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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:

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

The inverse Laplace transform can be done using the Bromwich integral, but this integral requires that you evaluate residues at $pm iomega$, 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

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

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

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

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