# Fractional derivatives of delta function $\delta (x)$

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 $g(x)$.

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You might want to explain applying formal integration by parts $\alpha$ times when $\alpha$ is not a nonnegative integer. – Did Sep 24 '12 at 13:27
Hint: Look at the Fourier transform. – Nate Eldredge Sep 24 '12 at 14:18
@NateEldredge can you elaborate on that? – Navin Sep 26 '12 at 22:54
A general principle is that taking derivatives in regular space is equivalent to multiplying by powers of $i \xi$ in frequency space. Additionally, the Fourier transform of the delta distribution is constant. So, one would reasonably define $D^\alpha \delta := \frac{1}{\sqrt{2 \pi}}\check{(i \xi)^\alpha}$. This inverse Fourier transform certainly exists as a tempered distribution, but it's not clear (to me) if there's any simpler form for it. – Nick Alger Oct 27 '12 at 12:58
Up to a constant, the inverse Fourier transform mentioned by @Nick Alger is a linear combination of $|x|^{-1-\alpha}$ and $\hbox{sign}(x)\cdot |x|^{-1-\alpha}$, computed by analytic continuation. – paul garrett Nov 26 '12 at 14:08

I don't know whether the current integral representation of the fractional derivative works on delta function or not. If it works:

$D^\alpha\delta(x)=\begin{cases}\dfrac{1}{\Gamma(\lceil\alpha\rceil-\alpha)}\dfrac{d^{\lceil\alpha\rceil}}{dx^{\lceil\alpha\rceil}}\int_0^x(x-t)^{\lceil\alpha\rceil-\alpha-1}\delta(t)~dt&\text{when}~\alpha>0~\text{and}~\alpha~\text{is not an integer}\\\dfrac{1}{\Gamma(-\alpha)}\int_0^x(x-t)^{-\alpha-1}\delta(t)~dt&\text{when}~\alpha<0\end{cases}=\begin{cases}\dfrac{1}{\Gamma(\lceil\alpha\rceil-\alpha)}\dfrac{d^{\lceil\alpha\rceil}}{dx^{\lceil\alpha\rceil}}(x^{\lceil\alpha\rceil-\alpha-1}H(x))&\text{when}~\alpha>0~\text{and}~\alpha~\text{is not an integer}\\\dfrac{x^{-\alpha-1}H(x)}{\Gamma(-\alpha)}&\text{when}~\alpha<0\end{cases}$

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$$\delta(x)=\frac{ e^{-(x/\varepsilon)^2}}{\varepsilon\sqrt{\pi }}$$

where $\varepsilon$ is infinitesmal.

Now take fractional derivative of it. By substitution it comes to the fractional derivative of normal distribution.

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