Inverse Laplace transform of $\frac1{s^{1/2}(s^2 + 1)}$

Question

I was studying about Caputo Fractional Derivative for a scientific project and I was trying determine the 1/2 order derivative in Caputo-Sense of $\sin(\omega t)$. Throughout the development of the expression, I've found the following expression to calculate:

$\cos(\omega t)*t^{1/2} $

Where $*$ is the convolution product.

I thought that applying Laplace Transform it would be easy to solve, but I got the following expression after applying Laplace Transform:

$$\frac{\sqrt{\pi} \cdot s^{1/2}}{(s^2 + \omega ^2)}$$

I've tried to use the matlab symbolic math toolbox to solve this, but it can't solve. I want to know a way to solve this. Or solve the original expression of the fractional derivative, which is:

$$(D^\alpha _{0} \sin(\omega t))(t) = \frac{-\omega}{\Gamma(1/2)}\int_0^x \cos(\omega t)(x-t)^{-1/2}\mathrm dt$$

Answer 1

The inverse Laplace transform can be done using the Bromwich integral, but this integral requires that you evaluate residues at $\pm i\omega$, and the contour needs to go around the branch cut on the negative real axis (keyhole contour). Alternatively, we can do the inverse transform term by term. First write

$$ \frac{\sqrt{\pi}\sqrt{s}}{s^{2} + \omega^{2}} = \frac{\sqrt{\pi}}{s^{3/2}}\frac{1}{1+\omega^{2}/s^{2}} = \sqrt{\pi}\sum_{k=0}^{\infty}\frac{(-1)^{k}\omega^{2k}}{s^{2k+3/2}}. $$

Laplace transform of the power function is

$$ \mathcal{L}[t^{n}] = \frac{\Gamma(n+1)}{s^{n+1}},$$

so identifying $n = 2k + 1/2$, we have

$$\begin{aligned} \mathcal{L}^{-1}\left[\frac{1}{s^{2k+3/2}}\right] &= \frac{t^{2k+1/2}}{\Gamma(2k+3/2)} = \frac{t^{2k+1/2}}{(2k+1/2)(2k-1/2)\cdots (1/2)\Gamma(1/2)} \\ &= \sqrt{\frac{t}{\pi}}\frac{t^{2k}}{2^{2k+1}(k+1/4)(k-1/4)\cdots(3/4)(1/4)} \\ &= \frac{1}{2}\sqrt{\frac{t}{\pi}}\frac{(t/2)^{2k}}{(1/4)(5/4)\cdots(k-1+5/4)\cdot(3/4)(7/4)\cdots(k-1+3/4)} \\ &= 2\sqrt{\frac{t}{\pi}}\frac{(1)_{k}}{(3/4)_{k}(5/4)_{k}}\frac{(t/2)^{2k}}{k!}.\end{aligned}$$

The $(a)_{k}$ are Pochhammer symbols (rising factorials). Therefore

$$\begin{aligned} \mathcal{L}^{-1}\left[\frac{\sqrt{\pi}\sqrt{s}}{s^{2} + \omega^{2}}\right] &= 2\sqrt{t}\sum_{k=0}^{\infty}\frac{(1)_{k}}{(3/4)_{k}(5/4)_{k}}\frac{(-1)^{k}(\omega^{2})^{k}(t^{2}/4)^{k}}{k!} \\ &= 2\sqrt{t}\sum_{k=0}^{\infty}\frac{(1)_{k}}{(3/4)_{k}(5/4)_{k}}\frac{(-\omega^{2}t^{2}/4)^{k}}{k!} \\ &= 2\sqrt{t}\,{}_{1}F_{2}\left(1;\frac{3}{4},\frac{5}{4};-\frac{\omega^{2}t^{2}}{4}\right).\end{aligned}$$

Answer 2

There is some confusion with the $\pm 1/2$ power in the title and in the formula with the convolution. I think you want $$D^{1/2} \sin \omega t = \frac \omega {\sqrt \pi} \int_0^t \frac {\cos \omega \tau} {\sqrt{t - \tau}} d\tau = \\ \frac {\omega \sin \omega t} {\sqrt \pi} \int_0^t \frac {\sin \omega \tau} {\sqrt \tau} d\tau + \frac {\omega \cos \omega t} {\sqrt \pi} \int_0^t \frac {\cos \omega \tau} {\sqrt \tau} d\tau = \\ 2 \sqrt{\frac \omega \pi} \,(S(\sqrt{\omega t}) \sin \omega t + C(\sqrt{\omega t}) \cos \omega t),$$ where $S$ and $C$ are the Fresnel integrals.