Open Access

On q-analogs of degenerate Bernoulli polynomials

Advances in Difference Equations20152015:194

https://doi.org/10.1186/s13662-015-0522-7

Received: 5 April 2015

Accepted: 27 May 2015

Published: 25 June 2015

Abstract

The degenerate Bernoulli polynomials were introduced by Carlitz and rediscovered later by Ustiniv under the name of Korobov polynomials of the second kind (see Carlitz in Arch. Math. (Basel) 7:28-33, 1956; Util. Math. 15:51-88, 1979). In this paper, we study q-analogs of degenerate Bernoulli polynomials and give some formulas related to these polynomials.

Keywords

q-analogs of degenerate Bernoulli polynomials q-analogs of λ-Daehee polynomials q-analogs of degenerate Bernoulli polynomials of the second kind

MSC

05A1905A3011B6811B83

1 Introduction

Let p be a fixed prime number. Throughout this paper, \(\mathbb{Z}_{p}\), \(\mathbb{Q}_{p}\), and \(\mathbb{C}_{p}\) will denote the ring of p-adic integers, the field of p-adic rational numbers, and the completion of the algebraic closure of \(\mathbb{Q}_{p}\). The p-adic norm is normalized as \(\vert p\vert _{p}=\frac{1}{p}\). Let \(UD (\mathbb{Z}_{p})\) be the space of all \(\mathbb{C}_{p}\)-valued uniformly differentiable functions on \(\mathbb{Z}_{p}\), and let q be an indeterminate such that \(\vert 1-q\vert _{p}< p^{-\frac{1}{p-1}}\). The q-extension of the number x is defined as \([x ]_{q}=\frac{1-q^{x}}{1-q}\). Note that \(\lim_{q\rightarrow1} [x ]_{q}=x\). For \(f\in UD (\mathbb{Z}_{p})\), the p-adic q-integral on \(\mathbb{Z}_{p}\) is defined by Kim to be
$$ I_{q} (f )=\int_{\mathbb{Z}_{p}}f (x )\,d\mu_{q} (x )=\lim_{N\rightarrow\infty}\frac{1}{ [p^{N} ]_{q}}\sum _{x=0}^{p^{N}-1}f (x )q^{x} \quad(\mbox{see [1]}). $$
(1.1)
The ordinary p-adic invariant integral on \(\mathbb{Z}_{p}\) is given by
$$ I_{0} (f )=\lim_{q\rightarrow1}I_{q} (f )=\int _{\mathbb{Z}_{p}}f (x )\,d\mu_{0} (x )=\lim _{N\rightarrow\infty } \frac {1}{p^{N}}\sum_{x=0}^{p^{N}-1}f (x ). $$
(1.2)
From (1.1), we can derive the following integral equation:
$$ qI_{q} (f_{1} )-I_{q} (f )= (q-1 )f (0 )+ \frac{q-1}{\log q}f^{\prime} (0 ) \quad(\mbox{see [1--9]}), $$
(1.3)
where \(f_{1} (x )=f (x+1 )\).
For \(\lambda,t\in\mathbb{C}\) with \(\vert \lambda t\vert _{p}< p^{-\frac {1}{p-1}}\), the degenerate Bernoulli polynomials are defined as
$$ \frac{t}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac{x}{\lambda}}=\sum _{n=0}^{\infty}\beta_{n} (x\mid \lambda ) \frac{t^{n}}{n!} \quad(\mbox{see [10, 11]}). $$
(1.4)
When \(x=0\), \(\beta_{n} (\lambda )=\beta_{n} (0\mid \lambda )\) are called the degenerate Bernoulli numbers. As is well known, the Bernoulli polynomials of the second kind are defined by the generating function:
$$ \frac{t}{\log (1+t )} (1+t )^{x}=\sum_{n=0}^{\infty }b_{n} (x )\frac{t^{n}}{n!} \quad(\mbox{see [3, 4]}). $$
(1.5)
When \(x=0\), \(b_{n}=b_{n} (0 )\) are called the Bernoulli numbers of the second kind. The Daehee polynomials are also given by the generating function:
$$ \frac{\log (1+t )}{t} (1+t )^{x}=\sum_{n=0}^{\infty }D_{n} (x )\frac{t^{n}}{n!} \quad(\mbox{see [3, 4, 12, 13]}). $$
(1.6)
Now, we define the q-analogs of Bernoulli polynomials of the second kind as follows:
$$ \frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1+t )} (1+t )^{x}=\sum_{n=0}^{\infty}b_{n,q} (x )\frac {t^{n}}{n!}. $$
(1.7)

Note that \(\lim_{q\rightarrow1}b_{n,q} (x )=b_{n} (x )\).

