I find the following identity in many special functions books without proof. This identity is called the Laplace transform of the Mittag-Leffler function with three parameters. The result is in the form of Fox-Wright function.
$$\int _0^{\infty }e^{-q t}t^{\rho -1}E_{\alpha ,\theta }^{\gamma }\left(w t^{\beta }\right)dt=\frac{q^{-\rho }}{\Gamma(\gamma )}\, _2\psi _1\left[\frac{w}{q^{\beta }}| \begin{array}{cc} (\gamma ,1) & (\rho ,\beta ) \\ (\theta ,\alpha ) & \\ \end{array} \right],$$ where $\Re(\alpha )$, $\Re(\theta )$, $\Re(\gamma )$,$\Re(\rho )$,$\Re(q)>0$ and $q>|w|^{\frac{1}{\Re(\alpha)}}$.
This identity is taken from $Eq.~(5.1 .30)$ in "Mittag - Leffler Functions, Related Topics and Applications", Springer. The series expression of three parameter Mittag-Leffler function is $$E_{\alpha ,\theta }^{\gamma }(z)=\sum _{k=0}^{\infty } \frac{(\gamma )_k}{k!\Gamma(\alpha k+\theta )}z^k,\alpha ,\theta ,\gamma >0.$$ You may other representation of this function in the same chapter. Definitions and convergence of the Fox-Wright function are in Appendix F of that book.
I suppose the proof is expanding the Mittag-Leffler function into series and then integrating term by term. Therefore, the last condition is for the uniformly convergence of the Fox-Wright function. However, what if $q\leq|w|^{\frac{1}{\Re(\alpha)}}$. Do we have any other way to deal with this case?