## Mellin Interpolation of Differential Ops and Associated Infinigens and Appell Polynomials: The Ordered, Laguerre, and Scherk-Witt-Lie Diff Ops

Interpolations of the derivative operator $D_x^n \; ,$ the fundamental ordered op $:xD_x:^n=x^nD_x^n \; ,$ the Laguerre op $:D_xx:^n = D^nx^n \; ,$ the shifted Laguerre op $(xD_xx)^n = x^nD_x^nx^n \; ,$ and the generalized Scherk-Witt Lie ops $(x^{1+y}D_x)^n$ to the fractional operators $D_x^s\; , \; :xD_x:^s = x^sD_x^s \; , \; :D_xx:^s = D_x^sx^s \; , \; (xD_xx)^s = x^sD_x^sx^s \; , \;$ and $(x^{1+y}D_x)^s$ are consistently achieved using the Mellin transform of the negated e.g.f.s of the differential ops. Associated infinitesimal generators (infinigens) are then determined for each fractional op and related to the raising ops for associated Appell sequences.

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 viewed as the fundamental modified Mellin transform of the exponential generating functions for the basic positive integer powers $p^n$ interpolates the integer power functions to $p^s$ through

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

This holds strictly for $Real(s) > 0$, but can be analytically continued to the left complex plane either by using a Hankel countour or subtracting out the singularities step-by-step for swathes of the left side (see Domain of the gamma function).

This fundamental modified Mellin transform (MT) and its inverse provide the scaffolding for similarly interpolating differential operators.

Interpolation of $D_x^n$ and the Riemann-Liouville fractional calculus

Let’s first show that a natural interpolation of positive integer powers of the derivative is the fractional integroderivative of fractional calculus by using the Mellin transform to interpolate the operator coefficients, $D^n_x = (d/dx)^n$ of the op e.g.f. $\displaystyle e^{tD_x} \;,$ i.e., the shift operator:

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

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

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 $s$, and so justifies the analytic continuation of the integral by taking the Hadamard finite part, i.e., by ignoring the singularities incurred at the lower limit $t =0$ for $Real(s+k) < 0$ by subtracting out the appropriates terms of the regular e.g.f., or, equivalently, shifting the contour of the inverse Mellin transform as demonstrated in the MSE question/answer referenced above. Another method of circumventing the singularities, equivalent to reversing a Hankel contour, is to blow up the integral from the real line into the complex plane, forming a Cauchy contour integral related to a Fourier transform of the binomial coefficients, leading again to the sinc function interpolation of the general binomial coefficient as demonstrated in other entries here.

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.

Looking at this op from another perspective enables us to more easily generalize the results to similar operators below. Expressing it in terms of the Euler (state / number) op,

$\displaystyle D_x^n x^\alpha= x^{-n} n! \binom{xD_x}{n} x^\alpha = x^{-n} \; :xD_x:^n x^\alpha = x^{-n}\frac{\alpha!}{(\alpha-n)!} \; x^\alpha\;,$

where an ordering notation for ops is defined by $:ABC:^n = A^nB^nC^n \; .$ (This implies the well-known relation for the Bell / Touchard / exponential polynomials $\binom{\phi(:xD_x:)}{n} = \binom{xD}{n} = \frac{:xD_x:^n}{n!} \; .)$ Then the op e.g.f. and its action is

$\displaystyle e^{-tD_x} H(x)x^\alpha = \sum_{k \ge 0} (-t/x)^n\binom{xD_x}{n} H(x)x^\alpha = (1-t/x)^{xD_x} H(x)x^\alpha$

$\displaystyle = e^{(t/u)\; :xD_x:} H(x) x^\alpha |_{u=x} = H(x(1-t/x))(x(1-t/x))^\alpha \; = H(x-t)(x-t)^\alpha.$

Note that for $s = 0,-1,-2,... \;$ that $D_x^{-s} = D_x^n\;$, so the MT becomes

$\displaystyle D_x^n \; g(x) = H(x)\; \int_{-\infty}^x H(t)\frac{t^{-n-1}}{(-n-1)!} \; g(x-t) \; dt$

$\displaystyle = H(x)\;\int_{-\infty}^x \delta^{(n)}(t)\; g(x-t) \; dt = H(x)\; (-D_t)^n |_{t=0} \; g(x-t) \;$

