ) = { 1 \over \Gamma ( \alpha ) } \int_0^x (x-t)^{\alpha-1} f(t) dt$$ ought to be holomorphic in $\alpha$ for a fixed $f$ and $x$ in the region where it converges ($\Re\alpha>0$ for well behaved $f$). Furthermore it should have a meromorphic extension to the entire complex plane (barring anything like branch cuts, which I suspect it can't have, or lacunary function type behavior, which I would be surprised if it had).
I know that the analytical continuation can't (always) be the Caputo fractional derivative because the Caputo derivative of $f(x)=x$ vanishes for all orders greater than 1, which would imply the derivative is 0 everywhere, which it isn't. (It's possible that different definitions are the correct analytical continuation for different $f$s but I would be very surprised if that were the case.)
The Riemann–Liouville fractional derivative seems reasonable but I don't know how to prove the derivative as a function of $\alpha$ should obey the derivative recurrence relation that extends Cauchy's formula to the Riemann–Liouville derivative on the whole complex plane. And if it turns out that the Riemann–Liouville fractional derivative isn't meromorphic, is there a formula for the analytical continuation of Cauchy's formula?