Open Access

Degenerate q-Euler polynomials

Advances in Difference Equations20152015:246

https://doi.org/10.1186/s13662-015-0563-y

Received: 5 May 2015

Accepted: 2 July 2015

Published: 8 August 2015

Abstract

Recently, some identities of degenerate Euler polynomials arising from p-adic fermionic integrals on \(\mathbb{Z}_{p}\) were introduced in Kim and Kim (Integral Transforms Spec. Funct. 26(4):295-302, 2015). In this paper, we study degenerate q-Euler polynomials which are derived from p-adic q-integrals on \(\mathbb{Z}_{p}\).

Keywords

degenerate Euler polynomials p-adic q-fermionic integral

MSC

11B6811S80

1 Introduction

Let p be a fixed odd 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 algebraic closure of \(\mathbb{Q}_{p}\), respectively. Let \(\nu_{p}\) be the normalized exponential valuation in \(\mathbb{C}_{p}\) with \(\vert p\vert _{p}=p^{-\nu_{p} (p )}=\frac{1}{p}\).

Let q be an indeterminate in \(\mathbb{C}_{p}\) such that \(\vert 1-q\vert _{p}< p^{-\frac{1}{p-1}}\). The q-extension of x is defined as \([x ]_{q}=\frac{1-q^{x}}{1-q}\). Note that \(\lim_{q\rightarrow1} [x ]_{q}=x\). For \(f\in C (\mathbb{Z}_{p})\) = {\(f\mid, f\) is a \(\mathbb{C}_{p}\)-valued continuous function on \(\mathbb{Z}_{p}\)}, the fermionic p-adic q-integral on \(\mathbb{Z}_{p}\) is defined by Kim to be
$$ I_{-q} (f )=\int_{\mathbb{Z}_{p}}f (x )\,d\mu_{-q} (x )=\lim_{N\rightarrow\infty}\frac{1}{ [p^{N} ]_{-q}}\sum _{x=0}^{p^{N}-1}f (x ) (-q )^{x} \quad(\mbox{see [1, 2]} ), $$
(1.1)
where \([x ]_{-q}=\frac{1- (-q )^{x}}{1+q}\).
By (1.1), we easily get
$$ qI_{-q} (f_{1} )+I_{-q} (f )= [2 ]_{q}f (0 )\quad \bigl(f_{1} (x )=f (x+1 ) \bigr), $$
(1.2)
and
$$ q^{n}I_{-q} (f_{n} )+ (-1 )^{n-1}I_{-q} (f )= [2 ]_{q}\sum_{l=0}^{n-1} (-1 )^{n-1-l}q^{l}f (l )\quad (n\in\mathbb{N} ), $$
(1.3)
where \(f_{n} (x )=f (x+n )\) (see [116]).
The ordinary fermionic p-adic integral on \(\mathbb{Z}_{p}\) is defined as
$$ \lim_{q\rightarrow1}I_{-q} (f )=I_{-1} (f )=\int _{\mathbb{Z}_{p}}f (x )\,d\mu_{-1} (x )=\lim_{N\rightarrow\infty} \sum_{x=0}^{p^{N}-1}f (x ) (-1 )^{x} \quad(\mbox{see [2]}). $$
(1.4)
The degenerate Euler polynomials of order r (\(\in\mathbb{N}\)) are defined by the generating function to be
$$ \biggl(\frac{2}{ (1+\lambda t )^{\frac{1}{\lambda}}+1} \biggr)^{r} (1+\lambda t )^{\frac{x}{\lambda}}= \sum_{n=0}^{\infty }\mathcal{E}_{n}^{ (r )} (x\mid\lambda )\frac {t^{n}}{n!} \quad(\mbox{see [5, 6, 10]} ), $$
(1.5)
where \(\lambda,t\in \mathbb{Z}_{p}\) such that \(\vert \lambda t\vert _{p}< p^{-\frac {1}{p-1}}\).
From (1.5), we have
$$\begin{aligned} & \sum_{n=0}^{\infty}\lim_{\lambda\rightarrow0} \mathcal{E}_{n}^{ (r )} (x\mid\lambda )\frac{t^{n}}{n!} \\ &\quad= \lim_{\lambda\rightarrow0} \biggl(\frac{2}{ (1+\lambda t )^{\frac{1}{\lambda}}+1} \biggr)^{r} (1+\lambda t )^{\frac {x}{\lambda}} \\ &\quad= \biggl(\frac{2}{e^{t}+1} \biggr)^{r}e^{xt} \\ &\quad= \sum_{n=0}^{\infty}E_{n}^{ (r )} (x )\frac {t^{n}}{n!}, \end{aligned}$$
(1.6)
where \(E_{n}^{ (r )} (x )\) are the higher-order Euler polynomials.
Thus, by (1.6), we get
$$ \lim_{\lambda\rightarrow0}\mathcal{E}_{n}^{ (r )} (x\mid \lambda )=E_{n}^{ (r )} (x )\quad(n\geq 0 ). $$
(1.7)

