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 ??
 A: 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.
A: 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 
$$
\frac{f(x)}{\Gamma(1-a)}
$$
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
$=\begin{cases}0&\text{when}~x\leq-1\\\dfrac{1}{(x+1)\sqrt{\pi x}}&\text{when}~x>-1\end{cases}$
$=\dfrac{H(x+1)}{(x+1)\sqrt{\pi x}}$
A: 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]} 
$
