# 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.

• I'm also interested in the use of the laplace transform method. About the formula, the derivarive of the integral in the final expression is equal to (thanks to Mathematica) $$\frac{\Gamma(1-a)\Gamma(1+m)}{\Gamma(1-a+m)}t^{-a+m}$$
– NN2
Commented Apr 14, 2021 at 19:04
• This could be seen without Mathematica (substitute $z=tx$). Commented Apr 14, 2021 at 19:55

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: $$$$\int_{0}^{1}(1-x)^{-\alpha}x^{m}\,dx=B(1-\alpha,1+m) =\frac{\Gamma(1-\alpha)\Gamma(1+m)}{\Gamma(2-\alpha+m)}$$$$ and thus, \begin{align} \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) &=t^{-\alpha}\sum_{m=0}^{\infty}\frac{(-1)^m}{(m!)^2}\left( m+1-\alpha \right)\left(\frac{\sqrt t}{2}\right)^{2m}\frac{\Gamma(1+m)}{\Gamma(2-\alpha+m)}\\ &=t^{-\alpha}\sum_{m=0}^{\infty}\frac{(-1)^m}{m!\Gamma(m-\alpha+1)}\left(\frac{\sqrt t}{2}\right)^{2m} \end{align} This series can be rearranged to nearly match a Bessel function expansion: $$J_\nu(x)\sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+\nu+1)}\left(\frac x2\right)^{2m+\nu}$$ with $$\nu=-\alpha$$, \begin{align} \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) &=2^{-\alpha} t^{-\frac{\alpha}{2}} \sum_{m=0}^{\infty}\frac{(-1)^m}{m!\Gamma(m-\alpha+1)}\left(\frac{\sqrt t}{2}\right)^{2m-\alpha}\\ &=2^{-\alpha} t^{-\frac{\alpha}{2}} J_{-\alpha}\left(\sqrt{t}\right) \end{align}
Alternatively, the integral $$$$I(t)=\int_{0}^{t}(t-z)^{-\alpha}J_0(\sqrt z)\,dz$$$$ can be considered as a convolution operation for the Laplace transform for the functions $$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)}$$ (see Ederlyi TI 4.14(25) p. 185). Then $$$$I(t)=\mathcal L^{-1}\left[\Gamma(1-\alpha) p^{\alpha-2}e^{-1/(4p)}\right]$$$$ This inverse transform can be found in Ederlyi TI 5.6(40) p.245: $$$$I(t)=\Gamma(1-\alpha)2^{1-\alpha}t^{\frac{1-\alpha}{2}}J_{1-\alpha}\left( \sqrt{t} \right)$$$$ Finally, after differentiation, $$$$\mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) =2^{-\alpha} t^{-\frac{\alpha}{2}} J_{-\alpha}\left(\sqrt{t}\right)$$$$ as expected from the above calculation.