How does Mathematica calculate a fractional derivative?

Question

The Riemann-Liouville fractional derivative of order $1/2$ of $x^{-1}$ is defined as
$$ \mathcal{D}_x^{1/2} x^{-1} = \frac{1}{\Gamma(1/2)} \frac{d\;}{dx} \left[\int_0^x (x-\tau)^{-1/2}\tau^{-1} \,d\tau\right] . $$
Since the above integral does not converge, I understand that this fractional derivative does not exists. However, when I ask Mathematica to calculate it using the command FractionalD, I obtain that the result is
$$ \frac{-\log{x}+ \gamma+\psi^{(0)}(-1/2)}{2\sqrt{\pi} x^{3/2}} $$
where $\gamma$ is the Euler-Mascheroni constant and $\psi^{(0)}$ is the digamma function. How did Mathematica arrive to this result?

Answer 1

This is rather a math than Mathematica question. As I understand it, the formula

 FractionalD[f[x], {x, a}]==FractionalD[Integrate[f[x], x], {x, a + 1}] 

which holds onder certain assumptions is used to this end i.e. to define the fractional derivative of order $1/2$ of $1/x$ (I don't have a reference at hand.). Indeed,

FractionalD[Integrate[1/x, x], {x, 3/2}

(EulerGamma - Log[x] + PolyGamma[0, -(1/2)])/(2 Sqrt[\[Pi]] x^(3/2))

where the integral converges. Hope that gap in the documentation will be filled in next versions of Mathematica.