When \(x=0\), \(\mathcal{E}_{n}^{ (r )} (\lambda )=\mathcal{E}_{n}^{ (r )} (0\mid\lambda )\) are called the higher-order degenerate Euler numbers, while \(\lim_{\lambda\rightarrow0}\mathcal{E}_{n}^{ (r )} (\lambda )=E_{n}^{ (r )}\) are called the higher-order Euler numbers.

In [10], it was shown that
$$ \mathcal{E}_{n}^{ (r )} (x\mid\lambda )=\int _{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (x_{1}+x_{2}+ \cdots+x_{r}+x\mid\lambda )_{n}\,d\mu_{-1} (x_{1} )\cdots \,d\mu_{-1} (x_{r} ), $$
(1.8)
where \((x )_{n}=x (x-1 )\cdots (x-n+1 )\) and \(n\in\mathbb{Z}_{\ge0}\).

In this paper, we study q-extensions of the degenerate Euler polynomials and give some formulae and identities of those polynomials which are derived from the fermionic p-adic q-integrals on \(\mathbb{Z}_{p}\).

2 Some identities of q-analogues of higher-order degenerate Euler polynomials

In this section, we assume that \(\lambda,t\in \mathbb{Z}_{p}\) with \(\vert \lambda t\vert _{p}< p^{-\frac{1}{p-1}}\). From (1.2), we have
$$\begin{aligned} &\int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (1+\lambda t )^{ (x_{1}+\cdots +x_{r}+x )/\lambda}\,d\mu_{-q} (x_{1} )\cdots \,d\mu_{-q} (x_{r} ) \\ &\quad= \biggl(\frac{ [2 ]_{q}}{q (1+\lambda t )^{1/\lambda }+1} \biggr)^{r} (1+\lambda t )^{\frac{x}{\lambda}}. \end{aligned}$$
(2.1)
Now, we define a q-analogue of degenerate Euler polynomials of order r as follows:
$$ \biggl(\frac{ [2 ]_{q}}{q (1+\lambda t )^{1/\lambda }+1} \biggr)^{r} (1+\lambda t )^{\frac{x}{\lambda}}= \sum_{n=0}^{\infty}\mathcal{E}_{n,q}^{ (r )} (x\mid\lambda )\frac{t^{n}}{n!}. $$
(2.2)
Thus, by (2.2), we easily get
$$\begin{aligned} & \sum_{n=0}^{\infty}\lim_{\lambda\rightarrow0} \mathcal{E}_{n,q}^{ (r )} (x\mid\lambda )\frac{t^{n}}{n!} \\ &\quad= \lim_{\lambda\rightarrow0} \biggl(\frac{ [2 ]_{q}}{q (1+\lambda t )^{1/\lambda}+1} \biggr)^{r} (1+\lambda t )^{\frac{x}{\lambda}} \\ &\quad= \biggl(\frac{ [2 ]_{q}}{qe^{t}+1} \biggr)^{r}e^{xt} \\ &\quad= \sum_{n=0}^{\infty}E_{n,q}^{ (r )} (x )\frac {t^{n}}{n!}, \end{aligned}$$
(2.3)
where \(E_{n,q}^{ (r )} (x )\) are called the higher-order q-Euler polynomials (see [1517]). Thus, by (2.3), we get
$$ \lim_{\lambda\rightarrow0}\mathcal{E}_{n,q}^{ (r )} (x\mid \lambda )=E_{n,q}^{ (r )} (x )\quad (n\geq 0 ). $$
