Theory and Modern Applications

# Barnes-type Daehee polynomials

## Abstract

In this paper, we consider Barnes-type Daehee polynomials of the first kind and of the second kind. From the properties of the Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

MSC:05A15, 05A40, 11B68, 11B75, 65Q05.

## 1 Introduction

In this paper, we consider the polynomials ${D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)$ and ${\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)$ called the Barnes-type Daehee polynomials of the first kind and of the second kind, whose generating functions are given by

$\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{n}}{n!},$
(1)
$\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{n}}{n!},$
(2)

respectively, where ${a}_{1},\dots ,{a}_{r}\ne 0$. When $x=0$, ${D}_{n}\left({a}_{1},\dots ,{a}_{r}\right)={D}_{n}\left(0|{a}_{1},\dots ,{a}_{r}\right)$ and ${\stackrel{ˆ}{D}}_{n}\left({a}_{1},\dots ,{a}_{r}\right)={\stackrel{ˆ}{D}}_{n}\left(0|{a}_{1},\dots ,{a}_{r}\right)$ are called the Barnes-type Daehee numbers of the first kind and of the second kind, respectively.

Recall that the Daehee polynomials of the first kind and of the second kind of order r, denoted by ${D}_{n}^{\left(r\right)}\left(x\right)$ and ${\stackrel{ˆ}{D}}_{n}^{\left(r\right)}\left(x\right)$, respectively, are given by the generating functions to be

$\begin{array}{c}{\left(\frac{ln\left(1+t\right)}{t}\right)}^{r}{\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{D}_{n}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{n!},\hfill \\ {\left(\frac{\left(1+t\right)ln\left(1+t\right)}{t}\right)}^{r}{\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{n!},\hfill \end{array}$

respectively. If ${a}_{1}=\cdots ={a}_{r}=1$, then ${D}_{n}^{\left(r\right)}\left(x\right)={D}_{n}\left(x|\underset{r}{\underset{⏟}{1,\dots ,1}}\right)$ and ${\stackrel{ˆ}{D}}_{n}^{\left(r\right)}\left(x\right)={\stackrel{ˆ}{D}}_{n}\left(x|\underset{r}{\underset{⏟}{1,\dots ,1}}\right)$. Daehee polynomials were defined by the second author  and have been investigated in .

In this paper, we consider Barnes-type Daehee polynomials of the first kind and of the second kind. From the properties of the Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

## 2 Umbral calculus

Let be the complex number field and let be the set of all formal power series in the variable t:

$\mathcal{F}=\left\{f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{a}_{k}}{k!}{t}^{k}|{a}_{k}\in \mathbb{C}\right\}.$
(3)

Let $\mathbb{P}=\mathbb{C}\left[x\right]$ and let ${\mathbb{P}}^{\ast }$ be the vector space of all linear functionals on . $〈L|p\left(x\right)〉$ is the action of the linear functional L on the polynomial $p\left(x\right)$, and we recall that the vector space operations on ${\mathbb{P}}^{\ast }$ are defined by $〈L+M|p\left(x\right)〉=〈L|p\left(x\right)〉+〈M|p\left(x\right)〉$, $〈cL|p\left(x\right)〉=c〈L|p\left(x\right)〉$, where c is a complex constant in . For $f\left(t\right)\in \mathcal{F}$, let us define the linear functional on by setting

$〈f\left(t\right)|{x}^{n}〉={a}_{n}\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(4)

In particular,

$〈{t}^{k}|{x}^{n}〉=n!{\delta }_{n,k}\phantom{\rule{1em}{0ex}}\left(n,k\ge 0\right),$
(5)

where ${\delta }_{n,k}$ is the Kronecker symbol.

For ${f}_{L}\left(t\right)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|{x}^{k}〉}{k!}{t}^{k}$, we have $〈{f}_{L}\left(t\right)|{x}^{n}〉=〈L|{x}^{n}〉$. That is, $L={f}_{L}\left(t\right)$. The map $L↦{f}_{L}\left(t\right)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto . Henceforth, denotes both the algebra of formal power series in t and the vector space of all linear functionals on , and therefore an element $f\left(t\right)$ of will be thought of as both a formal power series and a linear functional. We call the umbral algebra and the umbral calculus is the study of umbral algebra. The order $O\left(f\left(t\right)\right)$ of a power series $f\left(t\right)$ (≠0) is the smallest integer k for which the coefficient of ${t}^{k}$ does not vanish. If $O\left(f\left(t\right)\right)=1$, then $f\left(t\right)$ is called a delta series; if $O\left(f\left(t\right)\right)=0$, then $f\left(t\right)$ is called an invertible series. For $f\left(t\right),g\left(t\right)\in \mathcal{F}$ with $O\left(f\left(t\right)\right)=1$ and $O\left(g\left(t\right)\right)=0$, there exists a unique sequence ${s}_{n}\left(x\right)$ ($deg{s}_{n}\left(x\right)=n$) such that $〈g\left(t\right)f{\left(t\right)}^{k}|{s}_{n}\left(x\right)〉=n!{\delta }_{n,k}$, for $n,k\ge 0$. Such a sequence ${s}_{n}\left(x\right)$ is called the Sheffer sequence for $\left(g\left(t\right),f\left(t\right)\right)$, which is denoted by ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$.