The q-analogs of Daehee polynomials are defined by the generating function to be
$$ \frac{ (q-1 )+\frac{q-1}{\log q}\log (1+t )}{t} (1+t )^{x}=\sum_{n=0}^{\infty}D_{n,q} (x )\frac {t^{n}}{n!}. $$
(1.8)

When \(x=0\), \(b_{n,q}=b_{n,q} (0 )\) are called the q-analogs of Bernoulli numbers of the second kind and \(D_{n,q}=D_{n,q} (0 )\) are called the q-analogs of Daehee numbers.

From (1.7) and (1.8), we have
$$ b_{n,q} (x )=\sum_{l=0}^{n} \binom{n}{l} (x )_{n-l}b_{l,q},\qquad D_{n,q} (x )=\sum _{l=0}^{n}\binom {n}{l} (x )_{n-l}D_{l,q}, $$
(1.9)
where \((x )_{n}=x (x-1 )\cdots (x-n+1 )=\sum_{l=0}^{n}S_{1} (n,l )x^{l}\).

In this paper, we study q-analogs of degenerate Bernoulli polynomials and give some formulas related to these polynomials.

2 q-Analogs of degenerate Bernoulli polynomials

In this section, we assume that \(\lambda,t\in\mathbb{C}_{p}\) with \(\vert \lambda t\vert < p^{-\frac{1}{p-1}}\). Let us take \(f (y )= (1+\lambda t )^{\frac{y}{\lambda}}\). Then by (1.3), we get
$$\begin{aligned} & \int_{\mathbb{Z}_{p}} (1+\lambda t )^{\frac{x+y}{\lambda}}\,d \mu _{q} (y ) = \frac{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}}{q (1+\lambda t )^{\frac{1}{\lambda }}-1} (1+\lambda t )^{\frac{x}{\lambda}} = \sum_{n=0}^{\infty}D_{n,q} (x \mid \lambda )\frac {t^{n}}{n!}, \end{aligned}$$
(2.1)
where \(D_{n,q} (x\mid\lambda )\) are called the q-analogs of λ-Daehee polynomials. When \(x=0\), \(D_{n,q} (\lambda )=D_{n,q} (0\mid\lambda )\) are called the q-analogs of λ-Daehee numbers.
From (1.3), we can easily derive the following equation:
$$ q^{n}I_{q} (f_{n} )-I_{q} (f )= (q-1 )\sum_{l=0}^{n-1}q^{l}f (l )+ \frac{q-1}{\log q}\sum_{l=0}^{n-1}f^{\prime} (l )q^{l}, $$
(2.2)
where \(f_{n} (x )=f (x+n )\).
Thus, by (2.2), we get
$$\begin{aligned} & q^{n}\frac{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}}{q (1+\lambda t )^{\frac {1}{\lambda}}-1} (1+\lambda t )^{\frac{n}{\lambda}} \\ &\qquad{} -\frac{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}}{q (1+\lambda t )^{\frac{1}{\lambda }}-1} \\ &\quad= (q-1 )\sum_{l=0}^{n-1}q^{l} (1+\lambda t )^{\frac {l}{\lambda}}+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac {1}{\lambda}}\sum_{l=0}^{n-1} (1+\lambda t )^{\frac {l}{\lambda }}q^{l}. \end{aligned}$$
(2.3)
By (2.3), we get
$$\begin{aligned} & q^{n}\frac{t}{q (1+\lambda t )^{\frac{1}{\lambda }}-1} (1+\lambda t )^{\frac{n}{\lambda}}- \frac{t}{q (1+\lambda t )^{\frac{1}{\lambda}}-1} \\ &\quad= t\sum_{l=0}^{n-1}q^{l} (1+ \lambda t )^{\frac {l}{\lambda }}. \end{aligned}$$
(2.4)
It is easy to see that
$$\begin{aligned} (1+\lambda t )^{\frac{x}{\lambda}}= \sum_{l=0}^{\infty } (x\mid\lambda )_{l}\frac{t^{l}}{l!}, \end{aligned}$$
where \((x\mid\lambda )_{l}=x (x-\lambda )\cdots (x- (l-1 )\lambda )\) (see [117]).
Now, we define the q-analogs of degenerate Bernoulli polynomials as follows:
$$ \frac{t}{q (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac{x}{\lambda}}=\sum _{n=0}^{\infty}\beta _{n,q} (x\mid\lambda ) \frac{t^{n}}{n!}. $$
(2.5)