For \(\lambda\in\mathbb{C}_{p}\) with \(\lambda\neq1\), the Frobenius-Euler polynomials of order r are defined by the generating function to be
$$ \biggl(\frac{1-\lambda}{e^{t}-\lambda} \biggr)^{r}e^{xt}=\sum _{n=0}^{\infty}H_{n}^{ (r )} (x\mid \lambda )\frac {t^{n}}{n!} \quad(\mbox{see [3, 18]} ). $$
(2.4)
By replacing λ by \(-q^{-1}\), we get
$$ \biggl(\frac{1+q^{-1}}{e^{t}+q^{-1}} \biggr)^{r}e^{xt}=\sum _{n=0}^{\infty }H_{n}^{ (r )} \bigl(x \mid{-}q^{-1} \bigr)\frac{t^{n}}{n!}. $$
(2.5)
Now, we define the degenerate Frobenius-Euler polynomials of order r as follows:
$$ \biggl(\frac{1-u}{ (1+\lambda t )^{\frac{1}{\lambda}}-u} \biggr)^{r} (1+\lambda t )^{\frac{x}{\lambda}}= \sum_{n=0}^{\infty }h_{n}^{ (r )} (x,u\mid\lambda )\frac {t^{n}}{n!}. $$
(2.6)
From (2.6), we note that
$$\begin{aligned} \sum_{n=0}^{\infty}\lim_{\lambda\rightarrow0}h_{n}^{ (r)} (x,u\mid\lambda )\frac{t^{n}}{n!} &= \lim_{\lambda\rightarrow0} \biggl(\frac{1-u}{ (1+\lambda t )^{\frac{1}{\lambda}}-u} \biggr)^{r} (1+\lambda t )^{\frac {x}{\lambda}} \\ &= \biggl(\frac{1-u}{e^{t}-u} \biggr)^{r}e^{xt} = \sum_{n=0}^{\infty}H_{n} (x\mid u )\frac {t^{n}}{n!}. \end{aligned}$$
(2.7)
Thus, by (2.7), we get
$$\lim_{\lambda\rightarrow0}h_{n}^{ (r )} (x,u\mid\lambda )=H_{n} (x\mid u )\quad (n\ge0 ). $$
By (2.2) and (2.6), we get
$$ \mathcal{E}_{n,q}^{ (r )} (x\mid\lambda )=h_{n}^{ (r )} \bigl(x,-q^{-1}\mid\lambda \bigr)\quad (n\ge0 ). $$
(2.8)
From (2.1) and (2.2), we have
$$\begin{aligned} & \sum_{n=0}^{\infty}\int_{\mathbb{Z}_{p}} \cdots\int_{\mathbb{Z}_{p}} \biggl(\frac{x_{1}+\cdots +x_{r}+x}{\lambda} \biggr)_{n}\,d\mu_{-q} (x_{1} )\cdots \,d\mu _{-q} (x_{r} )\frac{\lambda^{n}t^{n}}{n!} \\ &\quad= \sum_{n=0}^{\infty}\mathcal{E}_{n,q}^{ (r )} (x\mid \lambda )\frac{t^{n}}{n!}. \end{aligned}$$
(2.9)
Now, we define
$$\begin{aligned} &(x\mid\lambda )_{n}= x (x-\lambda )\cdots \bigl(x- (n-1 )\lambda \bigr)\quad (n>0 ),\\ &(x\mid\lambda )_{0}= 1. \end{aligned}$$
(2.10)
By (2.9) and (2.10), we get
$$ \int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (x+x_{1}+\cdots+x_{r}\mid\lambda )_{n}\,d\mu_{-q} (x_{1} )\cdots \,d\mu_{-q} (x_{r} )=\mathcal{E}_{n,q}^{ (r )} (x\mid\lambda )\quad (u\ge0 ). $$
(2.11)

Therefore, by (2.6) and (2.11), we obtain the following theorem.

