How to find half derivative of $x^{-\frac{1}{2}}$?

Question

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?

Answer 1

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.

Answer 2

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} $