Suppose $0<\alpha<1$, I am wondering whether there is a closed form expression to Caputo derivative to log Mittag-Leffler function, i.e., \begin{align*} \frac{\partial^\alpha}{\partial x^\alpha}\log E_\alpha(x^\alpha)=&\frac{1}{\Gamma[1-\alpha]}\int_0^x(x-t)^{-\alpha}\left(\frac{d}{dt}\log E_\alpha(t^\alpha)\right)dt\\ =&\frac{1}{\Gamma[1-\alpha]}\int_0^x(x-t)^{-\alpha}\frac{1}{E_\alpha(t^\alpha)}\left(\frac{d}{dt}E_\alpha(t^\alpha)\right)dt\\ =&\frac{1}{\Gamma[1-\alpha]}\int_0^x(x-t)^{-\alpha}\frac{1}{E_\alpha(t^\alpha)}\left(\frac{d}{dt}\sum_{n=0}^\infty\frac{(t^\alpha)^n}{\Gamma[\alpha n+1]}\right)dt\\ =&\frac{1}{\Gamma[1-\alpha]}\int_0^x(x-t)^{-\alpha}\frac{1}{E_\alpha(t^\alpha)}\left(\frac{d}{dt}\sum_{n=1}^\infty\frac{(t^\alpha)^n}{\Gamma[\alpha n+1]}\right)dt\\ =&\frac{1}{\Gamma[1-\alpha]}\int_0^x(x-t)^{-\alpha}\frac{1}{E_\alpha(t^\alpha)}\sum_{n=1}^\infty\frac{t^{\alpha n-1}}{\Gamma[\alpha n]}dt, \end{align*} where we may assume $x>0$ to avoid well defined issues. Since there is an infinite summation or a complex integral in the denominator, can we further simplify this integral?
The reason I am asking is that we know the solution to \begin{align*} \frac{d}{dx}f(x)=g(x)f(x), \ f(x)\geq 0, \end{align*} is \begin{align*} f(x)=f(0)\exp{\int_0^xg(s)ds}. \end{align*} then what is the solution to \begin{align*} \frac{\partial^\alpha}{\partial x^\alpha}f(x)=g(x)f(x), \ f(x)\geq 0, \end{align*} I know one special case when $g(x)=1$, then $f(x)=E_\alpha(x^\alpha)$. But do we have a formula for a general $g$?
Any reference or idea is appreciated. Thanks in advance.
