Theory and Modern Applications

# Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type polynomials

## Abstract

In this paper, by considering Barnes-type Daehee polynomials of the first kind as well as poly-Cauchy polynomials of the first kind, we define and investigate the mixed-type polynomials of these polynomials. From the properties of 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 ${D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)$ called the Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type polynomials, whose generating function is given by

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

where ${a}_{1},\dots ,{a}_{r}\ne 0$. Here, ${Lif}_{k}\left(x\right)$ ($k\in \mathbb{Z}$) is the polyfactorial function  defined by

${Lif}_{k}\left(x\right)=\sum _{m=0}^{\mathrm{\infty }}\frac{{x}^{m}}{m!{\left(m+1\right)}^{k}}.$

When $x=0$, ${D}_{n}^{\left(k\right)}\left({a}_{1},\dots ,{a}_{r}\right)={D}_{n}^{\left(k\right)}\left(0|{a}_{1},\dots ,{a}_{r}\right)$ is called Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type number.

Recall that the Barnes-type Daehee polynomials of the first kind, denoted by ${D}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)$, are given by the generating function

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

If ${a}_{1}=\cdots ={a}_{r}=1$, then ${D}_{n}^{\left(r\right)}\left(x\right)={D}_{n}\left(x|\underset{r}{\underset{⏟}{1,\dots ,1}}\right)$ are the Daehee polynomials of the first kind of order r. Dahee polynomials were defined by the second author  and have been investigated in [3, 4].

The poly-Cauchy polynomials of the first kind, denoted by ${c}_{n}^{\left(k\right)}\left(x\right)$ [5, 6], are given by the generating function as

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

In this paper, by considering Barnes-type Daehee polynomials of the first kind as well as poly-Cauchy polynomials of the first kind, we define and investigate the mixed-type polynomials of these polynomials. From the properties of 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:

$\mathcal{F}=\left\{f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{a}_{k}}{k!}{t}^{k}|{a}_{k}\in \mathbb{C}\right\}.$
(2)

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

$〈f\left(t\right)|{x}^{n}〉={a}_{n}\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(3)

In particular,

$〈{t}^{k}|{x}^{n}〉=n!{\delta }_{n,k}\phantom{\rule{1em}{0ex}}\left(n,k\ge 0\right),$
(4)

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 so 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 the 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

$〈f\left(t\right)g\left(t\right)|p\left(x\right)〉=〈f\left(t\right)|g\left(t\right)p\left(x\right)〉=〈g\left(t\right)|f\left(t\right)p\left(x\right)〉$
(5)

and

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}〈f\left(t\right)|{x}^{k}〉\frac{{t}^{k}}{k!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}〈{t}^{k}|p\left(x\right)〉\frac{{x}^{k}}{k!}$
(6)

, Theorem 2.2.5]. Thus, by (6), we get

${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{e}^{yt}p\left(x\right)=p\left(x+y\right).$
(7)

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]:

$f\left(t\right){s}_{n}\left(x\right)=n{s}_{n-1}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right),$
(8)
${s}_{n}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈g{\left(\overline{f}\left(t\right)\right)}^{-1}\overline{f}{\left(t\right)}^{j}|{x}^{n}〉{x}^{j},$
(9)
${s}_{n}\left(x+y\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){s}_{j}\left(x\right){p}_{n-j}\left(y\right),$
(10)

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]

${C}_{n,m}=\frac{1}{m!}〈\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}l{\left(\overline{f}\left(t\right)\right)}^{m}|{x}^{n}〉.$
(11)

## 3 Main results

From the definition (1), ${D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)$ is the Sheffer sequence for the pair

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

So,

${D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)\sim \left(\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t}\right)\frac{1}{{Lif}_{k}\left(t\right)},{e}^{t}-1\right).$
(12)

### 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

$\frac{{t}^{r}}{{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-1\right)}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{n}}{n!},$
(13)

where ${a}_{1},\dots ,{a}_{r}\ne 0$ [8, 9]. 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

$\begin{array}{r}{D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\sum _{l=0}^{m}{S}_{1}\left(n,m\right)\frac{\left(\genfrac{}{}{0}{}{m}{l}\right)}{{\left(m-l+1\right)}^{k}}{B}_{l}\left(x|{a}_{1},\dots ,{a}_{r}\right)\end{array}$
(14)
$\phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\sum _{l=0}^{n-j}\sum _{i=0}^{l}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right){S}_{1}\left(n-l,j\right){c}_{i}^{\left(k\right)}{D}_{l-i}\left({a}_{1},\dots ,{a}_{r}\right){x}^{j}$
(15)
$\phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){D}_{n-l}\left({a}_{1},\dots ,{a}_{r}\right){c}_{l}^{\left(k\right)}\left(-x\right)$
(16)
$\phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{n-l}^{\left(k\right)}{D}_{l}\left(x|{a}_{1},\dots ,{a}_{r}\right).$
(17)

Proof Since

$\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-1}{t}\right)\frac{1}{{Lif}_{k}\left(t\right)}{D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)\sim \left(1,{e}^{t}-1\right)$
(18)

and

${\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right),$
(19)

we have

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

Thus, we get (14).