Theorem 2.1

For \(n\ge0\), we have
$$\begin{aligned} \mathcal{E}_{n,q}^{ (r )} (x\mid\lambda ) & =\int _{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (x_{1}+ \cdots+x_{r}+x\mid\lambda )_{n}\,d\mu _{-q} (x_{1} )\cdots \,d\mu_{-q} (x_{r} ) \\ & =h_{n}^{ (r )} \bigl(x,-q^{-1}\mid\lambda \bigr)\quad (n\ge0 ), \end{aligned}$$
where \(h_{n}^{ (r )} (x,u\mid\lambda )\) are called the degenerate Frobenius-Euler polynomials of order r.
It is not difficult to show that
$$\begin{aligned} & (x_{1}+\cdots+x_{r}+x\mid\lambda )_{n} \\ &\quad= (x_{1}+\cdots+x_{r}+x ) (x_{1}+\cdots+x_{r}+x-\lambda )\cdots \bigl(x_{1}+\cdots +x_{r}+x- (n-1 )\lambda \bigr) \\ &\quad= \lambda^{n} \biggl(\frac{x_{1}+\cdots+x_{r}+x}{\lambda} \biggr)_{n} \\ &\quad= \lambda^{n}\sum_{l=0}^{n}S_{1} (n,l ) \biggl(\frac {x_{1}+\cdots+x_{r}+x}{\lambda} \biggr)^{l} \\ &\quad= \sum_{l=0}^{n}\lambda^{n-l}S_{1} (n,l ) (x_{1}+\cdots +x_{r}+x )^{l}, \end{aligned}$$
(2.12)
where \(S_{1} (n,l )\) is the Stirling number of the first kind.
We observe that
$$ \int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}}e^{ (x_{1}+\cdots+x_{r}+x )t} \,d\mu _{-q} (x_{1} )\cdots \,d\mu_{-q} (x_{r} )= \biggl(\frac { [2 ]_{q}}{qe^{t}+1} \biggr)^{r}e^{xt}. $$
(2.13)
Thus, by (2.13), we get
$$\begin{aligned} & \sum_{n=0}^{\infty}\int_{\mathbb{Z}_{p}} \cdots\int_{\mathbb{Z}_{p}} (x_{1}+\cdots +x_{r}+x )^{n}\,d\mu_{-q} (x_{1} )\cdots \,d\mu_{-q} (x_{r} )\frac{t^{n}}{n!} \\ &\quad= \biggl(\frac{ [2 ]_{q}}{qe^{t}+1} \biggr)^{r}e^{xt} = \sum_{n=0}^{\infty}E_{n,q}^{ (r )} (x )\frac {t^{n}}{n!}. \end{aligned}$$
(2.14)
By comparing the coefficients on both sides of (2.14), we get
$$ E_{n,q}^{ (r )} (x )=\int_{\mathbb{Z}_{p}}\cdots\int _{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{r}+x )^{n}\,d\mu_{-q} (x_{1} )\cdots \,d\mu _{-q} (x_{r} ). $$
(2.15)
From Theorem 2.1, (2.12) and (2.15), we note that
$$\begin{aligned} h_{n}^{ (r )} \bigl(x,-q^{-1}\mid\lambda \bigr) &= \int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{r}+x\mid\lambda )_{n}\,d\mu_{-q} (x_{1} )\cdots \,d\mu_{-q} (x_{r} ) \\ &= \sum_{l=0}^{n}\lambda^{n-l}S_{1} (n,l )\int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{r}+x )^{l}\,d\mu_{-q} (x_{1} )\cdots \,d\mu_{-q} (x_{r} ) \\ &= \sum_{l=0}^{n}\lambda^{n-l}S_{1} (n,l )E_{l,q}^{ (r )} (x ) \\ & = \sum_{l=0}^{n}\lambda^{n-l}S_{1} (n,l )H_{l}^{ (r )} \bigl(x\mid-q^{-1} \bigr). \end{aligned}$$
(2.16)

Therefore, by (2.16), we obtain the following theorem.

Theorem 2.2

For \(n\geq0\), we have
$$h_{n}^{ (r )} \bigl(x,-q^{-1}\mid\lambda \bigr)=\sum _{l=0}^{n}\lambda^{n-l}S_{1} (n,l )H_{l}^{ (r )} \bigl(x\mid-q^{-1} \bigr). $$
In particular,
$$\mathcal{E}_{n,q}^{ (r )} (x\mid\lambda )=\sum _{l=0}^{n}\lambda^{n-l}S_{1} (n,l )E_{l,q}^{ (r )} (x ). $$
By replacing t by \((e^{\lambda t}-1 )/\lambda\) in (2.2), we get
$$\begin{aligned} & \biggl(\frac{ [2 ]_{q}}{qe^{t}+1} \biggr)^{r}e^{xt} \\ &\quad= \sum_{n=0}^{\infty}\mathcal{E}_{n,q}^{ (r )} (x\mid \lambda )\frac{1}{n!}\frac{1}{\lambda^{n}} \bigl(e^{\lambda t}-1 \bigr)^{n} \\ &\quad= \sum_{n=0}^{\infty}\mathcal{E}_{n,q}^{ (r )} (x\mid \lambda )\frac{1}{\lambda^{n}}\sum_{m=n}^{\infty}S_{2} (m,n )\frac{\lambda^{m}}{m!}t^{m} \\ &\quad= \sum_{m=0}^{\infty} \Biggl(\sum _{n=0}^{m}\mathcal{E}_{n,q}^{ (r )} (x\mid\lambda )\lambda^{m-n}S_{2} (m,n ) \Biggr) \frac{t^{m}}{m!}, \end{aligned}$$
(2.17)
where \(S_{2} (m,n )\) is the Stirling number of the second kind.

