# fractional derivative of a heaviside function

given the function

$$f(x)= \frac{H(x+1)}{\sqrt{x+1}}$$

how can i evaluate the fractional derivative

$$\frac{d^{1/2}}{dx^{1/2}}f(x)$$

if i use the standar definition for powers of 'x' i get a coefficient $\frac{\Gamma(1/2)}{\Gamma(0)}$ so apparently the derivative would be 0 but i think it should be something about $\delta (x+1)$ since the half derivative applied two times is just the ordinary derivative so what is the answer ??

• Where does that problem come from? – draks ... Jan 10 '13 at 12:03
• it is a doubt i have.. for example for the Heaviside step function $H(x+1)$ i know its fractional derivative isproportional to $\frac{H(x+1)}{\sqrt{x+1}}$ – Jose Garcia Jan 10 '13 at 13:07
• Would tell us why the $\frac{d^{1/2}}{dx^{1/2}}H(x+1)= \frac{H(x+1)}{\sqrt{x+1}}$? What about arbitrary fractionals $\frac{d^{a}}{dx^{a}}H(x+1)$? – draks ... Jan 10 '13 at 13:18
• if $x>-1$ then the Heaviside function is just a constant so .. the fractional derivative of a constant is $Ax^{-1/2}$ – Jose Garcia Jan 10 '13 at 13:20

Hint: Consider the Fourier transform of that function. The $p$th derivative is the inverse F.T. of $(i 2 \pi x)^p$ times the F.T. of the function.

EDIT

More detail: for $f(x) = \frac{\mathrm{H}(x+1)}{\sqrt{x+1}}$, the F.T. is very simple:

$$\hat{f}(v) = |v|^{-\frac{1}{2}} \exp{(i 2 \pi v)}$$

so that the $p$th derivative $f^{(p)}(x)$ is

$$f^{(p)}(x) = (i 2 \pi)^p \int_{-\infty}^{\infty} dv \: v^p |v|^{-\frac{1}{2}} \exp{[i 2 \pi (x+1) v]}$$

You can do this last integral analytically (this is where I run out of time), and this is your fractional derivative of your function. I am aware that $\Im{f^{(p)}(x)}$ should be 0, but it is not obvious until the integral is worked out that this will be so.

The fractional derivative is shift invariant. So, we can consider first the function $f(x)= x^{-a}H(x)$. I use the Grünwald-Letnikov definition. It is not very difficult to show that the derivative of order $a$ of the Heaviside function is $$\frac{f(x)}{\Gamma(1-a)}$$

$\dfrac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}\biggl(\dfrac{H(x+1)}{\sqrt{x+1}}\biggr)$

$=\begin{cases}\dfrac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}(0)&\text{when}~x\leq-1\\\dfrac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}\biggl(\dfrac{1}{\sqrt{x+1}}\biggr)&\text{when}~x>-1\end{cases}$

$=\begin{cases}0&\text{when}~x\leq-1\\\dfrac{1}{\Gamma\left(\dfrac{1}{2}\right)}\dfrac{d}{dx}\int_0^x\dfrac{1}{\sqrt{x-t}\sqrt{t+1}}dt&\text{when}~x>-1\end{cases}$

$=\begin{cases}0&\text{when}~x\leq-1\\\dfrac{1}{\sqrt{\pi}}\dfrac{d}{dx}(2\tan^{-1}\sqrt{x})&\text{when}~x>-1\end{cases}$ according to http://www.wolframalpha.com/input/?i=int1%2F%28%28x-t%29%5E%281%2F2%29%28t%2B1%29%5E%281%2F2%29%29%2Ct%2C0%2Cx

$=\begin{cases}0&\text{when}~x\leq-1\\\dfrac{1}{(x+1)\sqrt{\pi x}}&\text{when}~x>-1\end{cases}$

$=\dfrac{H(x+1)}{(x+1)\sqrt{\pi x}}$

I think the answer is $\delta(x+1) \frac{1 - i}{\sqrt{2 \pi}}$.

Extending the previous argument

$f^{(1/2)}(x) = \frac{1}{2 \pi} (i 2 \pi)^\frac{1}{2} \int_{-\infty}^{\infty} dv \: v^\frac{1}{2} |v|^{-\frac{1}{2}} \exp{[i 2 \pi (x+1) v]} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dv \: i^{1 - sgn(v)} \exp{[i 2 \pi (x+1) v]}$