An eigenfunction $f$ of a linear operator $T$ satisfies $T(f) = \lambda f$ for some $\lambda$, which we call its eigenvalue. If we let $T = \partial_x$ be the derivative, then $f(x) = e^{\lambda x}$ is an eigenfunction because $$\partial_x e^{\lambda x} = \lambda e^{\lambda x}$$ and more generally, $$\partial^n_x e^{\lambda x} = \lambda^n e^{\lambda x}$$ for all $n \in \mathbb{N}$. I understand that there are multiple definitions of the fractional derivative, but they seem to share the idea that if $k$ is a positive integer, the following natural extension can be made. $$\frac{d^a}{dx^a}x^k = \frac{k!}{(k-a)!}x^{k-a} \qquad \forall a \in \mathbb{N},\text{ can be extended to } \qquad \frac{d^a}{dx^a}x^k = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a} \qquad \forall a \in \mathbb{R}.$$
However, this seems to drop the eigen-property of the exponential function. In particular, if $\alpha \in (0,1)$ then $$\frac{d^\alpha}{dx^\alpha} e^{x} = \frac{d^\alpha}{dx^\alpha} \sum_{n \in \mathbb{N}} \frac{x^n}{n!} = \sum_{n \in \mathbb{N}} \frac{\Gamma(n+1)}{\Gamma(n-\alpha+1)} \frac{x^{n-\alpha}}{\Gamma(n+1)} = \sum_{n \in \mathbb{N}} \frac{x^{n-\alpha}}{\Gamma(n-\alpha+1)}$$ which is certainly not any multiple of the exponential function; it's even undefined at $0$. My question is really the following: is there any theory of fractional calculus that preserves the exponential map as an eigenfunction. It would make me very happy.