Are there solutions for the fractional differential equation $$\frac{\mathrm{d}^\alpha f}{\mathrm{d}x^\alpha} = xe^x$$ for $0 < \alpha < 1$? I have been using the Caputo definition $$ \frac{\mathrm{d}^\alpha f}{\mathrm{d}x^\alpha} = \frac{1}{\Gamma(1-\alpha)} \int_0^x \frac{f'(x')}{(x-x')^\alpha} \mathrm{d} x' $$ but I'm interested in solutions also with other definitions.
With integer order derivatives $$ \frac{\mathrm{d}^n f}{\mathrm{d}x^n} = xe^x$$ the solution is $f = (x-n)e^x$, so I guessed that for fractional order the solution might be $f \propto (x-\alpha)e^x$. However, plugging this into the Caputo definition doesn't really work, as one ends up with a lower incomplete Gamma function $\gamma$. I now realize that this is due to the exponential function having the property $$\frac{\mathrm{d}^\alpha e^x}{\mathrm{d}x^\alpha} \propto e^x \gamma(\alpha, x) \neq e^x .$$
Is there a way to obtain a solution or does this property of the exponentials mean that there is no solution?
For the Caputo derivative using Taylor series, because $\alpha \in (0,1)$ we can swap the integral and sum and we can say
$$D^\alpha (xe^x) = \sum_{n=0}^\infty \frac{n+1}{n!\cdot\Gamma(1-\alpha)}\int_0^x\frac{(x-x')^n}{{x'}^\alpha}dx' = \sum_{n=0}^\infty \frac{n+1}{\Gamma(n+2-\alpha)}x^{n+1-\alpha}$$
by dominated convergence.
For other fractional derivatives like the Riemann-Liouville variant and its equivalents, one should be able to use Feynman's trick instead $xe^x = \partial_a(e^{ax})|_{a=1}$ to evaluate those integrals.