When \(x=0\), \(\beta_{n,q} (\lambda )=\beta_{n,q} (0\mid \lambda )\) are called the q-analogs of degenerate Bernoulli numbers.

From (2.4) and (2.5), we have
$$ \sum_{m=0}^{\infty} \bigl\{ q^{n} \beta_{m,q} (n\mid\lambda )-\beta_{m,q} \bigr\} \frac{t^{m}}{m!}=\sum_{m=1}^{\infty} \Biggl(m \sum_{l=0}^{n-1} (l\mid\lambda )_{m-1}q^{l} \Biggr)\frac {t^{m}}{m!}. $$
(2.6)

Therefore, by (2.6), we obtain the following theorem.

Theorem 2.1

For \(m\in\mathbb{N}\), we have
$$\frac{q^{n}\beta_{m,q} (n\mid\lambda )-\beta _{m,q}}{m}=\sum_{l=0}^{n-1} (l\mid \lambda )_{m-1}q^{l}. $$
We observe that
$$\begin{aligned} \frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}}\int_{\mathbb{Z}_{p}} (1+\lambda t )^{\frac {x+y}{\lambda}}\,d\mu_{q} (y )&= \frac{t}{q (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac {x}{\lambda}} \\ &= \sum_{n=0}^{\infty}\beta_{n,q} (x \mid\lambda )\frac {t^{n}}{n!}. \end{aligned}$$
(2.7)
Now, we define the q-analogs of degenerate Bernoulli polynomials of the second kind as follows:
$$ \frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}} (1+\lambda t )^{\frac{x}{\lambda }}=\sum _{n=0}^{\infty}b_{n,q} (x\mid\lambda ) \frac {t^{n}}{n!}. $$
(2.8)

When \(x=0\), \(b_{n} (\lambda )=b_{n} (0\mid\lambda )\) are called the q-analogs of degenerate Bernoulli numbers of the second kind.

Indeed, we note that \(\lim_{\lambda\rightarrow1}b_{n,q} (x\mid \lambda )=b_{n,q} (x )\).

By (2.1), we easily get
$$ \int_{\mathbb{Z}_{p}} (x+y\mid\lambda )_{n}\,d \mu_{q} (y )=D_{n,q} (x\mid\lambda )\quad (n\ge0 ). $$
(2.9)
From (2.1) and (2.8), we note that
$$\begin{aligned} & \frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}}\int_{\mathbb{Z}_{p}} (1+\lambda t )^{\frac {x+y}{\lambda}}\,d\mu_{q} (y ) \\ &\quad= \Biggl(\sum_{l=0}^{\infty}b_{l,q} ( \lambda )\frac {t^{l}}{l!} \Biggr) \Biggl(\sum_{m=0}^{\infty}D_{m,q} (x\mid \lambda )\frac{t^{m}}{m!} \Biggr) \\ &\quad= \sum_{n=0}^{\infty} \Biggl(\sum _{l=0}^{n}b_{l,q} (\lambda )D_{n-l,q} (x\mid\lambda )\binom{n}{l} \Biggr)\frac {t^{n}}{n!}. \end{aligned}$$
(2.10)

Therefore, by (2.7) and (2.10), we obtain the following theorem.