By (9) with (12), we get

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

On the other hand,

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

Thus, we obtain

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

which is the identity (15).

Next,

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

Thus, we obtain (16).

Finally, we obtain

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

Thus, we get the identity (17). □

### 3.2 Sheffer identity

Theorem 2

${D}_{n}^{\left(k\right)}\left(x+y|{a}_{1},\dots ,{a}_{r}\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){D}_{j}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right){\left(y\right)}_{n-j}.$
(20)

Proof By (12) with

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

using (10), we have (20). □

### 3.3 Difference relations

Theorem 3

${D}_{n}^{\left(k\right)}\left(x+1|{a}_{1},\dots ,{a}_{r}\right)-{D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)=n{D}_{n-1}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right).$
(21)

Proof By (8) with (12), we get

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

By (7), we have (21). □

### 3.4 Recurrence

Theorem 4

$\begin{array}{rl}{D}_{n+1}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)=& x{D}_{n}^{\left(k\right)}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ -\sum _{m=0}^{n}\sum _{j=1}^{r}\sum _{l=1}^{m+1}\sum _{i=0}^{l}\frac{\left(\genfrac{}{}{0}{}{m+1}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right)}{\left(m+1\right){\left(l-i+1\right)}^{k}}{S}_{1}\left(n,m\right)\\ ×{\left(-{a}_{j}\right)}^{m+1-l}{B}_{m+1-l}{B}_{i}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ +\sum _{m=0}^{n}\sum _{l=0}^{m}\frac{\left(\genfrac{}{}{0}{}{m}{l}\right)}{{\left(m+2-l\right)}^{k}}{S}_{1}\left(n,m\right){B}_{l}\left(x-1|{a}_{1},\dots ,{a}_{r}\right),\end{array}$
(22)

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

Proof By applying

${s}_{n+1}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{\prime }\left(t\right)}{s}_{n}\left(x\right)$
(23)

, Corollary 3.7.2] with (12), we get

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

Now,

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

Observe that

$\begin{array}{r}\frac{{\sum }_{j=1}^{r}{\prod }_{i\ne j}\left({e}^{{a}_{i}t}-1\right)\left({a}_{j}t{e}^{{a}_{j}t}-{e}^{{a}_{j}t}+1\right)}{{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-1\right)}\\ \phantom{\rule{1em}{0ex}}=\frac{\frac{1}{2}\left({\sum }_{j=1}^{r}{a}_{1}\cdots {a}_{j-1}{a}_{j}^{2}{a}_{j+1}\cdots {a}_{r}\right){t}^{r+1}+\cdots }{\left({a}_{1}\cdots {a}_{r}\right){t}^{r}+\cdots }\\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\left(\sum _{j=1}^{r}{a}_{j}\right)t+\cdots \end{array}$

is a series with order ≥1. Since

${D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)=\prod _{j=1}^{r}\left(\frac{t}{{e}^{{a}_{j}t}-1}\right){Lif}_{k}\left(t\right){\left(x\right)}_{n}=\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\prod _{j=1}^{r}\left(\frac{t}{{e}^{{a}_{j}t}-1}\right){Lif}_{k}\left(t\right){x}^{m},$

we have

$\begin{array}{rl}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)=& \sum _{m=0}^{n}{S}_{1}\left(n,m\right)\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\left(\prod _{j=1}^{r}\frac{t}{{e}^{{a}_{j}t}-1}\right){Lif}_{k}\left(t\right){x}^{m}\\ =& \sum _{m=0}^{n}{S}_{1}\left(n,m\right){Lif}_{k}\left(t\right)\left(\prod _{j=1}^{r}\frac{t}{{e}^{{a}_{j}t}-1}\right)\\ ×\frac{{\sum }_{j=1}^{r}{\prod }_{i\ne j}\left({e}^{{a}_{i}t}-1\right)\left({a}_{j}t{e}^{{a}_{j}t}-{e}^{{a}_{j}t}+1\right)}{t{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-1\right)}{x}^{m}\\ -\sum _{m=0}^{n}{S}_{1}\left(n,m\right){Lif}_{k}^{\prime }\left(t\right)\left(\prod _{j=1}^{r}\frac{t}{{e}^{{a}_{j}t}-1}\right){x}^{m}.\end{array}$
(24)

