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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 ??

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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.


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.

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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 $$ \frac{f(x)}{\Gamma(1-a)} $$

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$=\begin{cases}0&\text{when}~x\leq-1\\\dfrac{1}{\sqrt{\pi}}\dfrac{d}{dx}(2\tan^{-1}\sqrt{x})&\text{when}~x>-1\end{cases}$ according to

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

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

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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]} $

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