Showing $D^{-1/2}[D^{-1/2}t^2]=t^3/3$ I'm having trouble showing this using gamma functions, it doesn't seem to work out.  I'm trying to use the fact that 
$D^{-1/2}(f(t))=1/\sqrt{\pi}\int_0^t{(t-T)^{-1/2}f(T)dT}$
Any help is appreciated thanks
 A: First, 
$$D^{-1/2}t^2 = \pi^{-1/2}\int_0^t (t - u)^{-1/2}u^2\, du = \pi^{-1/2}t^{5/2}\int_0^1 (1 - v)^{-1/2}v^2\, dv,$$
using the substitution $u = tv$. Now
$$\pi^{-1/2}\int_0^1 (1 - v)^{-1/2}v^2\, dv = \pi^{-1/2}\frac{\Gamma(1/2)\Gamma(3)}{\Gamma(7/2)} = \frac{2}{\Gamma(7/2)},$$
Thus
$$D^{-1/2}[D^{-1/2}t^2] = \frac{2}{\Gamma(7/2)}D^{-1/2}(t^{5/2}).\tag{1}$$
Since
$$D^{-1/2}(t^{5/2}) = \pi^{-1/2}\int_0^t (t - u)^{-1/2}u^{5/2}\, du = \pi^{-1/2}t^{3}\int_0^1 (1 - t)^{-1/2}u^{5/2}\, du,$$
and 
$$\pi^{-1/2}\int_0^1 (1 - u)^{-1/2}u^{5/2}\, du = \pi^{-1/2}\frac{\Gamma(1/2)\Gamma(7/2)}{\Gamma(4)} = \frac{\Gamma(7/2)}{6},$$
we have 
$$D^{-1/2}(t^{5/2}) = \frac{t^3\Gamma(7/2)}{6}. \tag{2}$$
By $(1)$ and $(2)$, $D^{-1/2}[D^{-1/2}t^2] = t^3/3$.
A: You can use the formula for a monomial

$$ D^{-n} x^{m} = \frac{\Gamma(m+1)}{\Gamma(m+n+1)} x^{m+n} $$

which comes from the definition you are using for fractional derivative. In your case $n=1/2$ and you need to apply it twice. 
Note: 

$$ D^{-n}t^m = \frac{1}{\Gamma(n)}\int_0^t{(t-y)^{n-1}y^m dy} = \frac{\Gamma(m+1)}{\Gamma(m+n+1)} x^{m+n}. $$

You can use the beta function to evaluate the above integral.