Thus, by (2.17), we obtain the following theorem.

Theorem 2.3

For \(m\ge0\), we have
$$H_{m}^{ (r )} \bigl(x\mid-q^{-1} \bigr)=\sum _{n=0}^{m}h_{n}^{ (r )} \bigl(x,-q^{-1}\mid\lambda \bigr)\lambda ^{m-n}S_{2} (m,n ). $$
In particular,
$$E_{m,q}^{ (r )} (x )=\sum_{n=0}^{m} \mathcal {E}_{n,q}^{ (r )} (x\mid\lambda )\lambda ^{m-n}S_{2} (m,n ). $$

When \(r=1\), \(\mathcal{E}_{n,q} (x\mid\lambda )=\mathcal {E}_{n,q}^{ (1 )} (x\mid\lambda )\) are called the degenerate q-Euler polynomials. In particular, \(x=0\), \(\mathcal{E}_{n,q} (\lambda )=\mathcal{E}_{n,q} (0\mid \lambda )\) are called the degenerate q-Euler numbers. \(h_{n} (x,u\mid\lambda )=h_{n}^{ (1 )} (x,u\mid\lambda )\) are called the degenerate Frobenius-Euler polynomials. When \(x=0\), \(h_{n} (u\mid\lambda )=h_{n} (0,u\mid\lambda )\) are called the degenerate Frobenius-Euler numbers.

From (1.2), we have
$$\begin{aligned} & \int_{\mathbb{Z}_{p}} (1+\lambda t )^{\frac{x_{1}+x}{\lambda}}\,d\mu _{-q} (x_{1} ) \\ &\quad= \biggl(\frac{ [2 ]_{q}}{q (1+\lambda t )^{\frac {1}{\lambda}}+1} \biggr) (1+\lambda t )^{\frac{x}{\lambda }} \\ &\quad= \biggl(\frac{1+q^{-1}}{ (1+\lambda t )^{\frac{1}{\lambda }}+q^{-1}} \biggr) (1+\lambda t )^{\frac{x}{\lambda}} \\ &\quad= \sum_{n=0}^{\infty}h_{n} \bigl(x,-q^{-1}\mid\lambda \bigr)\frac {t^{n}}{n!}. \end{aligned}$$
(2.18)
Thus, by (2.18), we get
$$\begin{aligned} & h_{n} \bigl(x,-q^{-1}\mid\lambda \bigr) \\ &\quad= \int_{\mathbb{Z}_{p}} (x_{1}+x\mid\lambda )_{n}\,d\mu_{-q} (x_{1} ) \\ &\quad= \lambda^{n}\int_{\mathbb{Z}_{p}} \biggl(\frac{x_{1}+x}{\lambda} \biggr)_{n}\,d\mu _{-q} (x_{1} ) \\ &\quad= \sum_{l=0}^{n}S_{1} (n,l ) \lambda^{n-l}\int_{\mathbb{Z}_{p}} (x_{1}+x )^{l}\,d\mu_{-q} (x_{1} ) \\ &\quad= \sum_{l=0}^{n}S_{1} (n,l ) \lambda^{n-l}H_{l} \bigl(x\mid -q^{-1} \bigr) \end{aligned}$$
(2.19)
and
$$\begin{aligned} & h_{n} \bigl(-q^{-1}\mid\lambda \bigr) \\ &\quad= \int_{\mathbb{Z}_{p}} (x_{1}\mid\lambda )_{n}\,d\mu_{-q} (x_{1} ) \\ &\quad= \lambda^{n}\int_{\mathbb{Z}_{p}} \biggl(\frac{x_{1}}{\lambda} \biggr)_{n}\,d\mu _{-q} (x_{1} ) \\ &\quad= \sum_{l=0}^{n}S_{1} (n,l ) \lambda^{n-l}H_{l} \bigl(-q^{-1} \bigr). \end{aligned}$$
(2.20)
For \(d\in\mathbb{N}\), by (1.3), we get
$$\begin{aligned} & q^{d}\int_{\mathbb{Z}_{p}} (x_{1}+d\mid\lambda )_{n}\,d\mu_{-q} (x_{1} )+ (-1 )^{d-1}\int _{\mathbb{Z}_{p}} (x_{1}\mid\lambda )_{n}\,d\mu_{-q} (x_{1} ) \\ &\quad= [2 ]_{q}\sum_{l=0}^{d-1} (-1 )^{d-1-l}q^{l} (l\mid\lambda )_{n}. \end{aligned}$$
(2.21)
Let \(d\equiv1\ (\operatorname{mod}{2})\). Then we have
$$ [2 ]_{q}\sum_{l=0}^{d-1} (-1 )^{l}q^{l} (l\mid \lambda )_{n}=q^{d}h_{n} \bigl(d,-q^{-1}\mid\lambda \bigr)+h_{n} \bigl(-q^{-1} \mid\lambda \bigr). $$
(2.22)
For \(d\in\mathbb{N}\) with \(d\equiv0\ (\operatorname{mod}{2})\), we get
$$ [2 ]_{q}\sum_{l=0}^{d-1} (-1 )^{l-1}q^{l} (l\mid \lambda )_{n}=q^{d}h_{n} \bigl(d,-q^{-1}\mid\lambda \bigr)-h_{n} \bigl(-q^{-1} \mid\lambda \bigr). $$
(2.23)