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

Infinigen for the Fractional Integroderivative $D_x^s$

In the previous entry, some methods for determining the operadic infinitesimal generator (infinigen) for $D_x^\beta$ are presented. Another, direct method is available by expressing the operators in terms of the Euler (state / number operator) $xD_x$:

$\displaystyle D_x^{-s} x^\alpha = x^s \; \frac{(\alpha)!}{(\alpha+s)!} \; x^\alpha = x^s \; \frac{(xD_x)!}{(xD_x+s)!} \; x^\alpha .$

And, the group properties of the operator imply that it can be expressed as the exponential of an infinigen:

$\displaystyle D_x^{-s} = e^{-s\; R_x} = x^s \; \frac{(xD_x)!}{(xD_x+s)!} \; .$

Then

$\displaystyle R_x = -\frac{d}{ds} \; |_{s=0} \; x^s \; \frac{(xD_x)!}{(xD_x+s)!} = -\log(x) + \psi(1+xD_x)= \log(D_x)$

since

$\displaystyle \frac{d}{ds} \; |_{s=0} \; \frac{(xD_x)!}{(xD_x+s)!} = -\frac{d}{d(xD_x)} \; \log((xD_x)!) = -\psi(1+xD_x) \; ,$

where $\psi(1+x)=D_x \log(\Gamma(1+x))=D_x\log(x!)$ is the digamma, or Psi, function, which can be expressed in terms of the Riemann zeta function as $\psi(x+1) = -\gamma - \sum_{k \ge 1} \zeta(k+1)(-x)^k \;$ with $\gamma=-\psi(1)$, the Euler-Mascheroni constant.

Changing variables $x = e^{z}$ and letting the op act on unity, or more precisely $H(z)$, gives the e.g.f. for a Sheffer Appell sequence $p_n(z)$ with raising op $R_z$ and lowering op $D_z$ :

$\displaystyle D_x^{-s} H(x) = e^{-s\;R_x} 1 = \frac{x^s}{s!} = \frac{e^{sz}}{s!} = e^{s \; R_z} 1 = e^{s \; p.(z)}$

with $R_z = z - \psi(1+D_z) .$

This is consistent with derivations noted in the previous entry, including my related MSE and MO questions, and for Appell sequences in general as presented in the entry “Bernoulli Appells” here.

Another perspective is to rearrange terms

$\displaystyle D_x^{-s} \; \frac{x^\alpha}{\alpha!}= \frac{x^{s+\alpha}}{(s+\alpha)!} = e^{-sR_x} \; \frac{x^\alpha}{\alpha!} = e^{sD_\alpha} \; \frac{x^\alpha}{\alpha!}$

and regard $e^{-sR_x}$ as an evolution/shift operator defined by $(R_x + d/d\alpha) \frac{x^\alpha}{\alpha!}=0$. Note that applying $d/dx$ to this last equation generates the recursion $\psi(1+\alpha) = \psi(\alpha) + 1/\alpha \; .$

Interpolation of the ordered diff op $:xD_x:^n = x^nD_x^n$

Now interpolate the ordered op defined by $:xD_x:^n = x^nD_x^n = n! \binom{xD_x}{n}$. The associated op e.g.f. acting on $H(x)x^\alpha$ gives

$\displaystyle e^{-t\; :xD_x:} H(x) \; x^\alpha = H(x(1-t)) \; (x(1-t))^\alpha = H(1-t) \; (1-t)^\alpha \; x^\alpha$

$\displaystyle = (1-t)^{xD_x}\; H(x)x^\alpha \; ,$

so

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

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

Summarizing,

$\displaystyle :xD_x:^{-s} = x^{-s}D_x^{-s} = \frac{(xD_x)!}{(xD_x+s)!} \; ,$

and we see again that the MT is interpolating the coefficients of the op e.g.f. $:xD:^n = x^nD^n$ to $:xD:^s = x^s D^s\; ,$ or, equivalently, the coefficients of the regular e.g.f. $\binom{\alpha}{\alpha-n}$ to $\binom{\alpha}{\alpha-s} \; .$

Pseudo-infinigen for the fractional ordered operator $:xD_x: ^s$

Following the formal algorithm in the last part of the section on the infinigen for $D_x^{s}$, one might assume

