Are fractional derivatives defined analogously for a complex parameter?

Question

The Riemann-Liouville integral is defined for $q<0$ by $$ \left[ \frac{d^q f}{d(x-a)^q}\right]_{RL} = \frac{1}{\Gamma(-q)} \int _a^x (x-y)^{-q-1}f(y)\,dy $$and for $q\leq 0$ by analytic continuation if $f$ is $n$-times differentiable by analytic continuation using integration by parts. Since $\Gamma(z)\neq 0$ for $z\in \mathbb{C}$, it seems this is a valid way to define fractional derivatives for order of any complex number; is this correct, or do you need to do something else when passing to $\mathbb{C}$?

Answer 1

For $\Re(q)\ge0$, the integral diverges. In this case, the Riemann-Liouville fractional derivative is extended by differentiating $n=\lfloor\Re(q)\rfloor+1$ after "integrating" $n-q$ times.

$$^\mathrm{RL}_{~~~a}D_x^qf(x)=\frac1{\Gamma(n-q)}\frac{\mathrm d^n}{\mathrm dx^n}\int_a^x\frac{f(t)}{(x-t)^{n-q+1}}~\mathrm dt$$

Nothing such as analytic continuation is used.