## Fractional Calculus, Gamma Classes, the Riemann Zeta Function, and an Appell Pair of Sequences

The background info and comments for the MSE question Lie group heuristics for a raising operator for $\displaystyle(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|_{\beta=0}$ and the MO question Riemann zeta function at positive integers and an Appell sequence of poylnomials introduce an Appell sequence of polynomials containing the Euler-Mascheroni constant and the Riemann zeta function evaluated at the integers greater than one. The Appell sequence can be defined by its e.g.f.

$\displaystyle \exp(\beta \; p.(z)) = \exp(\beta \; z) / \beta! \; .$

The raising op for the Appell sequence

$\displaystyle R_z=z-\frac{\mathrm{d} }{\mathrm{d} \beta}ln[\beta!]\mid _{\beta=\frac{\mathrm{d} }{\mathrm{d} z}=D_z}=z-\Psi(1+D_z) \; ,$

where $\displaystyle \Psi(x)$ is the digamma or Psi function, is associated with the infinitesimal generator

$\displaystyle R_x = \log(d/dx)= \log(D_x)$

of a class of fractional integro-derivatives through a change of variables ($\displaystyle z=\log(x)$). Exponentiation gives

$\displaystyle e^{\beta \; R_x} = D_x^{\beta} \; .$

As discussed in previous entries here on Appell sequences, every Appell sequence of polynomials has an umbral compositional inverse sequence of polynomials, which is also an Appell sequence whose raising op involves only a simple change in a sign in the formula for the inverse sequence’s raising op. The e.g.f. of the umbral inverse sequence in this case is

$\displaystyle \exp(\beta \; \bar{p}.(z)) = \beta! \; exp(\beta \; z) \;,$

with the raising op

$\displaystyle \bar{R}_z=z+\frac{\mathrm{d} }{\mathrm{d} \beta}ln[\beta!]\mid _{\beta=\frac{\mathrm{d} }{\mathrm{d} z}=D_z}=z+\Psi(1+D_z) \; .$

The associated raising op through the transformation $\displaystyle z = \log(x)$ is

$\displaystyle \bar{R}_x = \log(xD_xx) \; ,$

which, when exponentiated, gives

$\displaystyle e^{\beta \; \bar{R}_x} = (xD_xx)^\beta = x^\beta \; D_x^{\beta} \; x^\beta = \beta! \; x^\beta \; L_{\beta}(-:xD_x:)$

$\displaystyle = \beta! \; x^\beta \; K(-\beta, 1, -:xD_x:) = \beta! \; x^\beta \; \binom{xD_x + \beta}{\beta}$

$\displaystyle = \beta! \; x^\beta \; \sum_{k \ge 0} (-1)^k \sum_{j=0}^k (-1)^j \binom{k}{j} \; \binom{j+\beta}{\beta} \; \frac{:xD_x:^k}{k!} = \beta! \; x^\beta \; \sum_{k \ge 0} \; \binom{\beta}{k} \; \frac{x^kD_x^k}{k!} \;,$

where $\displaystyle (:xD_x:)^n = x^n \; D_x^n \;$ (cf. Pochhammer symbol of a differential and hypergeometric polynomials for definitions of the symbols).

These relations are simple to derive given the property defining the raising ops:

$\displaystyle e^{\beta \; \bar{R}_z} \; 1 = e^{\beta \; \bar{p}.(z)} = \beta! \; e^{\beta \; z} \;,$ so

$\displaystyle e^{\alpha \; \bar{R}_z} e^{\beta \; \bar{R}_z} \; 1 = e^{(\alpha + \beta) \; \bar{R}_z} \; 1 = (\alpha + \beta)! \; e^{(\alpha+\beta) \; z} = e^{\alpha \; \bar{R}_z} \; \beta! \; e^{\beta \; z} \; .$

Then changing variables gives

$\displaystyle e^{\alpha \bar{R}_x} \beta! \; x^\beta = (\alpha + \beta)! \; x^{\alpha+\beta} \; ,$

implying

$\displaystyle e^{\alpha \bar{R}_x} \beta! \; x^\beta = x^{\alpha} \; D_x^{\alpha} \; x^{\alpha} \; \beta! \; x^\beta = (\alpha + \beta)! \; x^{\alpha+\beta} \; ,$ or

