How can I define the fractional derivative of the Delta function?
I mean $D^{\alpha}= \frac{d^{\alpha}}{dx^{\alpha}} $ where $\alpha$ can be any real number, then if we define $D^{\alpha} \delta (x) $ how can we define it in the sense of distribution?
Applying formal integration by parts $ \alpha $ times I guess that
$$ \int_{-\infty}^{+\infty}D^{\alpha}\delta (x) g(x)dx= (-1)^{[ \alpha]}\int_{-\infty}^{+\infty}D^{\alpha}g(x)\delta(x)dx= (-1)^{[ \alpha]}D^{\alpha}g(0) $$
for any test function $ g(x) $.
I don't know whether the current integral representation of the fractional derivative works on delta function or not. If it works:
$D^\alpha\delta(x)=\begin{cases}\dfrac{1}{\Gamma(\lceil\alpha\rceil-\alpha)}\dfrac{d^{\lceil\alpha\rceil}}{dx^{\lceil\alpha\rceil}}\int_0^x(x-t)^{\lceil\alpha\rceil-\alpha-1}\delta(t)~dt&\text{when}~\alpha>0~\text{and}~\alpha~\text{is not an integer}\\\dfrac{1}{\Gamma(-\alpha)}\int_0^x(x-t)^{-\alpha-1}\delta(t)~dt&\text{when}~\alpha<0\end{cases}=\begin{cases}\dfrac{1}{\Gamma(\lceil\alpha\rceil-\alpha)}\dfrac{d^{\lceil\alpha\rceil}}{dx^{\lceil\alpha\rceil}}(x^{\lceil\alpha\rceil-\alpha-1}H(x))&\text{when}~\alpha>0~\text{and}~\alpha~\text{is not an integer}\\\dfrac{x^{-\alpha-1}H(x)}{\Gamma(-\alpha)}&\text{when}~\alpha<0\end{cases}$