For $f\left(t\right),g\left(t\right)\in \mathcal{F}$ and $p\left(x\right)\in \mathbb{P}$, we have

$〈f\left(t\right)g\left(t\right)|p\left(x\right)〉=〈f\left(t\right)|g\left(t\right)p\left(x\right)〉=〈g\left(t\right)|f\left(t\right)p\left(x\right)〉$
(6)

and

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}〈f\left(t\right)|{x}^{k}〉\frac{{t}^{k}}{k!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}〈{t}^{k}|p\left(x\right)〉\frac{{x}^{k}}{k!}$
(7)

, Theorem 2.2.5]. Thus, by (7), we get

${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{e}^{yt}p\left(x\right)=p\left(x+y\right).$
(8)

Sheffer sequences are characterized by the generating function , Theorem 2.3.4].

Lemma 1 The sequence ${s}_{n}\left(x\right)$ is Sheffer for $\left(g\left(t\right),f\left(t\right)\right)$ if and only if

$\frac{1}{g\left(\overline{f}\left(t\right)\right)}{e}^{y\overline{f}\left(t\right)}=\sum _{k=0}^{\mathrm{\infty }}\frac{{s}_{k}\left(y\right)}{k!}{t}^{k}\phantom{\rule{1em}{0ex}}\left(y\in \mathbb{C}\right),$

where $\overline{f}\left(t\right)$ is the compositional inverse of $f\left(t\right)$.

For ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, we have the following equations , Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

$f\left(t\right){s}_{n}\left(x\right)=n{s}_{n-1}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right),$
(9)
${s}_{n}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈g{\left(\overline{f}\left(t\right)\right)}^{-1}\overline{f}{\left(t\right)}^{j}|{x}^{n}〉{x}^{j},$
(10)
${s}_{n}\left(x+y\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){s}_{j}\left(x\right){p}_{n-j}\left(y\right),$
(11)

where ${p}_{n}\left(x\right)=g\left(t\right){s}_{n}\left(x\right)$.

Assume that ${p}_{n}\left(x\right)\sim \left(1,f\left(t\right)\right)$ and ${q}_{n}\left(x\right)\sim \left(1,g\left(t\right)\right)$. Then the transfer formula , Corollary 3.8.2] is given by

${q}_{n}\left(x\right)=x{\left(\frac{f\left(t\right)}{g\left(t\right)}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 1\right).$

For ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$ and ${r}_{n}\left(x\right)\sim \left(h\left(t\right),l\left(t\right)\right)$, assume that

${s}_{n}\left(x\right)=\sum _{m=0}^{n}{C}_{n,m}{r}_{m}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$

Then we have , p.132]

${C}_{n,m}=\frac{1}{m!}〈\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}l{\left(\overline{f}\left(t\right)\right)}^{m}|{x}^{n}〉.$
(12)

## 3 Main results

We now note that ${D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)$ is the Sheffer sequence for

$g\left(t\right)=\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(t\right)={e}^{t}-1.$

Therefore,

${D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\sim \left(\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t}\right),{e}^{t}-1\right).$
(13)

${\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)$ is the Sheffer sequence for

$g\left(t\right)=\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t{e}^{{a}_{j}t}}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(t\right)={e}^{t}-1.$

So,

${\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\sim \left(\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t{e}^{{a}_{j}t}}\right),{e}^{t}-1\right).$
(14)

### 3.1 Explicit expressions

Recall that Barnes’ multiple Bernoulli polynomials ${B}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)$ are defined by the generating function as

$\frac{{t}^{r}}{{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-1\right)}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{n}}{n!},$
(15)

where ${a}_{1},\dots ,{a}_{r}\ne 0$ . Let ${\left(n\right)}_{j}=n\left(n-1\right)\cdots \left(n-j+1\right)$ ($j\ge 1$) with ${\left(n\right)}_{0}=1$. The (signed) Stirling numbers of the first kind ${S}_{1}\left(n,m\right)$ are defined by

${\left(x\right)}_{n}=\sum _{m=0}^{n}{S}_{1}\left(n,m\right){x}^{m}.$

Theorem 1

${D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=\sum _{m=0}^{n}{S}_{1}\left(n,m\right){B}_{m}\left(x|{a}_{1},\dots ,{a}_{r}\right)$
(16)
$\phantom{{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)}=\sum _{j=0}^{n}\left(\sum _{l=j}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,j\right){D}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right)\right){x}^{j}$
(17)
$\phantom{{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)}=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right){D}_{n-m}\left({a}_{1},\dots ,{a}_{r}\right){\left(x\right)}_{m},$
(18)
${\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=\sum _{m=0}^{n}{S}_{1}\left(n,m\right){B}_{m}\left(x+{a}_{1}+\cdots +{a}_{r}|{a}_{1},\dots ,{a}_{r}\right)$
(19)
$\phantom{{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)}=\sum _{j=0}^{n}\left(\sum _{l=j}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,j\right){\stackrel{ˆ}{D}}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right)\right){x}^{j}$
(20)
$\phantom{{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)}=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right){\stackrel{ˆ}{D}}_{n-m}\left({a}_{1},\dots ,{a}_{r}\right){\left(x\right)}_{m}.$
(21)

Proof Since

$\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t}\right){D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\sim \left(1,{e}^{t}-1\right)$
(22)

and

${\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right),$
(23)

we have

$\begin{array}{rl}{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)& =\prod _{j=1}^{r}\left(\frac{t}{{e}^{{a}_{j}t}-1}\right){\left(x\right)}_{n}\\ =\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\prod _{j=1}^{r}\left(\frac{t}{{e}^{{a}_{j}t}-1}\right){x}^{m}\\ =\sum _{m=0}^{n}{S}_{1}\left(n,m\right){B}_{m}\left(x|{a}_{1},\dots ,{a}_{r}\right).\end{array}$

So, we get (16).

