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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 $\beta>0$.

share|cite|improve this question
See also… – Tom Copeland Nov 2 '12 at 12:16
up vote 2 down vote accepted

Confirming that $R$ can be exponentiated to give the general fractional integro-derivative:

With $\bigtriangledown^{s}_{n}c_n=\sum_{n=0}^{\infty}(-1)^n \binom{s}{n}c_n$,

$$\displaystyle\frac{d^{-\beta}}{dx^-\beta}\frac{x^{\alpha}}{\alpha!}=\left (1-\left (1-\frac{d^{-1}}{dx^{-1}}\right ) \right )^{\beta}\frac{x^{\alpha}}{\alpha!}$$ $$=\bigtriangledown^{\beta}_{n} \bigtriangledown^{n}_{j}\frac{d^{-j}}{dx^{-j}} \frac{x^{\alpha}}{\alpha!}=\bigtriangledown^{\beta}_{n} \bigtriangledown^{n}_{j} \frac{x^{j+\alpha}}{(j+\alpha)!}=\frac{x^{\alpha+\beta}}{(\alpha+\beta)!},$$

implying $$-R\frac{x^{\alpha}}{\alpha!}=\frac{d}{d\beta}\frac{x^{\alpha+\beta}}{(\alpha+\beta)!}|_{\beta=0}=\ln\left (1-\left (1-\frac{d^{-1}}{dx^{-1}}\right ) \right )\frac{x^{\alpha}}{\alpha!}$$

$$=-\sum_{n=1}^{\infty }\frac{\bigtriangledown^{n}_{j}\frac{x^{j+\alpha}}{(j+\alpha)!}}{n}=\ln\left ( \frac{d^{-1}}{dx^{-1}}\right )\frac{x^{\alpha}}{\alpha!}=\ln\left [R,x\right ]\frac{x^{\alpha}}{\alpha!}.$$

Then considering the first line in each set of equations, exponentiating gives

$$\displaystyle\exp(-\beta R)=\left (1-\left (1-\frac{d^{-1}}{dx^{-1}}\right ) \right )^{\beta}=\frac{d^{-\beta}}{dx^{-\beta}}.$$

Aside: Evaluating the complex contour integral as a Fourier transform on the circle of radius $x$ from $\theta=-\pi$ to $\pi$ confirms

$\displaystyle R\frac{x^{\alpha}}{\alpha!}=\left \{ [-\ln\left ( x\right )+ \lambda]\frac{x^{\alpha}}{\alpha!} +\int_{0}^{x}\frac{\frac{x^\alpha}{\alpha!}-\frac{u^\alpha}{\alpha!}}{x-u}du \right \}=\left \{ -\ln\left ( x\right )+ \lambda +\int_{0}^{1}\frac{1-u^\alpha}{1-u}du \right \}\frac{x^{\alpha}}{\alpha!}=\left \{ -\ln\left ( x\right )+ \lambda +H_{\alpha}\right \}\frac{x^{\alpha}}{\alpha!}=\frac{d}{d\beta}\frac{x^{\alpha-\beta}}{(\alpha-\beta)!}|_{\beta=0},$ which agrees with an integral expression for the digamma function and connects the operator with combinatorics of the generalized harmonic numbers $H_{\alpha}$.

Edit May 2015:

This is also consistent with the Pincherle derivative with $x$ as the raising operator and $D=\frac{d}{dx}$ as the lowering operator for the monomials $x^n$, noting

$$[R,x] = \frac{d\ln(D)}{dD} = D^{-1} \; .$$

share|cite|improve this answer
For more info on the raising op, see… – Tom Copeland May 23 at 18:44

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