Theorem 2.2

For \(n\ge0\), we have
$$\beta_{n,q} (x\mid\lambda )=\sum_{l=0}^{n} \binom {n}{l}b_{l,q} (\lambda )D_{n-l,q} (x\mid\lambda ). $$
As is well known, the Apostol-Bernoulli polynomials are defined by the generating function:
$$ \frac{t}{qe^{t}-1}e^{xt}=\sum_{n=0}^{\infty}B_{n} (x\mid q )\frac{t^{n}}{n!}. $$
(2.11)

By (2.5) and (2.11), we get \(\lim_{\lambda \rightarrow 0}\beta_{n,q} (x\mid\lambda )=B_{n} (x\mid q )\) (\(n\ge0\)).

From (2.8), we can derive the following equation:
$$\begin{aligned} & \frac{e^{\lambda t}-1}{ (q-1 )+\frac{q-1}{\log q}t}\frac {1}{\lambda} e^{xt} \\ &\quad= \sum_{n=0}^{\infty}b_{n,q} (x\mid \lambda )\frac {1}{n!}\frac {1}{\lambda^{n}} \bigl(e^{\lambda t}-1 \bigr)^{n} \\ &\quad= \sum_{m=0}^{\infty}b_{m,q} (x \mid \lambda )\sum_{n=m}^{\infty}S_{2} (n,m )\lambda^{n-m}\frac {t^{n}}{n!} \\ &\quad= \sum_{n=0}^{\infty} \Biggl(\sum _{m=0}^{n}b_{m,q} (x\mid \lambda )S_{2} (n,m ) \lambda^{n-m} \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
(2.12)
By replacing t by \(\frac{1}{\lambda} (e^{\lambda t}-1 )\) in (2.7), we get
$$\begin{aligned} & \sum_{n=0}^{\infty} \beta_{n,q} (x\mid \lambda )\frac {1}{\lambda^{n}}\frac{1}{n!} \bigl(e^{\lambda t}-1 \bigr)^{n} \\ & \quad= \frac{e^{\lambda t}-1}{ (q-1 )+\frac{q-1}{\log q}t}\frac {1}{\lambda}\int_{\mathbb{Z}_{p}}e^{ (x+y )t} \,d\mu_{q} (y ) \\ &\quad= \Biggl(\sum_{k=0}^{\infty} \Biggl(\sum _{m=0}^{k}b_{m,q} (\lambda )S_{2} (k,m )\lambda^{k-m} \Biggr)\frac {t^{k}}{k!} \Biggr) \Biggl(\sum_{l=0}^{\infty}\int _{\mathbb{Z}_{p}} (x+y )^{l}\,d\mu _{q} (y ) \frac{t^{l}}{l!} \Biggr) \\ &\quad= \sum_{n=0}^{\infty} \Biggl(\sum _{k=0}^{n}\sum_{m=0}^{k} \binom {n}{k}b_{m,q} (\lambda )S_{2} (k,m )\lambda ^{k-m}B_{n-k,q} (x ) \Biggr)\frac{t^{n}}{n!}, \end{aligned}$$
(2.13)
where \(B_{n,q} (x )\) are the q-Bernoulli polynomials which are given by the generating function:
$$ \frac{ (q-1 )+\frac{q-1}{\log q}t}{qe^{t}-1}e^{xt}=\sum_{n=0}^{\infty}B_{n,q} (x )\frac{t^{n}}{n!}. $$
(2.14)
On the other hand,
$$\begin{aligned} \sum_{m=0}^{\infty} \beta_{m,q} (x\mid \lambda )\frac {1}{\lambda ^{m}}\frac{1}{m!} \bigl(e^{\lambda t}-1 \bigr)^{m} & =\sum _{m=0}^{\infty } \beta_{m,q} (x\mid\lambda ) \lambda^{-m}\sum_{n=m}^{\infty }S_{2} (n,m ) \lambda^{n}\frac{t^{n}}{n!} \\ & =\sum_{n=0}^{\infty} \Biggl(\sum _{m=0}^{n}\beta_{m,q} (x\mid \lambda ) \lambda^{n-m}S_{2} (n,m ) \Biggr)\frac {t^{n}}{n!}. \end{aligned}$$
(2.15)

Therefore, by (2.13) and (2.15), we obtain the following theorem.