Similarly, by

$\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t{e}^{{a}_{j}t}}\right){\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\sim \left(1,{e}^{t}-1\right)$
(24)

and (23), we have

$\begin{array}{rl}{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)& =\prod _{j=1}^{r}\left(\frac{t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}\right){\left(x\right)}_{n}\\ =\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\prod _{j=1}^{r}\left(\frac{t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}\right){x}^{m}\\ =\sum _{m=0}^{n}{S}_{1}\left(n,m\right){e}^{\left({a}_{1}+\cdots +{a}_{r}\right)t}\prod _{j=1}^{r}\left(\frac{t}{{e}^{{a}_{j}t}-1}\right){x}^{m}\\ =\sum _{m=0}^{n}{S}_{1}\left(n,m\right){B}_{m}\left(x+{a}_{1}+\cdots +{a}_{r}|{a}_{1},\dots ,{a}_{r}\right).\end{array}$

Therefore, we get (19).

By (10) with (13), we get

${D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=\sum _{j=0}^{n}\frac{1}{j!}〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(ln\left(1+t\right)\right)}^{j}|{x}^{n}〉{x}^{j}.$

Since

$\begin{array}{c}〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(ln\left(1+t\right)\right)}^{j}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{\left(ln\left(1+t\right)\right)}^{j}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|j!\sum _{l=j}^{\mathrm{\infty }}{S}_{1}\left(l,j\right)\frac{{t}^{l}}{l!}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=j!\sum _{l=j}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,j\right)〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{x}^{n-l}〉\hfill \\ \phantom{\rule{1em}{0ex}}=j!\sum _{l=j}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,j\right)〈\sum _{i=0}^{\mathrm{\infty }}{D}_{i}\left({a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{i}}{i!}|{x}^{n-l}〉\hfill \\ \phantom{\rule{1em}{0ex}}=j!\sum _{l=j}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,j\right){D}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right),\hfill \end{array}$

we obtain (17).

Similarly, by (10) with (14), we get

${\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=\sum _{j=0}^{n}\frac{1}{j!}〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(ln\left(1+t\right)\right)}^{j}|{x}^{n}〉{x}^{j}.$

Since

$\begin{array}{c}〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(ln\left(1+t\right)\right)}^{j}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{\left(ln\left(1+t\right)\right)}^{j}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=j!\sum _{l=j}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,j\right)〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{x}^{n-l}〉\hfill \\ \phantom{\rule{1em}{0ex}}=j!\sum _{l=j}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,j\right)〈\sum _{i=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{i}\left({a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{i}}{i!}|{x}^{n-l}〉\hfill \\ \phantom{\rule{1em}{0ex}}=j!\sum _{l=j}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,j\right){\stackrel{ˆ}{D}}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right),\hfill \end{array}$

we obtain (20).

Next, we obtain

$\begin{array}{rl}{D}_{n}\left(y|{a}_{1},\dots ,{a}_{r}\right)& =〈\sum _{i=0}^{\mathrm{\infty }}{D}_{i}\left(y|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{i}}{i!}|{x}^{n}〉\\ =〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{y}|{x}^{n}〉\\ =〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|\sum _{m=0}^{\mathrm{\infty }}{\left(y\right)}_{m}\frac{{t}^{m}}{m!}{x}^{n}〉\\ =\sum _{m=0}^{n}{\left(y\right)}_{m}\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{x}^{n-m}〉\\ =\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right){D}_{n-m}\left({a}_{1},\dots ,{a}_{r}\right){\left(y\right)}_{m}.\end{array}$

Thus, we get the identity (18).

Similarly,

$\begin{array}{rl}{\stackrel{ˆ}{D}}_{n}\left(y|{a}_{1},\dots ,{a}_{r}\right)& =〈\sum _{i=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{i}\left(y|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{i}}{i!}|{x}^{n}〉\\ =〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{y}|{x}^{n}〉\\ =〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|\sum _{m=0}^{\mathrm{\infty }}{\left(y\right)}_{m}\frac{{t}^{m}}{m!}{x}^{n}〉\\ =\sum _{m=0}^{n}{\left(y\right)}_{m}\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{x}^{n-m}〉\\ =\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right){\stackrel{ˆ}{D}}_{n-m}\left({a}_{1},\dots ,{a}_{r}\right){\left(y\right)}_{m}.\end{array}$

Thus, we get the identity (21). □

### 3.2 Sheffer identity

Theorem 2

${D}_{n}\left(x+y|{a}_{1},\dots ,{a}_{r}\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){D}_{j}\left(x|{a}_{1},\dots ,{a}_{r}\right){\left(y\right)}_{n-j},$
(25)
${\stackrel{ˆ}{D}}_{n}\left(x+y|{a}_{1},\dots ,{a}_{r}\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){\stackrel{ˆ}{D}}_{j}\left(x|{a}_{1},\dots ,{a}_{r}\right){\left(y\right)}_{n-j}.$
(26)

Proof By (13) with

$\begin{array}{rl}{p}_{n}\left(x\right)& =\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t}\right){D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\\ ={\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right),\end{array}$

using (11), we have (25).

