Riemann-Liouville fractional derivative of order $\alpha$ of the Bessel function I'm trying to compute the Riemann-Liouville fractional derivative of order $\alpha$ of the Bessel function $J_{0}(\sqrt{t})$ with $0 < \alpha < 1.$
$\mathbf{Some\,\,terminology}$
The Riemann-Liouville fractional derivative of a function $f(t)$ is given by
$$I_t^{\alpha}(f(t)) = \frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-z)^{\alpha-1}f(z)dz, \text{ where}\, \alpha > 0.$$
The Riemann-Liouville fractional derivative is defined as the derivative of RL fractional integral, i.e. $$\mathcal{D}_t^{\alpha}(f(t)) = \frac{1}{\Gamma(1 - \alpha)}\frac{d}{dt}\int_{0}^{t}(t-z)^{-\alpha}f(z)dz, \text{where} \,\,0 < \alpha \leq 1. $$
In order to compute $\mathcal{D}_t^{\alpha}\big(J_0(\sqrt{t})\big),$ I can simply plug  $J_0(\sqrt{t}) = \sum_{m=0}^\infty \frac{(-1)^m}{(m!)^2}\left(\frac{\sqrt{t}}{2}\right)^{2m}$ into the above formula to obtain
$$\frac{1}{\Gamma(1 - \alpha)}\sum_{m=0}^{\infty}\bigg(\frac{1}{2}\bigg)^{2m}\frac{(-1)^m}{(m!)^2}\frac{d}{dt}\int_{0}^{t}(t-z)^{-\alpha}z^{m}dz.$$ I don't know how to simplify the above expression. Any help highly appreciated. Also is there any way to compute RL fractional derivative of bessel function using laplace transforms.
 A: Using the proposed series expansion:
\begin{align}
 \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) &=\frac{1}{\Gamma(1 - \alpha)}\sum_{m=0}^{\infty}\left(\frac{1}{2}\right)^{2m}\frac{(-1)^m}{(m!)^2}\frac{d}{dt}\int_{0}^{t}(t-z)^{-\alpha}z^{m}\,dz\\
 &=\frac{1}{\Gamma(1 - \alpha)}\sum_{m=0}^{\infty}\left(\frac{1}{2}\right)^{2m}\frac{(-1)^m}{(m!)^2}\frac{d}{dt}t^{m+1-\alpha}\int_{0}^{1}(1-x)^{-\alpha}x^{m}\,dx\\
 &=\frac{1}{\Gamma(1 - \alpha)}\sum_{m=0}^{\infty}\left(\frac{1}{2}\right)^{2m}\frac{(-1)^m}{(m!)^2}\left( m+1-\alpha \right)t^{m-\alpha}\int_{0}^{1}(1-x)^{-\alpha}x^{m}\,dx
\end{align}
where $z=tx$ was changed in the integral, as proposed by @metamorphy. This integral corresponds to the Beta function integral representation:
\begin{equation}
\int_{0}^{1}(1-x)^{-\alpha}x^{m}\,dx=B(1-\alpha,1+m) =\frac{\Gamma(1-\alpha)\Gamma(1+m)}{\Gamma(2-\alpha+m)}
\end{equation}
and thus,
\begin{align}
 \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) &=t^{-\alpha}\sum_{m=0}^{\infty}\frac{(-1)^m}{(m!)^2}\left( m+1-\alpha \right)\left(\frac{\sqrt t}{2}\right)^{2m}\frac{\Gamma(1+m)}{\Gamma(2-\alpha+m)}\\
 &=t^{-\alpha}\sum_{m=0}^{\infty}\frac{(-1)^m}{m!\Gamma(m-\alpha+1)}\left(\frac{\sqrt t}{2}\right)^{2m}
\end{align}
This series can be rearranged to nearly match a Bessel function expansion:
$$J_\nu(x)\sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+\nu+1)}\left(\frac x2\right)^{2m+\nu}$$
with $\nu=-\alpha$,
\begin{align}
 \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) &=2^{-\alpha} t^{-\frac{\alpha}{2}} \sum_{m=0}^{\infty}\frac{(-1)^m}{m!\Gamma(m-\alpha+1)}\left(\frac{\sqrt t}{2}\right)^{2m-\alpha}\\
 &=2^{-\alpha} t^{-\frac{\alpha}{2}} J_{-\alpha}\left(\sqrt{t}\right)
\end{align}

Alternatively, the integral
\begin{equation}
 I(t)=\int_{0}^{t}(t-z)^{-\alpha}J_0(\sqrt z)\,dz
\end{equation}
can be considered as a convolution operation for the Laplace transform for the functions $z^{-\alpha}$ and $J_0(\sqrt z)$. Their Laplace transforms are respectively $\Gamma(1-\alpha) p^{\alpha-1}$ (classical integral representation of the Gamma function) and $p^{-1}e^{-1/(4p)}$ (see Ederlyi TI 4.14(25) p. 185). Then
\begin{equation}
 I(t)=\mathcal L^{-1}\left[\Gamma(1-\alpha) p^{\alpha-2}e^{-1/(4p)}\right]
\end{equation}
This inverse transform can be found in Ederlyi TI 5.6(40) p.245:
\begin{equation}
 I(t)=\Gamma(1-\alpha)2^{1-\alpha}t^{\frac{1-\alpha}{2}}J_{1-\alpha}\left( \sqrt{t} \right)
\end{equation}
Finally, after differentiation,
\begin{equation}
 \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) =2^{-\alpha} t^{-\frac{\alpha}{2}} J_{-\alpha}\left(\sqrt{t}\right)
\end{equation}
as expected from the above calculation.