$\displaystyle :xD_x:^{-s} \frac{x^\alpha}{\alpha!}= x^{-s}D_x^{-s} \; \frac{x^\alpha}{\alpha!}= x^{-s}\frac{x^{\alpha+s}}{(\alpha+s)!} = x^{-s} e^{sD_{\alpha}} \; \frac{x^\alpha}{\alpha!}= e^{-sR_x} \; \frac{x^\alpha}{\alpha!} \;$

and infer that the associated infinigen is

$\displaystyle R_x = \psi(1+xD_x) \; ,$ satisfying $\displaystyle (R_x-\log(x)+D_\alpha) \; \frac{x^\alpha}{\alpha!}= 0 \; ,$

but this is incorrect, giving $\displaystyle e^{-sR_x} \frac{x^\alpha}{\alpha!} = e^{-s\psi(1+\alpha)} \frac{x^\alpha}{\alpha!} \; .$

In fact, there can be no infinigen of the type assumed because the ordered operator does not satisfy the addition of exponents upon multiplication; that is,

$\displaystyle :xD_x:^s \; :xD_x:^r \; \neq \; :xD_x:^{s+r} \; .$

Instead, note

$\displaystyle :xD_x:^{-s} H(x) = \frac{1}{s!} = \frac{x^s}{s!} |_{x=1} = e^{-s \hat{R}_x} |_{x=1} 1 = \frac{e^{sz}}{s!} |_{z=0}= e^{s \hat{R}_z}|_{z=0} \; H(z)$

$\displaystyle = e^{sp.(0)} \; ,$

where

$\displaystyle \hat{R}_x^n |_{x=1} = (-\log(x) + \psi(1+xD_x))^n |_{x=1}\; ,$

and

$\displaystyle \hat{R}_z^n |_{z=0} = (z - \psi(1+D_z))^n |_{z=0}\; ,$

with $\displaystyle \hat{R}_x$ and $\displaystyle \hat{R}_z$, the infinigens for $D_x^{-s}$. The e.g.f. $1/s! = e^{sp.(0)}$ encodes the fundamental base sequence for the Appell polynomial sequence associated with $D_x^{-s}$ above. Recall that, for an Appell sequence, $(p.(0)+z)^n = p_n(z) \; ,$ where $p_n(0)$ is the fundamental base sequence, e.g., the Bernoulli numbers, for the Bernoulli polynomials.

And, since

$\displaystyle x^s D^s = x^s e^{s\hat{R}_x} = e^{s(\log(u)+\hat{R}_x)}|_{u=x} \; ,$ we can express a pseudo-infinigen as

$:xD_x:^s = e^{sR_{u,x}}|_{u=x} = e^{s(\log(u)+\hat{R}_x)}|_{u=x} \; .$

Interpolation of the classic shifted Laguerre diff op $(xD_xx)^n = x^n D_x^n x^n$

Next let’s use the same method to interpolate

$\displaystyle (x \; D_x \;x)^n = x^{-1}(x^2D_x)^n x= x^n D_x^n x^n = x^n \; n!\; L_n(-:xD_x:) = n! x^n \binom{xD_x+n}{n},$

where $L$ denotes the Laguerre polynomials, and, again, $(:xD_x:)^k = x^kD_x^k$ by definition.

The op e.g.f. is

$\displaystyle e^{-txD_xx}= \sum_{n \ge 0} (-tx)^n \binom{xD_x + n}{n} = \sum_{n \ge 0} (tx)^n \binom{-xD-1}{n} = (1+tu)^{-xD_x-1}|_{u=x}$

$\displaystyle = \frac{1}{1+xt}\; (\frac{1}{1+xt})^{xD_x} \; ,$

so

$\displaystyle e^{-txD_xx}\; H(x) x^\alpha = \frac{1}{1+xt}\; (\frac{1}{1+xt})^{xD_x} \; H(x) x^\alpha = H(\frac{x}{1+xt}) \frac{x^\alpha}{(1+xt)^{\alpha+1}}$

consistent with the e.g.f. of the Laguerre polynomials giving

$\displaystyle e^{-txD_xx}\; = \sum_{n \ge 0} (-xt)^n L_n(-:xD_x:) =\frac{\exp[\frac{-ut}{1+ut} \; :xD_x:]}{1+ut} |_{u=x}\; .$

Then the Mellin transform interpolation gives

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

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

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

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

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

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 $(x D_x x)^s = x^s D_x^s x^s$. 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.