By (14) with

$\begin{array}{rl}{p}_{n}\left(x\right)& =\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t{e}^{{a}_{j}t}}\right){\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\\ ={\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right),\end{array}$

using (11), we have (26). □

### 3.3 Difference relations

Theorem 3

${D}_{n}\left(x+1|{a}_{1},\dots ,{a}_{r}\right)-{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=n{D}_{n-1}\left(x|{a}_{1},\dots ,{a}_{r}\right),$
(27)
${\stackrel{ˆ}{D}}_{n}\left(x+1|{a}_{1},\dots ,{a}_{r}\right)-{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=n{\stackrel{ˆ}{D}}_{n-1}\left(x|{a}_{1},\dots ,{a}_{r}\right).$
(28)

Proof By (9) with (13), we get

$\left({e}^{t}-1\right){D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=n{D}_{n-1}\left(x|{a}_{1},\dots ,{a}_{r}\right).$

By (8), we have (27).

Similarly, by (9) with (14), we get

$\left({e}^{t}-1\right){\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=n{\stackrel{ˆ}{D}}_{n-1}\left(x|{a}_{1},\dots ,{a}_{r}\right).$

By (8), we have (28). □

### 3.4 Recurrence

Theorem 4

$\begin{array}{c}{D}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)=x{D}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{{D}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)=}-\sum _{m=0}^{n}\left(\sum _{i=m}^{n}\sum _{l=i}^{n}\sum _{j=1}^{r}\frac{1}{i+1}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{i+1}{m}\right){S}_{1}\left(l,i\right)\hfill \\ \phantom{{D}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)=}×{B}_{i+1-m}{\left(-{a}_{j}\right)}^{i+1-m}{D}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right)\right){\left(x-1\right)}^{m},\hfill \end{array}$
(29)
$\begin{array}{c}{\stackrel{ˆ}{D}}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)=\left(x+\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{{\stackrel{ˆ}{D}}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)=}-\sum _{m=0}^{n}\left(\sum _{i=m}^{n}\sum _{l=i}^{n}\sum _{j=1}^{r}\frac{1}{i+1}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{i+1}{m}\right){S}_{1}\left(l,i\right)\hfill \\ \phantom{{\stackrel{ˆ}{D}}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)=}×{B}_{i+1-m}{\left(-{a}_{j}\right)}^{i+1-m}{\stackrel{ˆ}{D}}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right)\right){\left(x-1\right)}^{m},\hfill \end{array}$
(30)

where ${B}_{n}$ is the nth ordinary Bernoulli number.

Proof By applying

${s}_{n+1}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{\prime }\left(t\right)}{s}_{n}\left(x\right)$
(31)

, Corollary 3.7.2] with (13), we get

${D}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)=x{D}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)-{e}^{-t}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right).$

Now,

$\begin{array}{rl}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}& ={\left(lng\left(t\right)\right)}^{\prime }\\ ={\left(\sum _{j=1}^{r}ln\left({e}^{{a}_{j}t}-1\right)-rlnt\right)}^{\prime }\\ =\sum _{j=1}^{r}\frac{{a}_{j}{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-\frac{r}{t}\\ =\frac{{\sum }_{j=1}^{r}{\prod }_{i\ne j}\left({e}^{{a}_{i}t}-1\right)\left({a}_{j}t{e}^{{a}_{j}t}-{e}^{{a}_{j}t}+1\right)}{t{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-1\right)}.\end{array}$

Since

$\begin{array}{rl}\sum _{j=1}^{r}\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-r& =\frac{{\sum }_{j=1}^{r}{\prod }_{i\ne j}\left({e}^{{a}_{i}t}-1\right)\left({a}_{j}t{e}^{{a}_{j}t}-{e}^{{a}_{j}t}+1\right)}{{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-1\right)}\\ =\frac{\frac{1}{2}\left({\sum }_{j=1}^{r}{a}_{1}\cdots {a}_{j-1}{a}_{j}^{2}{a}_{j+1}\cdots {a}_{r}\right){t}^{r+1}+\cdots }{\left({a}_{1}\cdots {a}_{r}\right){t}^{r}+\cdots }\\ =\frac{1}{2}\left(\sum _{j=1}^{r}{a}_{j}\right)t+\cdots \end{array}$

is a series with order ≥1, by (17) we have

$\begin{array}{c}{D}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=x{D}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)-{e}^{-t}\frac{{\sum }_{j=1}^{r}\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-r}{t}{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=x{D}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)-{e}^{-t}\frac{{\sum }_{j=1}^{r}\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-r}{t}\left(\sum _{i=0}^{n}\sum _{l=i}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,i\right){D}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right){x}^{i}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=x{D}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)-\sum _{i=0}^{n}\sum _{l=i}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,i\right){D}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right){e}^{-t}\left(\sum _{j=1}^{r}\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-r\right)\frac{{x}^{i+1}}{i+1}.\hfill \end{array}$

