Theory and Modern Applications

# Some identities of Barnes-type special polynomials

## Abstract

In this paper, we consider Barnes-type special polynomials and give some identities of their polynomials which are derived from the bosonic p-adic integral or the fermionic p-adic integral on $$\mathbb{Z}_{p}$$.

## 1 Introduction

As is known, the Bernoulli polynomials of order r are defined by the generating function to be

$${\biggl(\frac{t}{e^{t}-1}\biggr) ^{r}}e^{xt} = \sum_{n=0}^{\infty }B_{n}^{(r)}(x) \frac{t^{n}}{n!} \quad (\mbox{see [1--29]}).$$
(1)

When $$x=0$$, $$B_{n}^{(r)}=B_{n}^{(r)}(0)$$ are called the Bernoulli numbers of order r.

For $$a_{1},a_{2},\ldots,a_{r}\neq0\in\mathbb{C}_{p}$$, the Barnes-Bernoulli polynomials are defined by the generating function to be

$$\prod_{i=1}^{r}\biggl(\frac{t}{e^{a_{i}t}-1} \biggr)e^{xt}=\sum_{n=0}^{\infty}B_{n}(x| a_{1},a_{2},\ldots,a_{r})\frac{t^{n}}{n!}.$$

When $$x=0$$, $$B_{n}(0|a_{1},a_{2},\ldots,a_{r})=B_{n}(a_{1},a_{2},\ldots,a_{r})$$ are called Barnes Bernoulli numbers (see ).

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 numbers and the completion of algebraic closure of $$\mathbb{Q}_{p}$$, respectively. The p-adic norm is defined as $$|p|_{p}=1/p$$. Let $$UD(\mathbb{Z}_{p})$$ be the space of uniformly differentiable function on $$\mathbb{Z}_{p}$$. For $$f\in UD(\mathbb{Z}_{p})$$, the bosonic p-adic integral on $$\mathbb{Z}_{p}$$ is defined by

\begin{aligned}[b] I_{0}(f)&=\int_{\mathbb{Z}_{p}}f(x)\,d\mu_{0}(x)=\lim_{N\to\infty} \sum_{x=0}^{p^{N}-1}f(x)\mu_{0} \bigl(x+p^{N}\mathbb{Z}_{p}\bigr)\\ &=\lim_{N\to\infty }\frac{1}{p^{N}}\sum _{x=0}^{p^{N}-1}f(x)\quad (\mbox{see [16, 21]}). \end{aligned}
(2)

From (2), we have

$$I_{0}(f_{1})=I_{0}(f)+f'(0),$$
(3)

where $$f_{1}(x)=f(x+1)$$.

By using iterative method, we get

$$I_{0}(f_{n})=I_{0}(f)+\sum _{i=0}^{n-1}f'(i),$$
(4)

where $$f_{n}(x)=f(x+n)$$ ($$n\in\mathbb{N}$$).

As is well known, the fermionic p-adic integral on $$\mathbb{Z}_{p}$$ is defined by Kim to be

\begin{aligned}[b] I_{-1}(f)&=\int _{\mathbb{Z}_{p}}f(x)\,d\mu_{-1}(x)=\lim_{N\to\infty} \sum_{x=0}^{p^{N}-1}f(x)\mu_{-1} \bigl(x+p^{N}\mathbb{Z}_{p}\bigr)\\ &=\lim_{N\to \infty}\sum_{x=0}^{p^{N}-1}f(x) (-1)^{x} \quad(\mbox{see [26, 27]}). \end{aligned}
(5)

From (5), we can derive

$$I_{-1}(f_{n})+(-1)^{n-1}I_{-1}(f)=2 \sum_{l=0}^{n-1}(-1)^{n-l-1}f(l).$$
(6)

In particular, $$n=1$$, we have

$$I_{-1}(f_{1})+I_{-1}(f)=2f(0) \quad(\mbox{see [26, 27]}).$$
(7)

The purpose of this paper is to investigate several special polynomials related to Barnes-type polynomials and give some identities including Witt’s formula of their polynomials.

Finally, we give some identities of mixed-type Bernoulli and Euler polynomials.

## 2 Barnes-type polynomials

Let $$a_{1},a_{2},\ldots,a_{r}\neq0\in\mathbb{C}_{p}$$. Then, by (3), we get

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb {Z}_{p}}e^{(a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{r}x_{r}+x)t}\,d\mu_{0}(x_{1})\cdots \,d \mu_{0}(x_{r}) \\ &\quad=\Biggl(\prod_{i=1}^{r}a_{i} \Biggr) \biggl(\frac {t^{r}}{(e^{a_{1}t}-1)(e^{a_{2}t}-1)\cdots(e^{a_{r}t}-1)}\biggr)e^{xt}\\ &\quad=\Biggl(\prod_{i=1}^{r}a_{i} \Biggr)\sum_{n=0}^{\infty}B_{n}(x|a_{1},a_{2},\ldots,a_{r})\frac{t^{n}}{n!}. \end{aligned}
(8)

From (8), we obtain the following Witt’s formula for the Barnes-Bernoulli polynomials.