Infinigen for the fractional shifted Laguerre operator $(xD_xx)^s$

The associated infinigen for this op is derived in the previous entry several ways, consistent with the method presented above for $D_x^{-s}$:

$\displaystyle (xD_xx)^{-s} x^\alpha = x^{-s}D_x^{-s}x^{-s} \; x^\alpha = x^{-s} \; \frac{(\alpha-s)!}{\alpha!}\; x^\alpha = x^{-s} \; \frac{(xD_x-s)!}{(xD_x)!}\; x^\alpha \;,$

so

$\displaystyle \bar{R}_x = -\frac{d}{ds}\; |_{s=0} \; x^{-s}\; \frac{(xD_x-s)!}{(xD_x)!}\; = \log(x) + \psi(1+xD_x)$

$\displaystyle = \log(xD_xx) = 2\log(x) + \log(D_x)\; .$

Equivalently,

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

with $\displaystyle e^{s\bar{R}_x}$ as an evolution/shift operator defined by $(\bar{R}_x - d/d\alpha) \; \alpha! \; x^\alpha =0$.

Action on the identity and a simple transformation give the associated Appell sequence $\bar{p}_n(z)$ and raising op $\bar{R}_z$:

$\displaystyle (xD_xx)^{s} 1 = e^{s \; \bar{R}_x} 1 = s! \; x^s = s! e^{sz} = e^{s \bar{R}_z} \; 1 = e^{s \bar{p}.(z)} \; ,$

with $x = e^z$ and

$\displaystyle \bar{R}_z = z + \psi(1+D_z) \; ,$

consistent with this Appell sequence of polynomials $\bar{p}_n(z)$ being the inverse under umbral composition of the Appell sequence $p_n(z)$ associated with $D_x^s$, i.e., $\bar{p}_n(p.(z)) = z^n = p_n(\bar{p}.(z))$. See the entry “Bernoulli Appells” for a discussion of the simple relations among the e.g.f.s and raising ops of a pair of umbrally inverse Appell sequences. See also the previous entry on gamma classes and Appell sequences involving the Riemann zeta values at positive integers $n > 1$ for some more details on this op and the relation to fractional calculus.

Interpolation of the Laguerre diff op $:D_xx:^n = D_x^nx^n$

Now interpolate the Laguerre op defined by

$\displaystyle :D_xx:^n = D_x^nx^n = x^{-n}(xD_xx)^n = n! \; L_n(-:xD_x:) = n! \; \binom{xD_x+n}{n}$.

The op e.g.f., from slightly modifying the derivation for the shifted Laguerre op, is

$\displaystyle e^{-t:D_xx:}= (1+t)^{-xD_x-1}$

so

$\displaystyle e^{-t:D_xx:}\; H(x) x^\alpha = \frac{1}{1+t}\; (\frac{1}{1+t})^{xD_x} \; H(x) x^\alpha = H(\frac{x}{1+t}) \frac{x^\alpha}{(1+t)^{\alpha+1}} \;,$

consistent with the e.g.f. of the Laguerre polynomials giving

$\displaystyle e^{-t:D_xx:}\; = \sum_{n \ge 0} (-t)^n L_n(-:xD_x:) =\frac{\exp[\frac{-t}{1+t} \; :xD_x:]}{1+t} \; .$

Then the Mellin transform interpolation gives

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

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

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

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

and it’s now clear by comparison that

$\displaystyle (:D_xx:)^{-s} = x^s(xD_xx)^{-s} = D_x^{-s}x^{-s} = \frac{(xD_x-s)!}{(xD_x)!}\;$

so that

$\displaystyle :D_xx:^{s} \;x^\alpha = D_x^s x^s x^\alpha = \frac{(xD_x+s)!}{(xD_x)!}\; x^\alpha = \frac{(\alpha+s)!}{\alpha!} x^\alpha\; .$

Pseudo-infinigen for the fractional Laguerre op $:D_xx:^{s}$

There can be no normal infinigen because the ordered operator does not satisfy the addition of exponents upon multiplication; that is,

$\displaystyle :Dx_x:^s \; :D_xx:^r \; \neq \; :D_xx:^{s+r} \; .$

But, proceeding as for the pseudo-infinigen for $:xD_x:^s$, note

$\displaystyle :D_xx:^s H(x) = s! = s! x^s |_{x=1} = e^{-s \hat{R}_x} |_{x=1} \; 1$