Since

$\begin{array}{rl}{e}^{-t}\left(\sum _{j=1}^{r}\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-r\right){x}^{i+1}& ={e}^{-t}\left(\sum _{j=1}^{r}\sum _{m=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{m}{B}_{m}{a}_{j}^{m}}{m!}{t}^{m}-r\right){x}^{i+1}\\ ={e}^{-t}\left(\sum _{j=1}^{r}\sum _{m=0}^{i+1}\left(\genfrac{}{}{0}{}{i+1}{m}\right){B}_{m}{\left(-{a}_{j}\right)}^{m}{x}^{i+1-m}-r{x}^{i+1}\right)\\ =\sum _{j=1}^{r}\sum _{m=1}^{i+1}\left(\genfrac{}{}{0}{}{i+1}{m}\right){B}_{m}{\left(-{a}_{j}\right)}^{m}{\left(x-1\right)}^{i+1-m}\\ =\sum _{j=1}^{r}\sum _{m=0}^{i}\left(\genfrac{}{}{0}{}{i+1}{m}\right){B}_{i+1-m}{\left(-{a}_{j}\right)}^{i+1-m}{\left(x-1\right)}^{m},\end{array}$
(32)

we have

$\begin{array}{rcl}{D}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)& =& x{D}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ -\sum _{i=0}^{n}\sum _{l=i}^{n}\sum _{j=1}^{r}\sum _{m=0}^{i}\frac{1}{i+1}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{i+1}{m}\right){S}_{1}\left(l,i\right)\\ ×{B}_{i+1-m}{\left(-{a}_{j}\right)}^{i+1-m}{D}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right){\left(x-1\right)}^{m}\\ =& x{D}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ -\sum _{m=0}^{n}\left(\sum _{i=m}^{n}\sum _{l=i}^{n}\sum _{j=1}^{r}\frac{1}{i+1}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{i+1}{m}\right){S}_{1}\left(l,i\right)\\ ×{B}_{i+1-m}{\left(-{a}_{j}\right)}^{i+1-m}{D}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right)\right){\left(x-1\right)}^{m},\end{array}$

which is the identity (29).

Next, by applying (31) with (14), we get

${\stackrel{ˆ}{D}}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)=x{\stackrel{ˆ}{D}}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)-{e}^{-t}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right).$

Now,

$\begin{array}{rl}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}& ={\left(lng\left(t\right)\right)}^{\prime }\\ ={\left(\sum _{j=1}^{r}ln\left({e}^{{a}_{j}t}-1\right)-rlnt-\left(\sum _{j=1}^{r}{a}_{j}\right)t\right)}^{\prime }\\ =\sum _{j=1}^{r}\frac{{a}_{j}{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-\frac{r}{t}-\sum _{j=1}^{r}{a}_{j}.\end{array}$

By (20) we have

$\begin{array}{c}{\stackrel{ˆ}{D}}_{n+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(x+\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-{e}^{-t}\frac{{\sum }_{j=1}^{r}\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-r}{t}{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(x+\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-{e}^{-t}\frac{{\sum }_{j=1}^{r}\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-r}{t}\left(\sum _{i=0}^{n}\sum _{l=i}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,i\right){\stackrel{ˆ}{D}}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right){x}^{i}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(x+\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{i=0}^{n}\sum _{l=i}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,i\right){\stackrel{ˆ}{D}}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right){e}^{-t}\left(\sum _{j=1}^{r}\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-r\right)\frac{{x}^{i+1}}{i+1}.\hfill \end{array}$

By (32), we have the identity (30). □

### 3.5 Differentiation

Theorem 5

$\frac{d}{dx}{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{D}_{l}\left(x|{a}_{1},\dots ,{a}_{r}\right),$
(33)
$\frac{d}{dx}{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{\stackrel{ˆ}{D}}_{l}\left(x|{a}_{1},\dots ,{a}_{r}\right).$
(34)

Proof We shall use

$\frac{d}{dx}{s}_{n}\left(x\right)=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l}\right)〈\overline{f}\left(t\right)|{x}^{n-l}〉{s}_{l}\left(x\right)$

(cf. , Theorem 2.3.12]). Since

$\begin{array}{rl}〈\overline{f}\left(t\right)|{x}^{n-l}〉& =〈ln\left(1+t\right)|{x}^{n-l}〉=〈\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{m-1}{t}^{m}}{m}|{x}^{n-l}〉\\ =\sum _{m=1}^{n-l}\frac{{\left(-1\right)}^{m-1}}{m}〈{t}^{m}|{x}^{n-l}〉\\ =\sum _{m=1}^{n-l}\frac{{\left(-1\right)}^{m-1}}{m}\left(n-l\right)!{\delta }_{m,n-l}\\ ={\left(-1\right)}^{n-l-1}\left(n-l-1\right)!,\end{array}$

with (13), we have

$\begin{array}{rl}\frac{d}{dx}{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)& =\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l}\right){\left(-1\right)}^{n-l-1}\left(n-l-1\right)!{D}_{l}\left(x|{a}_{1},\dots ,{a}_{r}\right)\\ =n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{D}_{l}\left(x|{a}_{1},\dots ,{a}_{r}\right),\end{array}$

which is the identity (33). Similarly, with (14), we have the identity (34). □

### 3.6 More relations

The classical Cauchy numbers ${c}_{n}$ are defined by

$\frac{t}{ln\left(1+t\right)}=\sum _{n=0}^{\mathrm{\infty }}{c}_{n}\frac{{t}^{n}}{n!}$

(see e.g. [9, 10]).