Theorem 2.3

For \(n\ge0\), we have
$$\sum_{m=0}^{n}\beta_{m,q} (x\mid \lambda )\lambda ^{n-m}S_{2} (n,m )=\sum _{k=0}^{n}\binom{n}{k}\sum _{m=0}^{k}b_{m,q} (\lambda )S_{2} (k,m )\lambda ^{k-m}B_{n-k,q} (x ). $$
From (2.7), we have
$$\begin{aligned} & \sum_{n=0}^{\infty} \beta_{n,q} (x\mid \lambda )\frac {t^{n}}{n!} \\ &\quad= \frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}}\int_{\mathbb{Z}_{p}} (1+\lambda t )^{\frac {x+y}{\lambda}}\,d\mu_{q} (y ) \\ &\quad= \Biggl(\sum_{l=0}^{\infty}b_{l,q} ( \lambda )\frac {t^{l}}{l!} \Biggr) \Biggl(\sum_{m=0}^{\infty} \int_{\mathbb{Z}_{p}} (x+y\mid \lambda )_{m}\,d \mu_{q} (y )\frac{t^{m}}{m!} \Biggr) \\ &\quad= \sum_{n=0}^{\infty} \Biggl(\sum _{m=0}^{n}\binom {n}{m}b_{n-m,q} (\lambda )\int_{\mathbb{Z}_{p}} (x+y\mid\lambda )_{m}\,d\mu _{q} (y ) \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
(2.16)
Note that
$$ (x+y\mid\lambda )_{m}=\lambda^{m}\sum _{l=0}^{m}S_{1} (m,l ) \lambda^{-l} (x+y )^{l} $$
(2.17)
and
$$ \int_{\mathbb{Z}_{p}} (x+y )^{l}\,d\mu_{q} (y )=B_{n,q} (x )\quad (n\ge0 ). $$
(2.18)
By (2.16), (2.17), and (2.18), we get
$$\begin{aligned} & \beta_{n,q} (x\mid\lambda ) \\ &\quad= \sum_{m=0}^{n} \binom{n}{m}b_{n-m,q} (\lambda )\lambda ^{m}\sum _{l=0}^{m}S_{1} (m,l ) \lambda^{-l}\int_{\mathbb{Z}_{p}} (x+y )^{l}\,d \mu_{q} (y ) \\ &\quad= \sum_{m=0}^{n} \binom{n}{m}b_{n-m,q} (\lambda )\sum_{l=0}^{m}S_{1} (m,l )\lambda^{m-l}B_{l,q} (x ) \\ &\quad= \sum_{m=0}^{n}\sum _{l=0}^{m}\binom{n}{m}b_{n-m,q} (\lambda )S_{1} (m,l )\lambda^{m-l}B_{l,q} (x ). \end{aligned}$$
(2.19)

Therefore, by (2.19), we obtain the following theorem.

Theorem 2.4

For \(n\ge0\), we have
$$\beta_{n,q} (x\mid\lambda )=\sum_{m=0}^{n} \sum_{l=0}^{m}\binom {n}{m}b_{n-m,q} (\lambda )S_{1} (m,n )\lambda ^{m-l}B_{l,q} (x ). $$
For \(k\in\mathbb{N}\), we define the q-analogs of degenerate Bernoulli polynomials of order k as follows:
$$ \biggl(\frac{t}{q (1+\lambda t )^{\frac{1}{\lambda }}-1} \biggr)^{k} (1+\lambda t )^{\frac{x}{\lambda}}= \sum_{n=0}^{\infty }\beta_{n,q}^{ (k )} (x\mid\lambda )\frac {t^{n}}{n!}. $$
(2.20)

When \(x=0\), \(\beta_{n,q} (\lambda )=\beta_{n,q} (0\mid \lambda )\) are called the q-analogs of degenerate Bernoulli numbers of order k.

