How to calculate $D^{1/2}\ln x$? Due to Wikipedia Fractional calculus there are ways to generalize derivatives $D^a$ to real $a$. The fractional derivatives for basic power functions is:
$$\frac{d^a}{dx^a}x^n=\frac{\Gamma(n+1)}{\Gamma(n-a+1)}x^{n-a},\;n\ge0$$
I would like to see an expression of $\;D^{1/2}\ln x\;$ but is unfortunately not able to do that calculation myself. Is it possible to calculate it?
 A: Using the definition 
$$ D^{\alpha}f(x)= \frac{1}{\Gamma(1-\alpha)} \frac{d}{dx} \int_{0}^{x} \frac {f(t)}{(x-t)^{\alpha}} \, dt $$
of the complete partial derivative, which is valid for $0 < \alpha < 1$, we find that
$$ D^{1/2}(\log x) = \frac{1}{\Gamma(\frac{1}{2})} \frac{d}{dx} \int_{0}^{x} \frac{\log t}{\sqrt{x-t}} \, dt = \frac{\log (4x)}{\sqrt{\pi x}}. $$

To see that this is compatible with $D(\log x) = \frac{1}{x}$, we compute $D^{1/2}D^{1/2}(\log x)$ using the formula above. Then we notice that
$$ D^{1/2}\left( \frac{\log (4x)}{\sqrt{\pi x}} \right)
= \frac{1}{\Gamma(\frac{1}{2})} \frac{d}{dx} \int_{0}^{x} \frac{\log (4t)}{\sqrt{\pi t (x-t)}} \, dt
= \frac{1}{\sqrt{\pi}} \frac{d}{dx} \sqrt{\pi} \log x = \frac{1}{x}. $$
Here, we utilized the fact that
\begin{align*}
\int_{0}^{x} \frac{\log (4t)}{\sqrt{\pi t (x-t)}} \, dt
&= \int_{0}^{1} \frac{\log (4x) + \log t}{\sqrt{\pi t (1-t)}} \, dt \\
&= \frac{1}{\sqrt{\pi}}\left( B\left(\frac{1}{2},\frac{1}{2}\right) \log (4x) + \partial_1 B\left(\frac{1}{2},\frac{1}{2}\right) \right)
\end{align*}
for the beta function $B(s, t)$ and $\partial_1 B(s, t) = \frac{\partial}{\partial s}B(s, t)$ is the partial derivative of $B(s,t)$ w.r.t. the first argument of $B$. Although only the value of $B(\frac{1}{2},\frac{1}{2}) = \sqrt{\pi}$ is relevant for our computation, it is rather surprising that $\partial_1 B\left(\frac{1}{2},\frac{1}{2}\right) = -\pi \log 4$ and hence
$$ \int_{0}^{x} \frac{\log (4t)}{\sqrt{\pi t (x-t)}} \, dt = \sqrt{\pi} \log x. $$
A: With the definition of the fractional derivative in term of convolution or the Fourier transform.

The Fourier transform of $\text{sign}(x)$ is $\frac{1}{i\pi}\frac{d}{d\xi}\log |\xi|$.
Thus the inverse FT of $\frac{1}{i\pi}D^{1/2}\log |\xi|$  is $(2i \pi x)^{-1/2}\text{sign}(x)$. 
Therefore, as distributions, for $\xi > 0$ $$\frac{1}{i\pi} D^{1/2}\log |\xi| = (2i \pi)^{-1/2}\int_0^\infty x^{-1/2}e^{-2i \pi \xi x}dx+(-2i \pi)^{-1/2})\int_0^\infty x^{-1/2}e^{2i \pi \xi x}dx$$
$$ = 2 \xi^{-1/2} \Gamma(1/2)$$
A: You can repeat the steps present in the wikipedia page. 
It is immediate to see that 
$$D(\ln x) = \frac 1x$$
$$D^2(\ln x) = -\frac 1{x^2}$$
$$D^3(\ln x) = 2 x^{-3}$$
This is easy to generalize to 
$$D^n(\ln x) = (-1)^{n+1}(n-1)!x^{-n}$$
which is valid for $n$ integer. For fractional $n$ we can just plug in, to get (with $n=1/2$ and the factorial replaced by the gamma function)
$$D^{1/2}(\ln x) = (-1)^{3/2}\Gamma\left(\frac 12\right)x^{-1/2}$$
To check that this is correct, we can take the fractinoal derivative again and check that the repeated application of $D^{1/2}$ is indeed equal to $D$; so we compute 
$$D^{1/2}D^{1/2}(\ln x) = D^{1/2}(-1)^{n-1}\Gamma\left(\frac 12\right)x^{-1/2} = 1/x = D(\ln x)$$
(using the formulas for the fractional derivative of the power function found on wikipedia).
This is what we wanted, so we are done!