Theorem 6

$\begin{array}{c}{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=x{D}_{n-1}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=}+\frac{r}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{l}{D}_{n-l}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=}-\frac{1}{n}\sum _{j=1}^{r}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){a}_{j}{c}_{l}{D}_{n-l}\left(x+{a}_{j}-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right),\hfill \end{array}$
(35)
$\begin{array}{c}{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=\left(x+\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n-1}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=}+\frac{r}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{l}{\stackrel{ˆ}{D}}_{n-l}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)=}-\frac{1}{n}\sum _{j=1}^{r}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){a}_{j}{c}_{l}{\stackrel{ˆ}{D}}_{n-l}\left(x-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right).\hfill \end{array}$
(36)

Proof For $n\ge 1$, we have

$\begin{array}{rcl}{D}_{n}\left(y|{a}_{1},\dots ,{a}_{r}\right)& =& 〈\sum _{l=0}^{\mathrm{\infty }}{D}_{l}\left(y|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{l}}{l!}|{x}^{n}〉\\ =& 〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{y}|{x}^{n}〉\\ =& 〈{\partial }_{t}\left(\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{y}\right)|{x}^{n-1}〉\\ =& 〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\left({\partial }_{t}{\left(1+t\right)}^{y}\right)|{x}^{n-1}〉\\ +〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\right){\left(1+t\right)}^{y}|{x}^{n-1}〉\\ =& y{D}_{n-1}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)\\ +〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\right){\left(1+t\right)}^{y}|{x}^{n-1}〉.\end{array}$

Observe that

$\begin{array}{rcl}{\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)& =& \sum _{j=1}^{r}\prod _{i\ne j}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)\frac{\frac{1}{1+t}\left({\left(1+t\right)}^{{a}_{j}}-1\right)-ln\left(1+t\right)\left({a}_{j}{\left(1+t\right)}^{{a}_{j}-1}\right)}{{\left({\left(1+t\right)}^{{a}_{j}}-1\right)}^{2}}\\ =& \frac{1}{1+t}\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)\sum _{j=1}^{r}\left(\frac{1}{ln\left(1+t\right)}-\frac{{a}_{j}{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\\ =& \frac{1}{1+t}\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)\frac{{\sum }_{j=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right)}{t}.\end{array}$

Since

$\sum _{j=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right)=-\frac{1}{2}\left(\sum _{j=1}^{r}{a}_{j}\right)t+\cdots$

is a series with order ≥1, we have

$\begin{array}{c}〈\left({\partial }_{t}\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)\right){\left(1+t\right)}^{y}|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\frac{{\sum }_{j=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right)}{t}{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\sum _{j=1}^{r}〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right){x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{r}{n}〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\frac{t}{ln\left(1+t\right)}{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n}\sum _{j=1}^{r}{a}_{j}〈\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y+{a}_{j}-1}|\frac{t}{ln\left(1+t\right)}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{r}{n}〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\sum _{l=0}^{\mathrm{\infty }}{c}_{l}\frac{{t}^{l}}{l!}{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n}\sum _{j=1}^{r}{a}_{j}〈\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y+{a}_{j}-1}|\sum _{l=0}^{\mathrm{\infty }}{c}_{l}\frac{{t}^{l}}{l!}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{r}{n}\sum _{l=0}^{n}{c}_{l}\left(\genfrac{}{}{0}{}{n}{l}\right)〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|{x}^{n-l}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n}\sum _{j=1}^{r}{a}_{j}\sum _{l=0}^{n}{c}_{l}\left(\genfrac{}{}{0}{}{n}{l}\right)〈\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y+{a}_{j}-1}|{x}^{n-l}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{r}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{l}{D}_{n-l}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)-\frac{1}{n}\sum _{j=1}^{r}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){a}_{j}{c}_{l}{D}_{n-l}\left(y+{a}_{j}-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right).\hfill \end{array}$

Therefore, we obtain

$\begin{array}{rcl}{D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)& =& x{D}_{n-1}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ +\frac{r}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{l}{D}_{n-l}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ -\frac{1}{n}\sum _{j=1}^{r}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){a}_{j}{c}_{l}{D}_{n-l}\left(x+{a}_{j}-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right),\end{array}$

which is the identity (35).

