Theory and Modern Applications

# On q-analogs of degenerate Bernoulli polynomials

## 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.

## 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)

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.

## 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 ).

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)

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