$\displaystyle = s! e^{sz} |_{z=0}= e^{s \hat{R}_z}|_{z=0} \; H(z) = e^{sp.(0)} \; ,$

where, from comparison with the infinigen for $(xD_xx)^s$,

$\displaystyle \hat{R}_x^n |_{x=1} = (\log(x) + \psi(1+xD_x))^n |_{x=1}\; ,$

and

$\displaystyle \hat{R}_z^n |_{z=0} = (z + \psi(1+D_z))^n |_{z=0}\; .$

This is consistent with the Appell sequence formalism, and $s! = e^{s\bar{p}(0)}$, where $\bar{p}_n(z)=(\bar{p}.(0)+z)^n$ are the Appell polynomials associated with the fractional shifted Laguerre op. And, since

$\displaystyle D^sx^s = x^{-s}(x^sD^sx^s) = x^{-s}e^{s\hat{R}_x} = e^{s(-\log(u)+\hat{R}_x)}|_{u=x} \; ,$ we can express the pseudo-infinigen as

$:D_xx:^s = e^{sR_{u,x}}|_{u=x} = e^{s(-\log(u)+\hat{R}_x)}|_{u=x} \; .$

Interpolation of the generalized Scherk / Witt Lie ops $(x^{1+y}D_x)^n$

The Scherk-Witt Lie operators $(x^{1+m}D_x)^n$ have a long history and play important intersecting roles in operator calculus, special functions, complex analysis, and Lie algebra / group theory. Several entries here, in MSE, MO, and OEIS make note of this. So, let’s generalize from integer values for $\displaystyle m,n$ to real or complex numbers (or even operators) as

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

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

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

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 $t = x^{|y|}/|y|$ . With this interpretation, consistent with the interpolation heuristic, the basic fractional integroderivative $D^s$ characterizes the Scherk-Witt ops for negative $y$ through simple transformations:

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

and, for $y < 0$,

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

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

$\displaystyle = x^{s|y|} |y|^{-s} \int_0^1 \frac{u^{s-1}}{(s-1)!} \; (1-u)^{\alpha/|y|} \; du = x^{s|y|} |y|^{-s} D_{x=1}^{-s} \; x^{\alpha/|y|}$

with a change of variable $t = u \; x^{|y|}/|y| \; .$

Similarly, the Scherk-Witt ops for $y > 0$ are characterized by the generalized Laguerre operator of the previous section since $x \; (xD_xx)^n \; x^{-1} = (x^2D_x)^n$ so that $x \; (xD_xx)^{-s} \; x^{-1} = (x^2D_x)^{-s}$.

The $y = 0$ case, can be obtained as limits of the other two cases or, of course, directly from the action of the op generating function:

$\displaystyle e^{-txD_x} x^\alpha= e^{(e^{-t}-1)\; :xD_x:}\; x^\alpha = (e^{-t}x)^\alpha , \;$

so

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

The S-W ops can be written in a normal ordering form, following the examples in the last two entries here, as

$\displaystyle (x^{1+y}D_x)^s = H(y) \; x^{sy} \; y^s s! \binom{s+xD_x/y-1}{s} + H(-y) x^{-s|y|} \; |y|^s \; s!\binom{xD_x/|y|}{s}$

$\displaystyle = H(y) \; x^{sy} \; y^s \sum_{n \ge 0}(-1)^n [\triangle_k^n \; \binom{s+k/y-1}{s}s!] \; \frac{x^nD_x^n}{n!}$

$\displaystyle + H(-y) x^{-s|y|} \; |y|^s \; \sum_{n \ge 0}(-1)^n [\triangle_k^n \; \binom{k/|y|}{s}s!] \; \frac{x^nD_x^n}{n!} \; ,$

where $\displaystyle \triangle_k^n a_k = \sum_{k=0}^n (-1)^k \binom{n}{k} a_k \; .$

Let’s look more closely at the derivation of the Mellin transform for this operator.
Given a compositional inverse pair of functions, $\omega = h(z)$ and $z = h^{-1}(\omega)$ analytic about $z$ and $\omega$ with $g(z)=1/[dh(z)/dz]$, then

$\displaystyle \exp \left[ {t \cdot g(z)\frac{d}{{dz}}} \right]f(z) = \exp \left[ {t\frac{d}{{d\omega }}} \right]f[{h^{ - 1}}(\omega )] = f[{h^{ - 1}}[t + \omega]]$

$\displaystyle = f[{h^{ - 1}}[t + h(z)]],$

and we have a flow map or integral curve associated with the vector field $g(x)\frac{\mathrm{d} }{\mathrm{d} x}$ (cf. OEIS A145271, A094638, “Important formulas in combinatorics”, “Infinigens, the Pascal triangle, and the Witt and Virasoro algebras”, and several other entries here).