Next, for $n\ge 1$ we have

$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}\left(y|{a}_{1},\dots ,{a}_{r}\right)& =& 〈\sum _{l=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{l}\left(y|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{l}}{l!}|{x}^{n}〉\\ =& 〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{y}|{x}^{n}〉\\ =& 〈{\partial }_{t}\left(\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{y}\right)|{x}^{n-1}〉\\ =& 〈\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\left({\partial }_{t}{\left(1+t\right)}^{y}\right)|{x}^{n-1}〉\\ +〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\right){\left(1+t\right)}^{y}|{x}^{n-1}〉\\ =& y{\stackrel{ˆ}{D}}_{n-1}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)\\ +〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\right){\left(1+t\right)}^{y}|{x}^{n-1}〉.\end{array}$

Observe that

$\begin{array}{c}{\partial }_{t}\prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\hfill \\ \phantom{\rule{1em}{0ex}}={\partial }_{t}\left(\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\prod _{j=1}^{r}{\left(1+t\right)}^{{a}_{j}}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\right)\prod _{j=1}^{r}{\left(1+t\right)}^{{a}_{j}}\hfill \\ \phantom{\rule{2em}{0ex}}+\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\left({\partial }_{t}\prod _{j=1}^{r}{\left(1+t\right)}^{{a}_{j}}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{1+t}\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)\frac{{\sum }_{j=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right)}{t}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{1+t}\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)\sum _{j=1}^{r}{a}_{j}.\hfill \end{array}$

Thus, we have

$\begin{array}{c}〈\left({\partial }_{t}\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)\right){\left(1+t\right)}^{y}|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\frac{{\sum }_{j=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right)}{t}{x}^{n-1}〉\hfill \\ \phantom{\rule{2em}{0ex}}+\left(\sum _{j=1}^{r}{a}_{j}\right)〈\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n-1}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{n}〈\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\sum _{j=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right){x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n-1}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)+\frac{r}{n}〈\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\frac{t}{ln\left(1+t\right)}{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n}\sum _{j=1}^{r}{a}_{j}〈\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\frac{t}{ln\left(1+t\right)}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n-1}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)+\frac{r}{n}〈\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\sum _{l=0}^{\mathrm{\infty }}{c}_{l}\frac{{t}^{l}}{l!}{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n}\sum _{j=1}^{r}{a}_{j}〈\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|\sum _{l=0}^{\mathrm{\infty }}{c}_{l}\frac{{t}^{l}}{l!}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n-1}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{r}{n}\sum _{l=0}^{n}{c}_{l}\left(\genfrac{}{}{0}{}{n}{l}\right)〈\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|{x}^{n-l}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n}\sum _{j=1}^{r}{a}_{j}\sum _{l=0}^{n}{c}_{l}\left(\genfrac{}{}{0}{}{n}{l}\right)〈\frac{{\left(1+t\right)}^{{a}_{j}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\prod _{i=1}^{r}\left(\frac{{\left(1+t\right)}^{{a}_{i}}ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{y-1}|{x}^{n-l}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n-1}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)+\frac{r}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{l}{\stackrel{ˆ}{D}}_{n-l}\left(y-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n}\sum _{j=1}^{r}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){a}_{j}{c}_{l}{\stackrel{ˆ}{D}}_{n-l}\left(y-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right).\hfill \end{array}$

Therefore, we obtain

$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)& =& \left(x+\sum _{j=1}^{r}{a}_{j}\right){\stackrel{ˆ}{D}}_{n-1}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ +\frac{r}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{l}{\stackrel{ˆ}{D}}_{n-l}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ -\frac{1}{n}\sum _{j=1}^{r}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){a}_{j}{c}_{l}{\stackrel{ˆ}{D}}_{n-l}\left(x-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right),\end{array}$

which is the identity (36). □

### 3.7 Relations including the Stirling numbers of the first kind

Theorem 7 For $n-1\ge m\ge 1$, we have

$\begin{array}{c}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){D}_{l}\left({a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m-1\right){D}_{l}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){S}_{1}\left(n-l-1,m\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\left(r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0}{}{l+1}{i}\right){c}_{i}{D}_{l+1-i}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{j=1}^{r}\sum _{i=0}^{l+1}\left(\genfrac{}{}{0}{}{l+1}{i}\right){a}_{j}{c}_{i}{D}_{l+1-i}\left({a}_{j}-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right)\right),\hfill \end{array}$
(37)
$\begin{array}{c}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){\stackrel{ˆ}{D}}_{l}\left({a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m-1\right){\stackrel{ˆ}{D}}_{l}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){S}_{1}\left(n-l-1,m\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\left(r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0}{}{l+1}{i}\right){c}_{i}{\stackrel{ˆ}{D}}_{l+1-i}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{j=1}^{r}\sum _{i=0}^{l+1}\left(\genfrac{}{}{0}{}{l+1}{i}\right){a}_{j}{c}_{i}{\stackrel{ˆ}{D}}_{l+1-i}\left(-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m\right)\sum _{j=1}^{r}{a}_{j}{\stackrel{ˆ}{D}}_{l}\left(-1|{a}_{1},\dots ,{a}_{r}\right).\hfill \end{array}$
(38)

