Consider the fractional integro-derivative
$\displaystyle\frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=FP\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!}\frac{\beta!}{(z-x)^{\beta+1}}dz=FP\displaystyle\int_{0}^{x}\frac{z^{\alpha}}{\alpha!}\frac{(x-z)^{-\beta-1}}{(-\beta-1)!} dz$
$= \displaystyle\frac{x^{\alpha-\beta}}{(\alpha-\beta)!}$
where FP denotes a Hadamard-type finite part, $x>0$, and $\alpha$ and $\beta$ are real.
Identify Lie group elements and multiplication as
$\displaystyle(\frac{x^{\alpha}}{\alpha!},\frac{x^{\beta}}{\beta!})=FP \displaystyle\int_{0}^{\infty}\frac{z^{\alpha}}{\alpha!}\frac{d}{dx}H(x-z)\frac{(x-z)^{\beta}}{\beta!}dz= \frac{x^{\alpha+\beta}}{(\alpha+\beta)!}$
with $H(x)$ as the Heaviside step function.
The complex contour integral gives a continuation of the multiplication rule to the identity element $\beta=0$, so assume taking its derivative w.r.t. $\beta$ at $\beta=0$ gives a convolutional “infinitesimal generator” $R$ leading to
$\displaystyle(1-\epsilon R)\frac{x^{\alpha}}{\alpha!}=\frac{x^{\alpha}}{\alpha!}-\epsilon\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=|x|}\frac{-ln(z-x)+\lambda}{z-x}\frac{z^{\alpha}}{\alpha!} dz$ approximating $\frac{x^{\alpha+\epsilon }}{(\alpha+\epsilon)!}$ for small $\epsilon$ where $\lambda=d\beta!/d\beta|_{\beta=0}$.
Then, analogous to $(1+tA/n)^n$ tending to $exp(tA)$ as n tends to infinity, assume (letting $\alpha=0$ and $tA/n =-\beta R/n=-\epsilon R$)
$\displaystyle\frac{x^\beta}{\beta!} = exp(-\beta R) 1$.
Here $R^n$ represents repeated convolution initially acting on 1.
If this is true, then, equivalently, $R$ represents a raising operator for $\psi_{n}(x)=(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|_{\beta=0}$; that is,
$\psi_{n+1}(x)=R\psi_{n}(x)=\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=|x|}\frac{-ln(z-x)+\lambda}{z-x}\psi_{n}(z) dz$.
Update: The contour can be collapsed to the real line to obtain
$\psi_{n+1}(x)=R\psi_{n}(x)=(-ln(x)+\lambda)\psi_{n}(x)+\displaystyle\int_{0}^{x}\frac{\psi_{n}\left ( x\right )-\psi_n(u)}{x-u}du$.
I have basically two questions about the validity of these relations: A) Can the Lie group argument be made rigorous, and if so, how? B) Can anyone provide a proof of the raising operation independent of group theoretic arguments?
Any history on these relations would be appreciated also.
PS: By considering the limit of $\displaystyle\frac{1}{2}[\frac{(-1+a)!}{(z-x)^a}+\frac{(-1-a)!}{(z-x)^{-a}}]$
as $a$ tends to zero, using $\displaystyle\frac{sin(\pi u)}{\pi u}=\frac{1}{u!(-u)!}$, you can show that
$I_x=[R,x]=Rx-xR$ is the raising operator for $\displaystyle\frac{x^{\alpha}}{\alpha!}$ ; i.e.,
$\displaystyle I_x\frac{x^{\alpha}}{\alpha!}=\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=|x|}(-ln(z-x)+\lambda)\frac{z^{\alpha}}{\alpha!}dz=\frac{x^{\alpha+1}}{(\alpha+1)!}$.
So, $\displaystyle\frac{x^\beta}{\beta!} = \frac{1}{1+I_{\beta}R} 1$ also, eliminating all factorials, for $\beta>0$.