$\displaystyle e^{\alpha \bar{R}_x} = (xD_xx)^\alpha = x^{\alpha} \; D_x^{\alpha} \; x^{\alpha} \; .$

As a sanity check, confirm

$\displaystyle e^{\alpha \; \bar{R}_x} \; e^{\beta \; \bar{R}_x} x^s = \alpha! \; x^\alpha \; \binom{xD_x + \alpha}{\alpha} \; \beta! \; x^\beta \; \binom{xD_x + \beta}{\beta} x^s = (\alpha+\beta)! \; x^{\alpha+\beta} \; \binom{xD_x + \alpha + \beta}{\alpha+\beta} \; x^s \; .$

The relation to $R_x$ can be seen from the derivative

$\displaystyle D_{\beta=0} \; x^\beta D_x^\beta x^\beta = 2 \; \log(x) + \log(D_x) = D_{\beta=0} \; e^{\beta \; \bar{R}_x} = \bar{R}_x \; ,$

implying, from the notes attached to the MSE and MO questions,

$\displaystyle \bar{R}_x = 2 \; \log(x) + \log(D_x) = 2 \; \log(x) + R_x \; ,$

where

$\displaystyle R_xf(x)=\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=|x|}\frac{-\log(z-x)+\lambda}{z-x}f(z)dz$

$\displaystyle =(-\log(x)+\lambda)f(x)+\displaystyle\int_{0}^{x}\frac{f\left ( x\right )-f(u)}{x-u}du.$

with $\displaystyle \lambda=d\beta!/d\beta|_{\beta=0}$. (Note the integrand is related to the q (Jackson) derivative, and the Pincherle derivative / commutator is $\displaystyle [\bar{R}_x,x]=D_x^{-1}$.)

Additionally, using well-known formulas for the digamma function with $\displaystyle \gamma =- \frac{d\beta!}{d\beta} |_{\beta=0} = -\lambda$ ,

$\displaystyle R_x = -\log(x) + \psi(1 + xD_x) = -(\log(x) + \gamma) - \sum_{n \ge 1} (-1)^n \zeta(n+1) \; (xD_x)^n$

$\displaystyle = -(\log(x) + \gamma) - \sum_{n \ge 1} (-1)^n \; \frac{1}{n} \; \binom{xD_x}{n} = -(\log(x) + \gamma) - \sum_{n \ge 1} (-1)^n \; \frac{1}{n} \; \frac{x^n D_x^n}{n!} \; .$

(The last expression is consistent with the derivative of the formulas for $\displaystyle e^{\beta \; \bar{R}_x}$ above.)

Explicit formulas for the first few Appell polynomials associated with $\displaystyle R_z$ are

$\displaystyle p_{0}(z)=1 \; ,$
$\displaystyle p_{1}(z)=z+\gamma$,
$\displaystyle p_2(z)=(z+\gamma)^2-\zeta(2)$
$\displaystyle p_3(z)=(z+\gamma)^3-3\zeta(2)(z+\gamma)+2\zeta(3)$
$\displaystyle p_4(z)=(z+\gamma)^4-6\zeta(2)(z+\gamma)^2+8\zeta(3)(z+\gamma)+3[\zeta^2(2)-2\zeta(4)]$
$\displaystyle p_5=p_1^5-10\zeta(2)p_1^3+20\zeta(3)p_1^2+15[\zeta^2(2)-2\zeta(4)]p_1+4[-5\zeta(2)\zeta(3)+6\zeta(5)] \; ,$

where $\displaystyle \gamma=-\frac{\mathrm{d} }{\mathrm{d} \beta }\beta !\mid_{\beta =0 }$, the Euler-Mascheroni constant, and $\displaystyle \zeta(s)$ is the Riemann zeta function.

