Problem with the application of the fractional integral (as in wikipedia) , example $f(x)=\exp(x)-1$ I am still fiddling with the understanding and application of the fractional integration/differentation.
I've tried the wikipedia-formula for the Cauchy's iterated integration:
$$ (J^{\alpha} f)(x) = \frac 1{\Gamma(\alpha)} \cdot \int_0^x((x-t)^{\alpha-1}f(t)) dt$$ and inserted
$$ (J^{1/2} f)(x) = \frac 1{\Gamma(1/2)} \cdot \int_0^x (x-t)^{-1/2} f(t) dt$$
applied to the function $f(x) = \exp(x)-1$ .    
Two time the semi-integral for some interval $0 \ldots x$ should reproduce the usual integral, but I get a difference for any $x$ that I try.      
Q: Where is the error in the understanding/implementation of the half-integral?

This is, what I've done in Pari/GP: 
semiInt(x,d=1e-64)= 1/gamma(1/2) * intnum(t=0,x-d,(x-t)^(1/2-1)*(exp(t)-1))
        \\ introduce a delta-deviation "d" for the upper bound to prevent 
        \\  evaluation  with 0^(-1/2) when the formal parameter t=x

x1=0.5                          \\  = 0.500000000000      
x2 = semiInt(semiInt(x1))       \\  = 0.161464182768
x3 = intnum(t=0,x1,exp(t)-1)    \\  = 0.148721270700
x2-x3                           \\  = 0.0127429120679

 A: Symbolically...  Let $f(x) = e^x-1$.  Then I get
$$
J^1f(x) = e^x-x-1, \\
f_2(x):= J^{1/2}f(x) = \frac{e^x \sqrt{\pi}\;\text{erf}\sqrt{x} - 2\sqrt{x}}{\sqrt{\pi}}\\
J^{1/2}f_2(x) = \frac{1}{\pi}\int_0^x\frac{e^t \sqrt{\pi}\;\text{erf}\sqrt{t} - 2\sqrt{t}}{\sqrt{x-t}}\;dt .
$$
If we did not know where this integral came from, perhaps we would be amazed by the assertion
$$
\frac{1}{\pi}\int_0^x\frac{e^t \sqrt{\pi}\;\text{erf}\sqrt{t} - 2\sqrt{t}}{\sqrt{x-t}}\;dt = e^x-x-1
$$
But this is simply $J^1 f = J^{1/2} J^{1/2} f$, which is how the fractional integral is supposed to act.
A: This is a comment for G.Edgar's answer, but too long for a comment-box. 
Ahh, I see now: what I did was iterating the half-integral over its result, in the sense of $$ \int_0^{\int_0^x f(u) du} f(t) dt $$ instead of $$ \int_0^x \left( \int_0^t f(u) du \right) dt $$ 
(Of course, I should write here the semi-integral-notation, but I only want to point out the conceptional aspect)
Thanks, Gerald! I think I can now proceed on my own... I'll have some more questions another day, though.

So I should write:
$$ \begin{array}{lllllr} f(x) &=& \exp(x)-1  & &&&(1)\\
 f_1(x,x_0=0) &=& (J^{1/2} f){\Large |}_{x_0}^x  &=& \frac 1{\sqrt{\pi}} \int_{x_0}^x (x-t)^{-1/2} f(t) dt  &&(2)\\
 f_2(x,x_0=0) &=& (J^{1/2} f_1){\Large |}_{x_0}^x &=& {1 \over \sqrt{\pi}} \int_{x_0}^x (x-t)^{-1/2} f_1(t,x_0) dt &&(3) \end{array} $$
to arrive, of course, at equality:     
$ \displaystyle \begin{array}{lllllrrr} \phantom{aaaaa}&f_2(x,x_0=0) &=& (J^1 f){\Large |}_{x_0}^x &=& \int_{x_0}^x  f(t) dt &\phantom{aaaaaaaaaa}&\phantom{aaaaa}&(4) \end{array}$
Moreover, (according to G.Edgar's answer) we have some shortcuts (or "more closed forms") for the integrals:
$$\begin{array} {llll}(J^1 f)(x) &=& \int f(t) dt &=&  (e^x-1)  - x &&&(5) \\
 (J^{1/2} f)(x) &=&  &&  e^x\text{erf}(\sqrt{x})  - 2{\sqrt{x}\over\sqrt{\pi}} &&&(6) \\
 \text{ semi-derivative} \\
 (J^{-1/2} f)(x) &=& \frac d{dx}(J^{1/2}f)(x)  &=&  e^x\text{erf}(\sqrt{x}) &&&(7, \text{ WA }) \\
\end{array}$$