Therefore, by (2.22) and (2.23), we obtain the following theorem.

Theorem 2.4

Let \(d\in\mathbb{N}\) and \(n\geq0\).
  1. (i)
    For \(d\equiv1\ (\operatorname{mod}{2})\), we have
    $$q^{d}h_{n} \bigl(d,-q^{-1}\mid\lambda \bigr)+h_{n} \bigl(-q^{-1}\mid \lambda \bigr)= [2 ]_{q}\sum_{l=0}^{d-1} (-1 )^{l}q^{l} (l\mid\lambda )_{n}. $$
     
  2. (ii)
    For \(d\equiv0\ (\operatorname{mod}{2})\), we have
    $$q^{d}h_{n} \bigl(d,-q^{-1}\mid\lambda \bigr)-h_{n} \bigl(-q^{-1}\mid \lambda \bigr)= [2 ]_{q}\sum_{l=0}^{d-1} (-1 )^{l-1}q^{l} (l\mid\lambda )_{n}. $$
     

Corollary 2.5

Let \(d\in\mathbb{N}\) and \(n\geq0\).
  1. (i)
    For \(d\equiv1\ (\operatorname{mod}{2})\), we have
    $$q^{d}E_{n,q} (d\mid\lambda )+E_{n,q} (\lambda )= [2 ]_{q}\sum_{l=0}^{d-1} (-1 )^{l}q^{l} (l\mid\lambda )_{n}. $$
     
  2. (ii)
    For \(d\equiv0\ (\operatorname{mod}{2})\), we have
    $$q^{d}E_{n,q} (d\mid\lambda )-E_{n,q} (\lambda )= [2 ]_{q}\sum_{l=0}^{d-1} (-1 )^{l-1}q^{l} (l\mid \lambda )_{n}. $$
     
From (1.1), we note that
$$ \int_{\mathbb{Z}_{p}}f (x )\,d\mu_{-q} (x )=\frac{ [2 ]_{q}}{ [2 ]_{q^{d}}} \sum_{l=0}^{d-1} (-q )^{a}\int _{\mathbb{Z}_{p}}f (a+dx )\,d\mu_{-q^{d}} (x ), $$
(2.24)
where \(d\in\mathbb{N}\) with \(d\equiv1\ (\operatorname{mod}{2})\).
By (2.24), we get
$$\begin{aligned} & \int_{\mathbb{Z}_{p}} (x_{1}\mid\lambda )_{n}\,d\mu_{-q} (x_{1} ) \\ &\quad= \frac{ [2 ]_{q}}{ [2 ]_{q^{d}}}\sum_{a=0}^{d-1} (-q )^{a}\int_{\mathbb{Z}_{p}} (a+dx_{1}\mid\lambda )_{n}\,d\mu_{-q^{d}} (x_{1} ) \\ &\quad= \frac{ [2 ]_{q}}{ [2 ]_{q^{d}}}d^{n}\sum_{a=0}^{d-1} (-q )^{a}\int_{\mathbb{Z}_{p}} \biggl(\frac {a}{d}+x_{1}\Bigm| \frac{\lambda}{d} \biggr)_{n}\,d\mu_{-q^{d}} (x_{1} ) \\ &\quad= d^{n}\frac{ [2 ]_{q}}{ [2 ]_{q^{d}}}\sum_{a=0}^{d-1} (-q )^{a}\mathcal{E}_{n,q^{d}} \biggl(\frac {a}{d} \Bigm| \frac{\lambda}{d} \biggr), \end{aligned}$$
(2.25)
where \(d\in\mathbb{N}\) with \(d\equiv1\ (\operatorname{mod}{2})\) and \(n\geq0\).

Therefore, by (2.25), we obtain the following theorem.

