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
  • $\begingroup$ You want to compute a function $f_2$, say, as the semi-integral of $f$, then compute the semi-integral of $f_2$. You have only $f(t) = \exp(t)-1$ in your formula. $\endgroup$ – GEdgar Oct 9 '13 at 14:30
  • $\begingroup$ @GEdgar: I just assumed that "semi-integral" gives a function, which composed with itself gives the ordinary integral: $b=semiInt(0,a);c=semiInt(0,b); c=?=Int(0,a) $. Did I misunderstand something here? $\endgroup$ – Gottfried Helms Oct 9 '13 at 14:36

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.

  • $\begingroup$ Hmm, I could reproduce the ordinary integral $(J^1f)(x)$ with this. Do I understand it correctly: the Cauchy's semi-integral-formalism should be applied to some (beforehand) unknown function $f_2(x)$ such that $(J^{1/2}f_2)(x) = (J^1 f)(x) $. Then find an appropriate function $f_2(x)$ ? - Well, I've difficulties to understand this because I'm much conditioned for functional composition which seems a completely different rationale... $\endgroup$ – Gottfried Helms Oct 9 '13 at 15:36
  • $\begingroup$ And more generally, $J^a J^b f = J^{a+b} f$. $\endgroup$ – GEdgar Oct 9 '13 at 16:27
  • $\begingroup$ Please see my comment in a new "own answer"-box. Thanks for your explanations! $\endgroup$ – Gottfried Helms Oct 9 '13 at 16:47
  • $\begingroup$ Perhaps in the 3rd formula it could be made more explicite, that inside the integral is the previously defined function $f_2(x)$ such that $J^{1/2}f_2(x) = \frac 1{\sqrt {\pi}} \int_0^x f_2(t) (x-t)^{-1/2} dt $ which occurs visibly after expanding/cancelling of $\sqrt {\pi}$ $\endgroup$ – Gottfried Helms Oct 9 '13 at 17:25

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:
$$\begin{array} {llll}(J^1 f)(x) &=& \int f(t) dt &=& (e^x-1) - x &&&(5) \\ (J^{1/2} f)(x) &=& && e^x\text{erf}(\sqrt{x}) - 2{\sqrt{x}\over\sqrt{\pi}} &&&(6) \\ \text{ semi-derivative} \\ (J^{-1/2} f)(x) &=& \frac d{dx}(J^{1/2}f)(x) &=& e^x\text{erf}(\sqrt{x}) &&&(7, \text{ WA }) \\ \end{array}$$


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