From (2.20), we note that
$$\begin{aligned} & \sum_{n=0}^{\infty}\lim _{\lambda\rightarrow0} \beta_{n,q}^{ (k )} (x\mid\lambda ) \frac{t^{n}}{n!} \\ &\quad= \lim_{\lambda\rightarrow0} \biggl(\frac{t}{q (1+\lambda t )^{\frac{1}{\lambda}}-1} \biggr)^{k} (1+\lambda t )^{\frac {x}{\lambda}} \\ &\quad= \biggl(\frac{t}{qe^{t}-1} \biggr)^{k}e^{xt} \\ &\quad= \sum_{n=0}^{\infty}B_{n}^{ (k )} (x\mid q )\frac {t^{n}}{n!}, \end{aligned}$$
(2.21)
where \(B_{n}^{ (k )} (x\mid q )\) are called the higher-order Apostol-Bernoulli polynomials.

Thus, by (2.21), we get \(\lim_{\lambda\rightarrow0}\beta _{n,q}^{ (k )} (x\mid\lambda )=B_{n}^{ (k )} (x\mid q )\) (\(n\geq0 \)).

For \(k\in\mathbb{N}\), by (2.20), we get
$$\begin{aligned} & \biggl(\frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}} \biggr)^{k}\int _{\mathbb{Z}_{p}} \cdots\int_{\mathbb{Z}_{p}} (1+\lambda t )^{\frac{x_{1}+\cdots +x_{k}+x}{\lambda }}\,d\mu_{q} (x_{1} )\cdots \,d \mu_{q} (x_{k} ) \\ &\quad= \biggl(\frac{t}{q (1+\lambda t )^{\frac{1}{\lambda }}-1} \biggr)^{k} (1+\lambda t )^{\frac{x}{\lambda }} \\ & \quad= \sum_{n=0}^{\infty} \beta_{n,q}^{ (k )} (x\mid \lambda )\frac{t^{n}}{n!}. \end{aligned}$$
(2.22)
Now, we define the q-analogs of higher-order degenerate Bernoulli polynomials of the second kind as follows:
$$ \biggl(\frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}} \biggr)^{k} (1+\lambda t )^{\frac {x}{\lambda}}= \sum_{n=0}^{\infty}b_{n,q}^{ (k )} (x\mid \lambda )\frac{t^{n}}{n!}. $$
(2.23)

When \(x=0\), \(b_{n,q}^{ (k )} (\lambda )=b_{n,q}^{ (k )} (0\mid\lambda )\) are called the q-analogs of higher-order degenerate Bernoulli numbers of the second kind. Note that \(\lim_{\lambda\rightarrow 1}b_{n,q}^{ (k )} (x\mid\lambda )=b_{n,q}^{ (k )} (x )\), and \(\lim_{q\rightarrow1}b_{n,q}^{ (k )} (x )=b_{n}^{ (k )} (x )\).