For $\displaystyle n>0 \; ,$

$\displaystyle p_{n+1}(z)=(z+\gamma)p_{n}(x)+\sum_{j=1}^{n}(-1)^j\binom{n}{j}j!\zeta (j+1)p_{n-j}(z) \;.$

See the MO question for more notes on the polynomials and a derivation of the recursion relation. See the entry here on Appell polynomials and cumulants for a general derivation of a recursion using the raising operator that is satisfied by all Appell sequences.

The sequence of polynomials associated to $\displaystyle \bar{R}_z \;$ are given by negating the Euler-Mascheroni constant and the zeta values in the $\displaystyle p_n(z) \;$ since the raising op has the same form as that for the cycle index partition polynomials for the symmetric groups (cf. the MO question and OEIS A036039):

the first few are

$\displaystyle \bar{p}_{0}(z)=1 \; ,$
$\displaystyle \bar{p}_{1}(z)=z-\gamma$,
$\displaystyle \bar{p}_2(z)=(z-\gamma)^2+\zeta(2)$
$\displaystyle \bar{p}_3(z)=(z-\gamma)^3+3\zeta(2)(z-\gamma)-2\zeta(3)$
$\displaystyle \bar{p}_4(z)=(z-\gamma)^4+6\zeta(2)(z-\gamma)^2-8\zeta(3)(z-\gamma)+3[\zeta^2(2)+2\zeta(4)]$
$\displaystyle \bar{p}_5=\bar{p}_1^5+10\zeta(2)\bar{p}_1^3-20\zeta(3)\bar{p}_1^2+15[\zeta^2(2)+2\zeta(4)]\bar{p}_1+4[-5\zeta(2)\zeta(3)-6\zeta(5)] \; .$

They comprise the umbral compositional inverses of $\displaystyle p_n(z) \; ;$ that is,

$\displaystyle p_n(\bar{p}.(z)) = z^n = \bar{p}_n(p.(z))\; .$

For example,

$\displaystyle p_2(\bar{p}.(z)) =(\bar{p}.(z) +\gamma)^2-\zeta(2) = (\bar{p}.(z))^2 + 2 \gamma(\bar{p}.(z))^1 + \gamma^2 - \zeta(2) \;$

$\displaystyle = \bar{p}_2(z) + 2 \gamma\; \bar{p}_1(z) + \gamma^2 - \zeta(2)$

$\displaystyle = (z-\gamma)^2 + \zeta(2) + 2 \gamma \; (z- \gamma) + \gamma^2 - \zeta(2) = z^2 \; .$

Relation to the Mellin transform, convolutions, and umbral calculus:

Given the Mellin transform of $\displaystyle g(x)$ as $\displaystyle \hat{g}(s) \;$ and using the Mellin transform convolution theorem,

$\displaystyle x^{\beta} \; D_x^{\beta} \; x^{\beta} g(x) = x^{\beta} \; D_x^{\beta} \; x^{\beta} \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} \hat{g}(s) \; x^{-s} \; ds = x^{\beta} \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} \hat{g}(s) \; \frac{(-s+\beta)!}{(-s)!}\; x^{-s} \; ds$

$\displaystyle = x^{\beta} \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} \hat{g}(s) \; \hat{h}(1-s) \; x^{-s} \; ds = x^{\beta} \int_0^\infty h(t) \; g(xt) \; dt \; ,$

where, using the Euler reflection formula for factorials,

$\displaystyle \hat{h}(s) = \frac{(\beta+s-1)!}{(s-1)!} = (\beta)_{\frac{s}{}} \; = \frac{\pi}{\sin(\pi(s+\beta))} \; \frac{1}{(-s-\beta)!} \; \frac{1}{(s-1)!}$

and, with $\displaystyle H(x)$ the Heaviside step function, for $\sigma > 0 \; \mathrm{amd} \; \beta < 0$,

$\displaystyle h(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} \hat{h}(s) \; x^{-s} \; ds = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} \frac{\pi}{\sin(\pi(s+\beta))} \; \frac{1}{(-s-\beta)!} \; \frac{1}{(s-1)!} \; x^{-s} \; ds$

