How to find half derivative of $x^{-\frac{1}{2}}$? I use this general definition to do fractional differentiation:
$$(D^nf)(t)=\frac{1}{\Gamma(1-n)}\frac{d}{dx}\int_a^x (x-t)^{-n}\space f(t)\space\space dt,\space\space 0<n<1$$
However, when I try to take half derivative of $x^{-\frac{1}{2}}$, $x$ is lost in the definite integral so it ends up with $0$. However, when I try to take half derivative of $x^{-\frac{1}{2}}$ by following the pattern of derivatives of $x^{-k}$ I end up with $$\frac{i}{-2\sqrt{\pi}}x^{-1}$$
Can't the first definition I made be used for such functions? How can I generalise it so that it can be used for such functions?
 A: Just substituting blindly.
In
$(D^nf)(t)=\frac{1}{\Gamma(1-n)}\frac{d}{dx}\int_a^x (x-t)^{-n}\space f(t)\space\space dt,\space\space 0<n<1
$
if you put 
$f(t)
=t^{-1/2}
$
we get
$\begin{array}\\
(D^n(x^{-1/2})(t)
&=\dfrac{1}{\Gamma(1-n)}\frac{d}{dx}\int_a^x (x-t)^{-n}t^{-1/2}\space\space dt\\
&=\dfrac{1}{\Gamma(1-n)}\frac{d}{dx}\int_0^{x-a} t^{-n}(x-t)^{-1/2}\space\space dt\\
\text{so}\\
(D^{1/2}(x^{-1/2})(t)
&=\dfrac{1}{\Gamma(\frac12)}\frac{d}{dx}\int_0^{x-a} t^{-\frac12}(x-t)^{-1/2}\space\space dt\\
&=\dfrac{1}{\sqrt{\pi}}\frac{d}{dx}2 \arctan(\sqrt{t}/\sqrt{x - t})|_0^{x-a}
\qquad\text{(thanks to Wolfy)}\\
&=\dfrac{2}{\sqrt{\pi}}\frac{d}{dx} (\arctan(\sqrt{x-a}/\sqrt{a}))\\
&=\dfrac{2}{\sqrt{\pi}}\dfrac{\sqrt{a}}{2 x \sqrt{x - a}}
\qquad\text{(again, via Wolfy)}\\
&=\dfrac{\sqrt{a}}{\sqrt{\pi}x \sqrt{x - a}}\\
\end{array}
$
A: The fractional derivative is working "as intended", since actually $x^{-1/2}$ is $D^{1/2} 1$ (times a constant), so
$$ D^{1/2} (x^{-1/2}) = c (D^{1/2} D^{1/2} 1)(x) = c D^1(1)(x) \equiv 0.$$
I'm not sure how you're following the pattern to get complex numbers. Shouldn't it be
$$\frac{d^n}{dx^n}x^\alpha =\frac{\Gamma(\alpha+1)}{\Gamma(\alpha-n+1)}x^{\alpha-n}$$
so for $\alpha=-1/2,n=1/2$, with $\frac1{\Gamma(0)}=0$, we still get $0$?
PS Still waiting for a response on your other question.