From (2.23), we can derive the following equation:
$$\begin{aligned} & \biggl(\frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac{1}{\lambda}}} \biggr)^{k}\int _{\mathbb{Z}_{p}} \cdots\int_{\mathbb{Z}_{p}} (1+\lambda t )^{\frac{x_{1}+\cdots +x_{k}+x}{\lambda }}\,d\mu_{q} (x_{1} )\cdots \,d \mu_{q} (x_{k} ) \\ & \quad= \Biggl(\sum_{m=0}^{\infty}b_{m,q}^{ (k )} (\lambda )\frac{t^{m}}{m!} \Biggr) \Biggl(\sum_{l=0}^{\infty} \int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{k}\mid\lambda )_{l}\,d\mu _{q} (x_{1} )\cdots \,d\mu_{q} (x_{k} ) \frac {t^{l}}{l!} \Biggr) \\ &\quad= \sum_{n=0}^{\infty} \Biggl(\sum _{l=0}^{n}\binom {n}{l}b_{n-l,q}^{ (k )} (\lambda )\int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{k}+x\mid\lambda )_{l}\,d \mu_{q} (x_{1} )\cdots \,d\mu_{q} (x_{k} ) \Biggr)\frac {t^{n}}{n!}. \end{aligned}$$
(2.24)
It is easy to show that
$$\begin{aligned} & \int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}}e^{ (x_{1}+\cdots+x_{k}+x )t} \,d\mu _{q} (x_{1} )\cdots \,d\mu_{q} (x_{k} ) \\ &\quad= \biggl(\frac{q-1+\frac{q-1}{\log q}t}{qe^{t}-1} \biggr)^{k}e^{xt} \\ &\quad= \sum_{n=0}^{\infty}B_{n,q}^{ (k )} (x )\frac {t^{n}}{n!}, \end{aligned}$$
(2.25)
where \(B_{n,q}^{ (k )} (x )\) are called the q-Bernoulli polynomials of order k.
Thus, by (2.25), we get
$$\begin{aligned} & \int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{k}+x\mid \lambda )_{l}\,d \mu_{q} (x_{1} )\cdots \,d\mu_{q} (x_{k} ) \\ &\quad= \lambda^{l}\sum_{m=0}^{l} \lambda^{-m}S_{1} (l,m )\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{k}+x )^{m}\,d\mu _{q} (x_{1} )\cdots d \mu_{q} (x_{k} ) \\ &\quad= \sum_{m=0}^{l} \lambda^{l-m}S_{1} (l,m )B_{m,q}^{ (k )} (x ). \end{aligned}$$
(2.26)

Therefore, by (2.22), (2.24), and (2.26), we obtain the following theorem.

Theorem 2.5

For \(n\ge0\), we have
$$\beta_{n,q}^{ (k )} (x\mid\lambda )=\sum _{l=0}^{n}\sum_{m=0}^{l} \binom{n}{l}b_{n-l,q}^{ (k )} (\lambda )\lambda^{l-m}S_{1} (l,m )B_{m,q}^{ (k )} (x ). $$

Remark

We define the q-analogs of λ-Daehee polynomials of order k as follows:
$$ \biggl(\frac{q-1+\frac{q-1}{\log q}\log (1+\lambda t )^{\frac {1}{\lambda}}}{q (1+\lambda t )^{\frac{1}{\lambda }}-1} \biggr)^{k} (1+\lambda t )^{\frac{x}{\lambda}}= \sum_{n=0}^{\infty }D_{n,q}^{ (k )} (x\mid\lambda )\frac {t^{n}}{n!}. $$
(2.27)
From (2.27), we have
$$ \int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (1+\lambda t )^{\frac {x_{1}+\cdots +x_{k}+x}{\lambda}}\,d\mu_{q} (x_{1} )\cdots \,d \mu_{q} (x_{k} )=\sum_{n=0}^{\infty}D_{n,q}^{ (k )} (x\mid \lambda )\frac{t^{n}}{n!}. $$
(2.28)
Thus, by (2.28), we get
$$ D_{n,q}^{ (k )} (x\mid\lambda )=\int_{\mathbb{Z}_{p}} \cdots\int_{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{k}+x \mid\lambda )_{n}\,d\mu _{q} (x_{1} )\cdots d \mu_{q} (x_{k} ) \quad(n\ge 0 ). $$
(2.29)
From (2.22), (2.24), and (2.29), we have
$$ \beta_{n,q}^{ (k )} (x\mid\lambda )=\sum _{l=0}^{n}\binom{n}{l}b_{n-l,q}^{ (k )} (\lambda )D_{l,q}^{ (k )} (x\mid\lambda ). $$
(2.30)
From (2.1), we can derive the following equation:
$$\begin{aligned} \int_{\mathbb{Z}_{p}} (1-\lambda t )^{-\frac{x+y}{\lambda}}\,d\mu _{q} (y )&= \frac{ (q-1 )+\frac{q-1}{\log q}\log (1-\lambda t )^{-\frac{1}{\lambda}}}{q (1-\lambda t )^{-\frac{1}{\lambda}}-1} (1-\lambda t )^{-\frac {x}{\lambda }} \\ &= \sum_{n=0}^{\infty}D_{n,q} (x\mid- \lambda )\frac {t^{n}}{n!}. \end{aligned}$$
(2.31)

Note that \(D_{n} (1-x\mid\lambda )= (-1 )^{n}D_{n} (x\mid-\lambda ) \) (\(n\ge0\)).