$\displaystyle = H(1-x) x^\beta \; \sum_{n \ge 0} (-1)^n \frac{1}{(-\beta-1-n)!} \; \frac{x^n}{n!} = H(1-x) \frac{x^\beta}{(-\beta-1)!} \; \sum_{n \ge 0} (-1)^n \binom{-\beta-1}{n} \; x^n$

$\displaystyle = H(1-x) \; \frac{(1-x)^{-\beta-1}}{(-\beta-1)!} \; x^\beta \; ,$

so

$\displaystyle x^{\beta} \; D_x^{\beta} \; x^{\beta} g(x) = x^{\beta} \int_0^\infty h(t) \; g(xt) \; dt = x^{\beta} \int_0^1 \; \frac{(1-t)^{-\beta-1}}{(-\beta-1)!} \; t^\beta \; g(xt) \; dt \; ,$

where now $g(x)$ can be extended beyond those functions that have Mellin transforms. Moreover, the last convolution integral can be analytically continued to positive values of $\beta$ as a Hadamard finite part integral for fractional differintegration that formally acts on a function analytic at $\displaystyle t=1$ that has been dilated by $\displaystyle t^\beta$ and $\displaystyle x \;$:

$\displaystyle x^{\beta} \; D_x^{\beta} \; x^{\beta} g(x) = x^\beta \; D_t^{\beta} \; t^{\beta} g(xt) \; |_{t=1} = x^\beta \sum_{n \ge 0} g_n \; x^n \; D_t^{\beta} \; t^{\beta} \frac{t^n}{n!} \; |_{t=1}$

$\displaystyle = x^\beta \sum_{n \ge 0} g_n \frac{(n+\beta)!}{n!} \; \frac{x^n}{n!} = \beta! \; x^\beta \sum_{n \ge 0} g_n \binom{n+\beta}{n} \; \frac{x^n}{n!} = \beta! \; x^\beta \sum_{n \ge 0} (-1)^n \; g_n \binom{-\beta-1}{n} \; \frac{x^n}{n!}$

$\displaystyle = x^\beta \frac{\pi}{\sin{\pi \beta}} \; \sum_{n \ge 0} (-1)^{n+1} \; \frac{g_n}{n!} \; \frac{1}{(-\beta-1-n)!} \; \frac{x^n}{n!} \; .$

Summarizing, the Mellin convolution theorem leads to the fractional differintegral

$\displaystyle x^{\beta} \; D_x^{\beta} \; x^{\beta} g(x) = \beta! \; x^\beta \sum_{n \ge 0} g_n \binom{n+\beta}{n} \; \frac{x^n}{n!} \; = \beta! \; x^\beta g(a.x) \;$

with the umbral compositional rep

$\displaystyle (a.x)^n = a_n \; x^n = \binom{n+\beta}{n} x^n = (-1)^n \; \binom{-\beta-1}{n} x^n\; .$

The generalized shift operator gives a diff op rep for the umbral composition as

$\displaystyle e^{-(1-a.)\;:xD_x:} \; g(x) = g(x-(1-a.)x) = g(a.x) \; ,$

with the definition $\displaystyle (:xD_x:)^n = x^n\;D_x^n \; ,$

or

$\displaystyle \sum_{n \ge 0} (-1)^n \; \triangle_{k}^n \; a_k \; \frac{:xD_x:^n}{n!} g(x) = g(a.x) \; ,$

with $\displaystyle (1-a.)^n = \triangle_{k}^n \; a_k = \sum_{k=0}^n (-1)^k \; \binom{n}{k} \; a_k\; .$

From the Chu-Vandermonde identity,

$\displaystyle \sum_{k=0}^n (-1)^k \; \binom{n}{k} \; \binom{k+\beta}{k} = \sum_{k=0}^n \; \binom{n}{k} \; \binom{-\beta-1}{k} = \binom{-\beta-1+n}{n} = (-1)^n \;\binom{\beta}{n} ,$

so

$\displaystyle x^{\beta} \; D_x^{\beta} \; x^{\beta} = \beta! \; x^\beta \; \sum_{n \ge 0} (-1)^n \; \triangle_{k}^n \; \binom{k+\beta}{k} \; \frac{x^nD_x^n}{n!} = \beta! \; x^\beta \; \sum_{n \ge 0} \; \binom{\beta}{n} \; \frac{x^nD_x^n}{n!} \;$

in agreement with the other derivations above.