Theorem 2.6

For \(n\geq0\), \(d\in\mathbb{N}\) with \(d\equiv1\ (\operatorname{mod}{2})\), we have
$$\mathcal{E}_{n,q} (\lambda )=d^{n}\frac{ [2 ]_{q}}{ [2 ]_{q^{d}}}\sum _{a=0}^{d-1} (-q )^{a} \mathcal{E}_{n,q^{d}} \biggl(\frac{a}{d}\Bigm| \frac{\lambda }{d} \biggr). $$
Moreover,
$$\mathcal{E}_{n,q} (x\mid\lambda )=d^{n}\frac{ [2 ]_{q}}{ [2 ]_{q^{d}}}\sum _{a=0}^{d-1} (-q )^{a} \mathcal{E}_{n,q^{d}} \biggl(\frac{a+x}{d}\Bigm| \frac{\lambda }{d} \biggr). $$
Now, we consider the degenerate q-Euler polynomials of the second kind as follows:
$$ \widehat{\mathcal{E}}_{n,q} (x\mid\lambda )=\int_{\mathbb{Z}_{p}} \bigl(- (x_{1}+x )\mid \lambda \bigr)_{n}\,d\mu_{-q} (x_{1} )\quad (n\ge0 ). $$
(2.26)
From (2.26), we note that
$$\begin{aligned} & \sum_{n=0}^{\infty}\hat{\mathcal{E}}_{n,q} (x\mid\lambda )\frac{t^{n}}{n!} \\ &\quad= \sum_{n=0}^{\infty}\lambda^{n} \int_{\mathbb{Z}_{p}}\binom{-\frac {x_{1}+x}{\lambda}}{n}\,d\mu_{-q} (x_{1} )t^{n} \\ &\quad= (1+\lambda t )^{-x/\lambda}\int_{\mathbb{Z}_{p}} (1+\lambda t )^{-x_{1}/\lambda}\,d\mu_{-q} (x_{1} ) \\ &\quad= \frac{ [2 ]_{q}}{ (1+\lambda t )^{1/\lambda }+q} (1+\lambda t )^{ (1-x )/\lambda}. \end{aligned}$$
(2.27)

When \(x=0\), \(\hat{\mathcal{E}}_{n,q} (\lambda )=\hat{\mathcal {E}}_{n,q} (0\mid\lambda )\) are called the degenerate q-Euler numbers of the second kind.

By (2.26), we get
$$\begin{aligned} & \hat{\mathcal{E}}_{n,q} (x\mid\lambda ) \\ &\quad= \lambda^{n}\int_{\mathbb{Z}_{p}} \biggl(- \frac{x_{1}+x}{\lambda} \biggr)_{n}\,d\mu _{-q} (x ) \\ &\quad= \lambda^{n}\sum_{l=0}^{n}S_{1} (n,l )\frac{ (-1 )^{l}}{\lambda^{l}}\int_{\mathbb{Z}_{p}} (x_{1}+x )^{l}\,d\mu_{-q} (x ) \\ &\quad= \sum_{l=0}^{n}S_{1} (n,l ) \lambda^{n-l} (-1 )^{l}E_{l,q} (x ). \end{aligned}$$
(2.28)
Thus, from (2.28), we have
$$\begin{aligned} & (-1 )^{n}\hat{\mathcal{E}}_{n,q} (x\mid\lambda) \\ &\quad= \sum_{l=0}^{n} (-1 )^{n-l}S_{1} (n,l )\lambda ^{n-l}E_{l,q} (x ) \\ &\quad= \sum_{l=0}^{n}\bigl\vert S_{1} (n,l )\bigr\vert \lambda ^{n-l}E_{l,q} (x ). \end{aligned}$$
(2.29)
We observe that
$$\begin{aligned} & \sum_{n=0}^{\infty}E_{n,q^{-1}} (1-x ) \frac{t^{n}}{n!} \\ &\quad= \frac{1+q^{-1}}{q^{-1}e^{t}+1}e^{ (1-x )t}=\frac {1+q}{qe^{-t}+1}e^{-xt} \\ &\quad= \frac{ [2 ]_{q}}{qe^{-t}+1}e^{-xt}=\sum_{n=0}^{\infty} (-1 )^{n}E_{n,q} (x )\frac{t^{n}}{n!}. \end{aligned}$$
(2.30)
From (2.30), we have
$$ E_{n,q^{-1}} (1-x )= (-1 )^{n}E_{n,q} (x )\quad (n\ge0 ). $$
(2.31)
By replacing t by \(\frac{e^{\lambda t}-1}{\lambda}\) in (2.27), we get
$$\begin{aligned} & \sum_{n=0}^{\infty}\hat{\mathcal{E}}_{n,q} (x\mid\lambda )\frac{1}{n!}\frac{1}{\lambda^{n}} \bigl(e^{\lambda t}-1 \bigr)^{n} \\ &\quad= \frac{1+q}{e^{t}+q}e^{ (1-x )t} \\ &\quad= \frac{ [2 ]_{q^{-1}}}{q^{-1}e^{t}+1}e^{ (1-x )t} \\ &\quad= \sum_{n=0}^{\infty}E_{n,q^{-1}} (1-x ) \frac{t^{n}}{n!}. \end{aligned}$$
(2.32)
On the other hand, we have
$$\begin{aligned} & \sum_{m=0}^{\infty}\hat{\mathcal{E}}_{m,q} (x\mid\lambda )\frac{1}{m!}\frac{1}{\lambda^{m}} \bigl(e^{\lambda t}-1 \bigr)^{m} \\ &\quad= \sum_{m=0}^{\infty}\hat{ \mathcal{E}}_{m,q} (x\mid\lambda )\frac{1}{\lambda^{m}}\sum _{n=m}^{\infty}S_{2} (n,m )\frac {\lambda^{n}t^{n}}{n!} \\ &\quad= \sum_{n=0}^{\infty} \Biggl(\sum _{m=0}^{n}\hat{\mathcal{E}}_{m,q} (x\mid \lambda )S_{2} (m,n )\lambda^{n-m} \Biggr)\frac {t^{n}}{n!}. \end{aligned}$$
(2.33)
From (2.32) and (2.33), we note that
$$ (-1 )^{n}E_{n,q^{-1}} (x )=\sum_{m=0}^{n} \hat {\mathcal{E}}_{m,q} (x\mid\lambda )S_{2} (n,m )\lambda ^{n-m}. $$
(2.34)