We are focusing on the triple

$\displaystyle \omega = h(z) = \frac{z^{-y}}{-y} \; \; , \; \; z = h^{-1}(\omega) = (-y\omega)^{\frac{1}{y}} \; \; \mathrm{and}, \;\; g(z) = z^{1+y},$

so

$\displaystyle h^{-1}(-t+h(z))= \frac{z}{(1+ty\;z^y)^{1/y}},$

and

$\displaystyle e^{-tx^{1+y}D_x}\; x^\alpha = [h^{-1}(-t+h(x))]^\alpha = x^\alpha (1+ty\;x^y)^{-\alpha/y}.$

So,

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

translates into

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

The (Ramanujan / Mellin) interpolation heuristic amounts to expanding $(g(x)D_x)^n \; x^m$ and then replacing $n \; \mathrm{and} \; m$ by $-s \; \mathrm{and} \; \alpha$,

and since

$\displaystyle (x^{1+y}D_x)^n \; x^m = x^{m+ny} (m)(m+y) \cdots (m+(n-1)y) = x^{m+ny} \; y^n \frac{(m/y-1 + n)!}{(m/y-1)!}$

$\displaystyle =x^{m+ny} (-y)^n \frac{(-m/y)!}{(-m/y-n)!} \;,$

we get the desired result, which can be corroborated (or not) quickly with a numerical integration of the Mellin transform and verified through the inverse Mellin transform.

Infinigens for the generalized Scherk-Witt operators $(x^{1+y}D_x)^s$

Following the algorithm presented for the other infinigens above, taking the derivative w.r.t. $s$ of the S-W op $(x^{1+y}D_x)^s = e^{sR_x(y)}$ and evaluating it at $s=0$ gives

$\displaystyle R_x(y)=\log(x^{1+y}D_x) = H(y) [\log(yx^y) + \psi(xD_x/y)]$

$\displaystyle + H(-y) [\log(|y| x^{-|y|}) + \psi(1+xD_x/|y|)] \; .$

Note that $\psi(x) = -1/x + \psi(x+1) \;,$ so $\psi(xD_x/y) = -y(xD_x)^{-1} + \psi(1 + xD_x/|y|) .$ However, only the infinigen for $y < 0$ gives a raising op for a related sequence of Appell polynomials/functions. Action of the S-W op for $y > 0$ gives zero, but, for $y < 0$,

$\displaystyle (x^{1+y}D_x)^{-s} \; 1 = x^{s|y|} \; |y|^{-s} \; (-s)!\binom{xD_x/|y|}{-s}\; 1 = \frac{(x^{|y|}/|y|)^{s}}{s!} = e^{-sR_x(y<0)} \; 1$

$\displaystyle = \frac{e^{zs}}{s!} = e^{sR_z} \; 1 = e^{sp.(z)}$

with $\displaystyle x^{|y|}/|y| = e^z \; ,$ or $\displaystyle z = |y|\log(x)-\log|y| \; ,$ and

$\displaystyle R_z = z - \psi(1+D_z) \;,$ same as for $D_z^{-s}.$

Related stuff:

“Exploring Visualization Methods for Complex Variables” by Hanson and Sha

“Gauss’ hypergeometric function” by Beukers

“Beta Function and its Applications” by Riddhi

“How does Euler’s beta function lead to string theory?” answered by Maimon

“Gabriela Veneziano, strong nuclear force, and the beta function” answered by Maimon

“The Euler Legacy to Modern Physics” by Dattoli

“New strings for old Veneziano ampltudes I: Analytical treatment “ by Kholodenko. Also see the sequels to this paper at arXiv..

“Traces of mirror symmetry in nature” by Kholodenko

“Fractional derivatives and special functions” by Lavoie, Osler, and Tremblay

“Transcendence of periods: State of the art” by Waldschmidt

“On Stanley’s reciprocity theorem for rational cones” by Beck and Develin

“Periods and Feynman integrals” by Bogner and Weinzierl

Visual Complex Functions: An Introduction with Phase Portraits by Wegert page 307

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