Note that when the diff op is acting on a function analytic about the origin, the simpler rep applies

$\displaystyle x^{\beta} \; D_x^{\beta} \; x^\beta = \beta! \; x^\beta \; e^{a. \; :xD_{x=0}:} = \beta! \; x^\beta \; \sum_{n \ge 0} \; \binom{n+\beta}{n} \; \frac{x^nD_{x=0}^n}{n!} \; ,$

but acting on the more general $\displaystyle x^s$ gives a Newton series interpolation, regenerating the generalized Chu-Vandermonde identity:

$\displaystyle x^{\beta} \; D_x^{\beta} \; x^\beta \; x^s = \frac{(s+\beta)!}{s!} \; x^{\beta} \; x^{s} = \beta! \; x^\beta \; \sum_{n \ge 0} (-1)^n \; \triangle_{k}^n \; \binom{k+\beta}{k} \; \binom{s}{n} \; x^s$

$\displaystyle = \beta! \; x^\beta \; \sum_{n \ge 0} \; \binom{\beta}{n} \; \binom{s}{n} \; x^s = \beta! \; x^\beta \; (a.x)^s = \beta! \; x^\beta \; a_s \; x^s \; .$

The infinigen $\displaystyle \hat{R}_x$ from a Cauchy contour integral rep:

$\displaystyle x^{\beta} \; D_x^{\beta} \; x^\beta \; g(x) = x^\beta F.P. \; \int_0^x \; \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} \; t^\beta \; g(t) \; dt = x^{\beta} \frac{1}{2 \pi i} \oint_{|z-x|=x} \frac{\beta!}{(z-x)^{\beta+1}}\; z^\beta \; g(z) \;dz \;$

is satisfied for $g(x)$ analytic on the interval of integration with no finite part required for $-1 < \beta \le 0$. For other values of $\beta$, whichever integral is convergent can be taken as the analytic continuation of the divergent one. Taking the derivative $\frac{d}{d\beta} \; |_{\beta=0}$ of the Cauchy contour formula gives

$\displaystyle \hat{R}_x g(x) = \log(x) g(x) \; + \; \frac{1}{2 \pi i} \oint_{|z-x|=x} \frac{-\log(z-x)-\gamma}{z-x} \; g(z) \;dz + \; \log(x)\; g(x) \; ,$

and collapsing the contour about the real axis gives

$\displaystyle \hat{R}_x \; g(x) = 2 \; \log(x)\; g(x) + (-\log(x)-\gamma) \; g(x) \; + \; \int_{0}^{x}\frac{g\left ( x\right )-g(t)}{x-t} \; dt$

$\displaystyle = (\log(x)-\gamma)\;g(x) + \int_{0}^{x}\frac{g\left ( x\right )-g(t)}{x-t} \; dt \; .$

Consistency with Mellin interpolation of operators:

Probably the mathematician to most ingeniously wield the power of the Mellin transform in some sense as an interpolation of the coefficients of exponential generating functions or Taylor series was Ramanujan with his Master Formula / Theorem. The iconic Euler integral for the gamma function for $Real(s) > 0$

$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-t\;p} \; dt = p^{-s}$

provides the scaffolding for similarly interpolating operators. Let’s first show that a natural interpolation of integer powers of the derivative is the fractional integroderivative of fractional calculus by using the Mellin transform to interpolate the op coefficients of the op e.g.f. $\displaystyle e^{tD_x} \;,$ i.e., the shift op, for the integer powers of the derivative:

$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; dt \; H(x) g(x) = D_x^{-s} H(x) g(x) = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; H(x) g(x)\; dt$

$\displaystyle = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; g(x-t) dt \; .$

Then specifically acting on the power function for $\displaystyle \alpha > -1$

$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; (x-t)^\alpha dt = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^\alpha \; dt$

