I wanted to calculate the Laplace Transform of Riesz Fractional Derivative. But got some troubles in the middle.
The Riesz Fractional Derivative is given by:
$R^{\alpha}f(t)=-\frac{1}{2\pi} \int_{-\infty}^{\infty}|\omega|^{\alpha}F(\omega)e^{j\omega t}d\omega$
So I tried to took the Laplace Transform of it:
$ \mathcal{L}[R^\alpha f(t)] = -\frac{1}{2\pi} \int_{-\infty}^{\infty} \left [ \int_{-\infty}^{\infty} |\omega|^\alpha F(\omega)e^{j\omega t} d\omega \right] e^{-st} dt $
Applying integration by parts, I assumed:
$ u = \int_{-\infty}^{\infty}|\omega|^\alpha F(\omega)e^{j\omega t} d\omega $
$\frac{du}{d\omega} = |\omega|^\alpha F(\omega) e^{j\omega t} $
And
$ dv = e^{-st}$
$v = -\frac{e^{-st}}{s}$
So, I got this:
$ \mathcal{L}[R^\alpha f(t)] = -\frac{1}{2\pi} \left [ \left( \int_{-\infty}^{\infty} |\omega|^\alpha F(\omega) e^{j\omega t} d\omega \right) \left ( - \frac{e^{-st}}{s} \right )_{-\infty}^{\infty} + \frac{1}{s} \int_{-\infty}^{\infty} |\omega|^\alpha F(\omega) e^{j\omega t} e^{-st} dt \right] $
And I can't find a way to solve this, the both parts, i.e. I can't understand a way to evaluate the first term so as the second. My guess is that the $|\omega|^{\alpha}F(\omega)e^{j\omega t}$ is my function which will be used to compute the Laplace Transform. If this is correct, the problem is the first term, how to compute it.
Thanks in advance.