Since

$\begin{array}{r}\frac{{\sum }_{j=1}^{r}{\prod }_{i\ne j}\left({e}^{{a}_{i}t}-1\right)\left({a}_{j}t{e}^{{a}_{j}t}-{e}^{{a}_{j}t}+1\right)}{t{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-1\right)}{x}^{m}\\ \phantom{\rule{1em}{0ex}}=\frac{{\sum }_{j=1}^{r}{\prod }_{i\ne j}\left({e}^{{a}_{i}t}-1\right)\left({a}_{j}t{e}^{{a}_{j}t}-{e}^{{a}_{j}t}+1\right)}{{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-1\right)}\frac{{x}^{m+1}}{m+1}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{m+1}\sum _{j=1}^{r}\left(\frac{{a}_{j}t{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-1}-1\right){x}^{m+1}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{m+1}\sum _{j=1}^{r}\left(\sum _{l=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{l}{B}_{l}{a}_{j}^{l}}{l!}{t}^{l}-1\right){x}^{m+1}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{m+1}\sum _{j=1}^{r}\left(\sum _{l=0}^{m+1}\left(\genfrac{}{}{0}{}{m+1}{l}\right){\left(-{a}_{j}\right)}^{l}{B}_{l}{x}^{m+1-l}-{x}^{m+1}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{m+1}\sum _{j=1}^{r}\sum _{l=1}^{m+1}\left(\genfrac{}{}{0}{}{m+1}{l}\right){\left(-{a}_{j}\right)}^{l}{B}_{l}{x}^{m+1-l}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{m+1}\sum _{j=1}^{r}\sum _{l=1}^{m+1}\left(\genfrac{}{}{0}{}{m+1}{l}\right){\left(-{a}_{j}\right)}^{m+1-l}{B}_{m+1-l}{x}^{l},\end{array}$

the first term in (24) is

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

Since

${Lif}_{k-1}\left(t\right)-{Lif}_{k}\left(t\right)=\left(\frac{1}{{2}^{k-1}}-\frac{1}{{2}^{k}}\right)t+\cdots ,$
(25)

the second term in (24) is

$\begin{array}{r}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\frac{{Lif}_{k-1}\left(t\right)-{Lif}_{k}\left(t\right)}{t}{B}_{m}\left(x|{a}_{1},\dots ,{a}_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\left({Lif}_{k-1}\left(t\right)-{Lif}_{k}\left(t\right)\right)\frac{{B}_{m+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)}{m+1}\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{m+1}\left({Lif}_{k-1}\left(t\right)-{Lif}_{k}\left(t\right)\right){B}_{m+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{m+1}\left(\sum _{l=0}^{m+1}\frac{{t}^{l}}{l!{\left(l+1\right)}^{k-1}}{B}_{m+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)-\sum _{l=0}^{m+1}\frac{{t}^{l}}{l!{\left(l+1\right)}^{k}}{B}_{m+1}\left(x|{a}_{1},\dots ,{a}_{r}\right)\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{m+1}\left(\sum _{l=0}^{m+1}\frac{\left(\genfrac{}{}{0}{}{m+1}{l}\right)}{{\left(l+1\right)}^{k-1}}{B}_{m+1-l}\left(x|{a}_{1},\dots ,{a}_{r}\right)-\sum _{l=0}^{m+1}\frac{\left(\genfrac{}{}{0}{}{m+1}{l}\right)}{{\left(l+1\right)}^{k}}{B}_{m+1-l}\left(x|{a}_{1},\dots ,{a}_{r}\right)\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{m+1}\sum _{l=1}^{m+1}\frac{\left(\genfrac{}{}{0}{}{m+1}{l}\right)l}{{\left(l+1\right)}^{k}}{B}_{m+1-l}\left(x|{a}_{1},\dots ,{a}_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\sum _{l=1}^{m+1}\left(\genfrac{}{}{0}{}{m}{l-1}\right){S}_{1}\left(n,m\right)\frac{1}{{\left(l+1\right)}^{k}}{B}_{m+1-l}\left(x|{a}_{1},\dots ,{a}_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\sum _{l=0}^{m}\frac{\left(\genfrac{}{}{0}{}{m}{l}\right)}{{\left(m+2-l\right)}^{k}}{S}_{1}\left(n,m\right){B}_{l}\left(x|{a}_{1},\dots ,{a}_{r}\right).\end{array}$