### Theorem 1

For $$a_{1},a_{2},\ldots,a_{r}\neq0\in\mathbb{C}_{p}$$, we have

$$B_{n}(x| a_{1},\ldots,a_{r}) =\Biggl(\prod _{i=1}^{r}a_{i} \Biggr)^{-1}\int_{\mathbb {Z}_{p}}\cdots\int_{\mathbb{Z}_{p}}(a_{1}x_{1}+ \cdots+a_{r}x_{r}+x)^{n} \,d \mu _{0}(x_{1}) \cdots \,d \mu_{0}(x_{r}).$$

Note that

\begin{aligned}[b] (a_{1}x_{1}+ \cdots+a_{r}x_{r})^{n}&=\sum _{l_{1}+\cdots+l_{r}=n} \binom {n}{l_{1},\ldots,l_{r}}a_{1}^{l_{1}}x_{1}^{l_{1}} \cdots a_{r}^{l_{r}}x_{r}^{l_{r}}\\ &=\sum_{l_{1}+\cdots+l_{r}=n}\binom{n}{l_{1},\ldots,l_{r}}\Biggl( \prod_{i=1}^{r}a_{i}^{l_{i}} \Biggr)x_{1}^{l_{1}}\cdots x_{r}^{l_{r}}. \end{aligned}
(9)

By (9) and Theorem 1, we obtain the following corollary.

### Corollary 2

For $$n\geq2$$, we have

$$B_{n}(a_{1},\ldots,a_{r}) =\sum _{l_{1}+\cdots+l_{r}=n} \binom {n}{l_{1},\ldots,l_{r}}\Biggl(\prod _{i=1}^{r}a_{i}^{l_{i}-1} \Biggr)B_{l_{1}}\cdots B_{l_{r}},$$

where $$B_{n}=B_{n}(1)$$ is the nth Bernoulli number.

From (2), we can easily derive the following integral equation:

$$\int_{\mathbb{Z}_{p}}f(x)\,d\mu_{0}(x)= \frac{1}{d} \sum_{a=0}^{d-1}\int _{\mathbb{Z}_{p}}f(a+dx)\,d\mu_{0}(x),$$
(10)

where $$d\in\mathbb{N}$$.

By (10), we get

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb {Z}_{p}}e^{(a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{r}x_{r}+x)t}\,d\mu_{0}(x) \\ &\quad=\frac{1}{d^{r}} \sum_{l_{1}=0}^{d-1}\cdots \sum_{l_{r}=0}^{d-1}\int_{\mathbb{Z}_{p}} \cdots\int_{\mathbb {Z}_{p}}e^{(l_{1}a_{1}+\cdots+l_{r}a_{r}+a_{1}\,dx_{1}+\cdots+a_{r}\,dx_{r}+x)t}\,d\mu _{0}(x_{1}) \cdots d \mu_{0}(x_{r}) \\ &\quad=\sum_{l_{1}=0}^{d-1}\cdots\sum _{l_{r}=0}^{d-1}\sum_{n=0}^{\infty} \frac{d^{n}}{d^{r}}\int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb {Z}_{p}} \biggl(\frac{l_{1}a_{1}+\cdots+l_{r}a_{r}}{d}\\ &\qquad{}+a_{1}x_{1}+\cdots+a_{r}x_{r}+ \frac {x}{d}\biggr)^{n}\,d\mu_{0}(x_{1})\cdots\,d \mu_{0}(x_{r}) \frac{t^{n}}{n!}. \end{aligned}
(11)

By Theorem 1 and (11), we get

$$B_{n}(x| a_{1},\ldots,a_{r})=d^{n-r} \sum_{l_{1}=0}^{d-1}\cdots\sum _{l_{r}=0}^{d-1}B_{n}\biggl( \frac{l_{1}a_{1}+\cdots+l_{r}a_{r}+x}{d}\Big|a_{1},\ldots,a_{r}\biggr).$$
(12)

Therefore, by (12), we obtain the following distribution relation for a Barnes-type Bernoulli polynomial.