By (2.31), we get
$$ \int_{\mathbb{Z}_{p}} \langle x+y| \lambda \rangle _{n}\,d \mu_{q} (y )=D_{n,q} (x\mid-\lambda ) \quad(n\ge0 ), $$
(2.32)
where \(\langle x\vert \lambda \rangle _{n}=x (x+\lambda )\cdots (x+ (n-1 )\lambda )\).
Note that
$$\begin{aligned} \langle x+y| \lambda \rangle _{n} &= \lambda^{n} \biggl(\frac{x+y}{\lambda} \biggr) \biggl( \frac {x+y}{\lambda }+1 \biggr)\cdots \biggl(\frac{x+y}{\lambda}+n-1 \biggr) \\ &= \sum_{l=0}^{n} \bigl\vert S_{1} (n,l ) \bigr\vert (x+y )^{l}\lambda^{n-l}. \end{aligned}$$
(2.33)
From (2.32) and (2.33), we have
$$\begin{aligned} D_{n,q} (x\mid-\lambda ) &= \sum_{l=0}^{n} \bigl\vert S_{1} (n,l ) \bigr\vert \lambda ^{n-l}\int _{\mathbb{Z}_{p}} (x+y )^{l}\,d\mu_{q} (y ) \\ &= \sum_{l=0}^{n} \bigl\vert S_{1} (n,l ) \bigr\vert \lambda ^{n-l}B_{l,q} (x ). \end{aligned}$$
(2.34)
By (2.7), we get
$$\begin{aligned} & \frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1-\lambda t )^{-\frac{1}{\lambda}}}\int_{\mathbb{Z}_{p}} (1-\lambda t )^{-\frac {x+y}{\lambda}}\,d\mu_{q} (y ) \\ &\quad= \frac{t}{q (1-\lambda t )^{-\frac{1}{\lambda }}-1} (1-\lambda t )^{-\frac{x}{\lambda}} \\ &\quad= \sum_{n=0}^{\infty}\beta_{n,q} (x \mid-\lambda )\frac {t^{n}}{n!}. \end{aligned}$$
(2.35)
By (2.8), we get
$$\begin{aligned} & \frac{t}{ (q-1 )+\frac{q-1}{\log q}\log (1-\lambda t )^{-\frac{1}{\lambda}}}\int_{\mathbb{Z}_{p}} (1-\lambda t )^{-\frac {x+y}{\lambda}}\,d\mu_{q} (y ) \\ &\quad= \Biggl(\sum_{m=0}^{\infty}b_{m,q} (-\lambda )\frac {t^{m}}{m!} \Biggr) \Biggl(\sum_{l=0}^{\infty} \int_{\mathbb{Z}_{p}} \langle x+y| \lambda \rangle _{l}\,d \mu_{q} (y )\frac{t^{l}}{l!} \Biggr) \\ & \quad= \sum_{n=0}^{\infty} \Biggl(\sum _{l=0}^{n}\binom {n}{l}b_{n-l,q} (-\lambda )\int_{\mathbb{Z}_{p}} \langle x+y| \lambda \rangle _{l}\,d\mu_{q} (y ) \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
(2.36)
Therefore, by (2.32), (2.33), (2.35), and (2.36), we get
$$\begin{aligned} & \beta_{n,q} (x\mid-\lambda ) \\ &\quad= \sum_{l=0}^{n} \binom{n}{l}b_{n-l,q} (-\lambda )D_{l,q} (x\mid-\lambda ) \\ &\quad= \sum_{l=0}^{n}\sum _{m=0}^{l}\binom{n}{l} \bigl\vert S_{1} (l,m ) \bigr\vert \lambda^{l-m}B_{m,q} (x )b_{n-l,q} (-\lambda ). \end{aligned}$$
(2.37)

Declarations

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Sogang University, Seoul, Republic of Korea
(2)
Department of Mathematics, Kwangwoon University, Seoul, Republic of Korea
(3)
Institute of Natural Sciences, Far Eastern Federal University, Vladivostok, Russia

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