Thus, we have

$\begin{array}{rl}{D}_{n+1}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)=& x{D}_{n}^{\left(k\right)}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ -\sum _{m=0}^{n}\sum _{j=1}^{r}\sum _{l=1}^{m+1}\sum _{i=0}^{l}\frac{\left(\genfrac{}{}{0}{}{m+1}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right)}{\left(m+1\right){\left(l-i+1\right)}^{k}}{S}_{1}\left(n,m\right)\\ ×{\left(-{a}_{j}\right)}^{m+1-l}{B}_{m+1-l}{B}_{i}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ +\sum _{m=0}^{n}\sum _{l=0}^{m}\frac{\left(\genfrac{}{}{0}{}{m}{l}\right)}{{\left(m+2-l\right)}^{k}}{S}_{1}\left(n,m\right){B}_{l}\left(x-1|{a}_{1},\dots ,{a}_{r}\right),\end{array}$

which is the identity (22). □

### 3.5 Differentiation

Theorem 5

$\frac{d}{dx}{D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{D}_{l}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right).$
(26)

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 (12), we have

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

which is the identity (26). □

### 3.6 One more relation

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. [1, 10]).

Theorem 6

$\begin{array}{rl}{D}_{n}^{\left(k\right)}\left(x|{a}_{1},\dots ,{a}_{r}\right)=& x{D}_{n-1}^{\left(k\right)}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ +\frac{1}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{l}{D}_{n-l}^{\left(k-1\right)}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ +\frac{r-1}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){c}_{l}{D}_{n-l}^{\left(k\right)}\left(x-1|{a}_{1},\dots ,{a}_{r}\right)\\ -\frac{1}{n}\sum _{j=1}^{r}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){a}_{j}{c}_{l}{D}_{n-l}^{\left(k\right)}\left(x+{a}_{j}-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right).\end{array}$
(27)

Proof For $n\ge 1$, we have

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

The third term is

$\begin{array}{r}y〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){Lif}_{k}\left(ln\left(1+t\right)\right){\left(1+t\right)}^{y-1}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=y{D}_{n-1}^{\left(k\right)}\left(y-1|{a}_{1},\dots ,{a}_{r}\right).\end{array}$

By (25), the second term is

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

Since

${\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)=\frac{1}{1+t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\frac{{\sum }_{i=1}^{r}\left(\frac{t}{ln\left(1+t\right)}-\frac{{a}_{i}t{\left(1+t\right)}^{{a}_{i}}}{{\left(1+t\right)}^{{a}_{i}}-1}\right)}{t},$

with

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

a series with order (≥1), the first term is

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

Therefore, we obtain

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

which is the identity (27). □

### 3.7 A relation including the Stirling numbers of the first kind

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

$\begin{array}{r}m\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){D}_{l}^{\left(k\right)}\left({a}_{1},\dots ,{a}_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{mr}{n}\sum _{l=0}^{n-m}\sum _{i=0}^{l}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right){S}_{1}\left(n-l,m\right){c}_{l-i}{D}_{i}^{\left(k\right)}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\\ \phantom{\rule{2em}{0ex}}-\frac{m}{n}\sum _{l=0}^{n-m}\sum _{i=0}^{l}\sum _{j=1}^{r}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right){S}_{1}\left(n-l,m\right){a}_{j}{c}_{l-i}{D}_{i}^{\left(k\right)}\left({a}_{j}-1|{a}_{1},\dots ,{a}_{r},{a}_{j}\right)\\ \phantom{\rule{2em}{0ex}}+\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m-1\right){D}_{l}^{\left(k-1\right)}\left(-1|{a}_{1},\dots ,{a}_{r}\right)\\ \phantom{\rule{2em}{0ex}}+\left(m-1\right)\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m-1\right){D}_{l}^{\left(k\right)}\left(-1|{a}_{1},\dots ,{a}_{r}\right).\end{array}$
(28)

Proof We shall compute

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

in two different ways. On the one hand,

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

On the other hand,

$\begin{array}{r}〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){Lif}_{k}\left(ln\left(1+t\right)\right){\left(ln\left(1+t\right)\right)}^{m}|{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=〈{\partial }_{t}\left(\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){Lif}_{k}\left(ln\left(1+t\right)\right){\left(ln\left(1+t\right)\right)}^{m}\right)|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\right){Lif}_{k}\left(ln\left(1+t\right)\right){\left(ln\left(1+t\right)\right)}^{m}|{x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right)\left({\partial }_{t}{Lif}_{k}\left(ln\left(1+t\right)\right)\right){\left(ln\left(1+t\right)\right)}^{m}|{x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈\prod _{j=1}^{r}\left(\frac{ln\left(1+t\right)}{{\left(1+t\right)}^{{a}_{j}}-1}\right){Lif}_{k}\left(ln\left(1+t\right)\right)\left({\partial }_{t}{\left(ln\left(1+t\right)\right)}^{m}\right)|{x}^{n-1}〉.\end{array}$
(29)

The third term of (29) is equal to

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

The second term of (29) is equal to

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

The first term of (29) is equal to

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