# Degenerate q-Euler polynomials

## 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 $$\mathbb{Z}_{p}$$.

## 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 ).

The ordinary fermionic p-adic integral on $$\mathbb{Z}_{p}$$ is defined as

$$\lim_{q\rightarrow1}I_{-q} (f )=I_{-1} (f )=\int _{\mathbb{Z}_{p}}f (x )\,d\mu_{-1} (x )=\lim_{N\rightarrow\infty} \sum_{x=0}^{p^{N}-1}f (x ) (-1 )^{x} \quad(\mbox{see }).$$
(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 , 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 $$\mathbb{Z}_{p}$$.

## 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 ). 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)

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## Acknowledgements

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

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