$\displaystyle = \int_0^x \frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \frac{\alpha!}{(\alpha-k)} \; \frac{t^k}{k!} \; dt = \frac{1}{(s-1)!} \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \binom{\alpha}{k} \; \frac{t^{s+k}}{s+k} \; |_{t=0}^{x}$

$\displaystyle = x^{\alpha + s} \; (-s)! \; \sum_{k \ge 0} \; \binom{\alpha}{k} \; \frac{sin(\pi (s+k))}{\pi (s+k)} = x^{\alpha +s} \frac{\alpha!}{(\alpha+s)!} \; = D_x^{-s} x^\alpha \; .$

The last summation converges with no restriction on $s$. So, we see that the Mellin transform does indeed interpolate the coefficients of the e.g.f. generated by the binomial theorem expansion $x^{\alpha-k} \frac{\alpha!}{(\alpha-k)}$ to $x^{\alpha+s} \frac{\alpha!}{(\alpha+s)}$ to give an interpolation of the coefficients of the shift op $D_x^k$ to $D_x^{-s}$ consistent with fractional calculus.

Next let’s use the same method to interpolate

$\displaystyle (x \; D_x \;x)^n = x^n D_x^n x^n = x^n \; n!\; L_n(-:xD_x:) ,$

where $L$ denotes the Laguerre polynomials, and, again, $(:xD_x:)^k = x^kD_x^k$ by definition. Following the algorithm above, we have from the generating function for the Laguerre polynomials the action of the generalized shift op

$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-txD_xx} \; dt \; H(x) x^\alpha = (xD_xx)^{-s}\; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{\exp[\frac{-xt}{1+xt} \; :uD_u:]}{1+xt} \; H(u)\; u^\alpha\; dt \; |_{u=x}$

$\displaystyle = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(\frac{x}{1+xt}) \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt$

$\displaystyle = \int_0^{1/x}\frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0}(-1)^k x^{\alpha+k} \frac{(\alpha+k)!}{\alpha!} \;\frac{t^k}{k!}\; dt + \int_{1/x}^{\infty}\frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0}(-1)^k x^{-k-1} \frac{(\alpha+k)!}{\alpha!} \;\frac{t^{-k-\alpha-1}}{k!}\; dt$

$\displaystyle = \frac{1}{(s-1)!} \; \sum_{k \ge 0}(-1)^k x^{\alpha+k} \binom{\alpha+k}{k} \; [\; \frac{t^{s+k}}{s+k}\; |_{t=0}^{1/x} \;+\; x^{-k-1} \; \frac{t^{s-\alpha-k-1}}{s-\alpha-k-1} \; |_{t=1/x}^\infty \;]$

$\displaystyle = \frac{x^{\alpha-s}}{(s-1)!} \; \sum_{k \ge 0}(-1)^k \binom{\alpha+k}{k} \; [ \; \frac{c^{s+k}}{s+k} \;+\; \frac{c^{-s+\alpha+k+1}}{-s+\alpha+k+1} \;]$

for $0 < Real(s) < \alpha +1$, where $c$ is a convergence factor approaching unity necessary when $\alpha > 0$. We can anticipate that this sum converges to an interpolation of the coefficients of the e.g.f. $x^{\alpha+k} \frac{(\alpha+k)!}{\alpha!}$ to $x^{\alpha-s} \frac{(\alpha-s)!}{\alpha!}$ in agreement with the interpretation above of $(x D_x x)^\beta = x^\beta D_x^\beta x^\beta$. With these interpolation heuristics, the summation formula can be verified by using the inverse Mellin transform and the singularities of the factorials.