Therefore, by (2.29) and (2.34), we obtain the following theorem.

Theorem 2.7

For \(n\geq0\), we have
$$(-1 )^{n}\hat{\mathcal{E}}_{n,q} (x\mid\lambda )=\sum _{l=0}^{n}\bigl\vert S_{1} (n,l )\bigr\vert \lambda^{n-l}E_{l,q} (x ) $$
and
$$(-1 )^{n}E_{n,q^{-1}} (x )=\sum_{l=0}^{n}S_{2} (n,l )\lambda^{n-l}\hat{\mathcal{E}}_{l,q} (x\mid\lambda ). $$
It is easy to show that
$$ \binom{x+y}{n}=\sum_{l=0}^{n} \binom{x}{l}\binom{y}{n-l}\quad(n\geq 0 ). $$
(2.35)
From (2.35), we have
$$\begin{aligned} & \frac{ (-1 )^{n}\mathcal{E}_{n,q} (\lambda )}{n!} \\ &\quad= \frac{ (-1 )^{n}}{n!}\int_{\mathbb{Z}_{p}} (x_{1}\mid\lambda )_{n}\,d\mu_{-q} (x_{1} ) \\ &\quad= \lambda^{n}\int_{\mathbb{Z}_{p}}\binom{-\frac{x_{1}}{\lambda}+n-1}{n}\,d\mu _{-q} (x_{1} ) \\ &\quad= \lambda^{n}\sum_{l=0}^{n} \binom{n-1}{n-l}\int_{\mathbb{Z}_{p}}\binom{-\frac {x_{1}}{\lambda}}{l}\,d\mu_{-q} (x_{1} ) \\ &\quad= \lambda^{n}\sum_{l=1}^{n} \binom{n-1}{l-1}\frac{1}{\lambda^{l}l!}\int_{\mathbb{Z}_{p}} (-x_{1}\mid\lambda )_{l}\,d\mu_{-q} (x_{1} ) \\ &\quad= \sum_{l=1}^{n}\binom{n-1}{l-1} \lambda^{n-l}\frac{1}{l!}\hat{\mathcal {E}}_{l,q} (\lambda) \end{aligned}$$
(2.36)
and
$$ \frac{ (-1 )^{n}}{n!}\hat{\mathcal{E}}_{n,q} (\lambda )=\sum _{l=1}^{n}\binom{n-1}{l-1}\lambda^{n-l} \frac{1}{l!}\mathcal {E}_{l,q} (\lambda ). $$
(2.37)

Declarations

Acknowledgements

This paper is supported by Grant No. 14-11-00022 of Russian Scientific Fund.

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

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