Proof We shall compute

$〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(ln\left(1+t\right)\right)}^{m}|{x}^{n}〉$

in two different ways. On the one hand,

$\begin{array}{c}〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(ln\left(1+t\right)\right)}^{m}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{\left(ln\left(1+t\right)\right)}^{m}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|\sum _{l=0}^{\mathrm{\infty }}\frac{m!}{\left(l+m\right)!}{S}_{1}\left(l+m,m\right){t}^{l+m}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m}\frac{m!}{\left(l+m\right)!}{S}_{1}\left(l+m,m\right){\left(n\right)}_{l+m}〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{x}^{n-l-m}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m}m!\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{1}\left(l+m,m\right){D}_{n-l-m}\left({a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m}m!\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){D}_{l}\left({a}_{1},\dots ,{a}_{r}\right).\hfill \end{array}$

On the other hand,

$\begin{array}{c}〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(ln\left(1+t\right)\right)}^{m}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\partial }_{t}\left(\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(ln\left(1+t\right)\right)}^{m}\right)|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\right){\left(ln\left(1+t\right)\right)}^{m}|{x}^{n-1}〉\hfill \\ \phantom{\rule{2em}{0ex}}+〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\partial }_{t}\left({\left(ln\left(1+t\right)\right)}^{m}\right)|{x}^{n-1}〉.\hfill \end{array}$
(39)

The second term of (39) is equal to

$\begin{array}{c}〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\partial }_{t}\left({\left(ln\left(1+t\right)\right)}^{m}\right)|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=m〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{-1}|{\left(ln\left(1+t\right)\right)}^{m-1}{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=m〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{-1}|\sum _{l=0}^{n-m}\frac{\left(m-1\right)!}{\left(l+m-1\right)!}{S}_{1}\left(l+m-1,m-1\right){t}^{l+m-1}{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=m\sum _{l=0}^{n-m}\frac{\left(m-1\right)!}{\left(l+m-1\right)!}{S}_{1}\left(l+m-1,m-1\right){\left(n-1\right)}_{l+m-1}\hfill \\ \phantom{\rule{2em}{0ex}}×〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){\left(1+t\right)}^{-1}|{x}^{n-l-m}〉\hfill \\ \phantom{\rule{1em}{0ex}}=m!\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l+m-1}\right){S}_{1}\left(l+m-1,m-1\right){D}_{n-l-m}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=m!\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m-1\right){D}_{l}\left(-1|{a}_{1},\dots ,{a}_{r}\right).\hfill \end{array}$

The first term of (39) is equal to

$\begin{array}{c}〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\right){\left(ln\left(1+t\right)\right)}^{m}|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{\left(ln\left(1+t\right)\right)}^{m}{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|\sum _{l=0}^{n-m-1}\frac{m!}{\left(l+m\right)!}{S}_{1}\left(l+m,m\right){t}^{l+m}{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m-1}\frac{m!}{\left(l+m\right)!}{S}_{1}\left(l+m,m\right){\left(n-1\right)}_{l+m}〈{\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)|{x}^{n-l-m-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m-1}m!\left(\genfrac{}{}{0}{}{n-1}{l+m}\right){S}_{1}\left(l+m,m\right)\hfill \\ \phantom{\rule{2em}{0ex}}×〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{-1}|\frac{{\sum }_{j=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right)}{t}{x}^{n-l-m-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=m!\sum _{l=0}^{n-m-1}\frac{1}{n-l-m}\left(\genfrac{}{}{0}{}{n-1}{l+m}\right){S}_{1}\left(l+m,m\right)\hfill \\ \phantom{\rule{2em}{0ex}}×〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{-1}|\sum _{j=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{j}t{\left(1+t\right)}^{{a}_{j}}}{{\left(1+t\right)}^{{a}_{j}}-1}\right){x}^{n-l-m}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{m!}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){S}_{1}\left(n-1-l,m\right)\left(r〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{-1}|\frac{t}{ln\left(1+t\right)}{x}^{l+1}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\sum _{j=1}^{r}{a}_{j}\right)〈\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}{\left(1+t\right)}^{{a}_{j}-1}\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)|\frac{t}{ln\left(1+t\right)}{x}^{l+1}〉\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{m!}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){S}_{1}\left(n-l-1,m\right)\left(r〈\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right){\left(1+t\right)}^{-1}|\sum _{i=0}^{l+1}{c}_{i}\frac{{t}^{i}}{i!}{x}^{l+1}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\sum _{j=1}^{r}{a}_{j}\right)〈\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}{\left(1+t\right)}^{{a}_{j}-1}\prod _{i=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{i}}-1}\right)|\sum _{i=0}^{l+1}{c}_{i}\frac{{t}^{i}}{i!}{x}^{l+1}〉\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{m!}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){S}_{1}\left(n-l-1,m\right)\left(r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0}{}{l+1}{i}\right){c}_{i}{D}_{l+1-i}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{j=1}^{r}{a}_{j}\sum _{i=0}^{l+1}\left(\genfrac{}{}{0}{}{l+1}{i}\right){c}_{i}{D}_{l+1-i}\left({a}_{j}-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right)\right).\hfill \end{array}$

Therefore, we have, for $n-1\ge m\ge 1$,

$\begin{array}{c}m!\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){D}_{l}\left({a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=m!\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m-1\right){D}_{l}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{m!}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){S}_{1}\left(n-l-1,m\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\left(r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0}{}{l+1}{i}\right){c}_{l+1-i}{D}_{i}\left(-1|{a}_{1},\dots ,{a}_{r}\hfill \end{array}$