So, analytically continued from $0 < Real(s) < \alpha + 1$,

$\displaystyle (x D_x x)^{-s} \; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt = x^{\alpha-s} \frac{(\alpha-s)!}{\alpha!} = x^{-s} D_x ^{-s} x^{-s} \; x^\alpha \; .$

To confirm the presumed Mellin transform evaluation from the interpolation heuristic applied either to the real coefficients of the e.g.f. in the integrand of the transform or the operator e.g.f. do the inverse Mellin transform. The modified MT pair is

$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; g_\sigma(t) \; dt = \hat{g}(s) \; \; \mathrm{and} \; \; \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} (s-1)! \; \hat{g}(s) \; t^{-s} \; ds \; ,$

and our presumed MT result gives

$\displaystyle (s-1)! \; \hat{g}(s) = x^{\alpha-s} \frac{(s-1)! \; (\alpha - s)!}{\alpha!} = x^{\alpha-s} \frac{\pi}{sin(\pi s)} \; \binom{\alpha-s}{-s}$

$\displaystyle = x^{\alpha-s} \; \frac{\pi}{sin(\pi(s-\alpha))} \; \binom{s-1}{\alpha} \; ,$

so, for $0 < \sigma < \alpha + 1$, the inverse Mellin transform to evaluate is

$\displaystyle \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} x^{\alpha} \frac{\pi}{sin(\pi s)} \; \binom{\alpha-s}{-s} \; (xt)^{-s} \; ds = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} x^{\alpha} \; \frac{\pi}{sin(\pi(s-\alpha))} \; \binom{s-1}{\alpha} \; (xt)^{-s} \; ds \; .$

Closing the first contour to the left for $x t < 1$ picks up the singularities at $s = 0,-1,-2, ..., -k, ....$, and closing the second contour to the right for $x t > 1$ picks up a sign for the clockwise path transversal and the singularities at $s = \alpha +1, \alpha + 2, ..., k+\alpha+1, ...$, giving

$\displaystyle \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} (s-1)! \; x^{\alpha-s} \frac{ (\alpha - s)!}{\alpha!} \; t^{-s} \; ds$

$\displaystyle = H(1-x t) \; x^\alpha \sum_{k \ge 0} (-1)^k \binom{\alpha+k}{k}(x t)^k + H(x t-1) \; x^\alpha \sum_{k \ge 0} (-1)^k \binom{\alpha+k}{k}(x t)^{-\alpha-k-1}$

$\displaystyle = \frac{x^\alpha}{(1+xt)^{\alpha + 1}}$

since $(-1)^k \binom{\alpha+k}{k} + \binom{-\alpha-1}{k} \;$ , confirming that the presumed solution is indeed correct.

On a more general note, the algorithm can be applied to differential operators encompassing the Witt Lie algebra to obtain

$\displaystyle (x^{1+y}D_x)^{-s} \; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H[\frac{x}{(1+y\;t\;x^y)^{1/y}}] \frac{x^\alpha}{(1+y\;t\;x^y)^{\alpha/y}} \; dt$

$\displaystyle =H(y) \; x^{\alpha-sy} y^{-s} \frac{(-s+\alpha/y-1)!}{(\alpha/y-1)!} \;+ \; H(-y) \; x^{\alpha+s|y|} |y|^{-s} \frac{(\alpha/|y|)!}{(\alpha/|y|+s)!} \;.$

$\displaystyle =H(y) \; x^{\alpha-sy} y^{-s} \frac{(-s+\alpha/y-1)!}{(\alpha/y-1)!} \;+ \; H(-y) \; x^{\alpha-sy} (-y)^{-s} \frac{(-\alpha/y)!}{(-\alpha/y+s)!} \;.$

For $y < 0$, the Heaviside step function in the integral is to be interpreted as setting the upper limit of the integral as the first zero from the origin of $\displaystyle (1-|y|tx^{-|y|})^{\alpha/|y|}$, i.e., the upper limit should be $t = x^{|y|}/|y|$ .

Related stuff:

For Appell sequences:

“Bernoulli Appells” entry below.

For differential ops:

“Diff. Ops, Special Polynomials, Binomial Transforms, Interpolation, and the Inverse Mellin Transform” entry below,

“Goin’ with the Flow: Log of the Derivative” entry below,

“Infinigens, the Pascal Triangle, and the Witt and Virasoro Algebra” entry below,

“A Generalized Dobinski Relation and the Confluent Hypergeometric Functions” entry below,

“Lagrange a la Lah: Part I” entry below.

For Ramanujan’s Master Theorem / Formula:

“Ramanajun’s Master Theorem and Duality of Symmetric Spaces” by Bertram