### Theorem 3

For $$n\geq0$$, we have

$$B_{n}(x| a_{1},\ldots,a_{r}) =d^{n-r} \sum_{l_{1}=0}^{d-1}\cdots\sum _{l_{r}=0}^{d-1}B_{n}\biggl( \frac{l_{1}a_{1}+\cdots+l_{r}a_{r}+x}{d}\Big|a_{1},\ldots,a_{r}\biggr).$$

From (4), we note that

$$\int_{\mathbb{Z}_{p}}e^{a_{1}(x_{1}+n)t}\,d\mu_{0}(x_{1})-\int_{\mathbb {Z}_{p}}e^{a_{1}x_{1}t}\,d\mu_{0}(x_{1})=a_{1}t\sum _{l=0}^{n-1}e^{a_{1}lt}.$$
(13)

By (13), we get

$$\int_{\mathbb{Z}_{p}}e^{a_{1}x_{1}t}\,d\mu_{0}(x_{1})=\frac {a_{1}t}{e^{a_{1}nt}-1}\sum _{l=0}^{n-1}e^{a_{1}lt}.$$
(14)

From (14), we can derive

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} e^{(a_{1}x_{1}+\cdots +a_{r}x_{r})t}\,d\mu_{0}(x_{1})\cdots \,d\mu_{0}(x_{r}) \\ &\quad=\biggl(\frac{a_{1}t}{e^{na_{1}t}-1}\biggr)\cdots \biggl(\frac{a_{r}t}{e^{na_{r}t}-1}\biggr)\sum _{l_{1},\ldots,l_{r}=0}^{n-1}e^{(a_{1}l_{1}+\cdots+a_{r}l_{r})t}\\ &\quad=\Biggl(\prod_{i=1}^{r}a_{i} \Biggr)\sum_{l_{1},\ldots,l_{r}=0}^{n-1}\Biggl(\sum _{k=0}^{\infty}B_{k}(na_{1}, \ldots,na_{r})\frac{t^{k}}{k!}\Biggr)\sum _{j=0}^{\infty}(a_{1}l_{1}+ \cdots+a_{r}l_{r})^{j}\frac{t^{j}}{j!} \\ &\quad=\Biggl(\prod_{i=1}^{r}a_{i} \Biggr)\sum_{m=0}^{\infty}\Biggl\{ \sum _{l_{1},\ldots,l_{r}=0}^{n-1}\sum_{j=0}^{m}(a_{1}l_{1}+ \cdots +a_{r}l_{r})^{j}B_{m-j}(na_{1}, \ldots,na_{r})\binom{m}{j}\Biggr\} \frac{t^{m}}{m!}. \end{aligned}
(15)

By (15), we get

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} (a_{1}x_{1}+\cdots +a_{r}x_{r})^{m} \,d\mu_{0}(x_{1}) \cdots \,d\mu_{0}(x_{r}) \\ &\quad=\Biggl(\prod_{i=1}^{r}a_{i} \Biggr)\sum_{l_{1},\ldots,l_{r}=0}^{n-1}\sum _{j=0}^{m}(a_{1}l_{1}+ \cdots+a_{r}l_{r})^{j}B_{m-j}(na_{1}, \ldots,na_{r})\binom{m}{j}, \end{aligned}
(16)

where $$n\in\mathbb{N}$$ and $$m\in\mathbb{Z}\geq0$$.

Therefore, by Theorem 1 and (16), we obtain the following theorem.

### Theorem 4

For $$n\in\mathbb{N}$$ and $$m\in\mathbb{Z}$$ with $$m\geq0$$, we have

$$B_{m}(a_{1},\ldots, a_{r})=\sum _{l_{1},\ldots,l_{r}=0}^{n-1}\sum_{j=0}^{m}(a_{1}l_{1}+ \cdots+a_{r}l_{r})^{j}B_{m-j}(na_{1}, \ldots,na_{r})\binom{m}{j}.$$

Moreover,

$$B_{m}(x|a_{1},\ldots, a_{r})=\sum _{l_{1},\ldots,l_{r}=0}^{n-1}\sum_{j=0}^{m}(a_{1}l_{1}+ \cdots+a_{r}l_{r}+x)^{j}B_{m-j}(na_{1}, \ldots ,na_{r})\binom{m}{j}.$$

From (15), we observe that

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} e^{(a_{1}x_{1}+\cdots +a_{r}x_{r})t}\,d\mu_{0}(x_{1}) \cdots \,d\mu_{0}(x_{r}) \\ &\quad=\biggl(\frac{a_{1}t}{e^{na_{1}t}-1}\biggr)\cdots\biggl(\frac{a_{r}t}{e^{na_{r}t}-1}\biggr)\sum _{l_{1},\ldots,l_{r}=0}^{n-1}e^{(a_{1}l_{1}+\cdots+a_{r}l_{r})t} \\ &\quad=\sum_{l_{1},\ldots,l_{r}=0}^{n-1}\frac{a_{1}t\cdots {a_{r}t}}{(e^{na_{1}t}-1)\cdots(e^{na_{r}t}-1)}e^{(\frac{a_{1}l_{1}+\cdots +a_{r}l_{r}}{n})nt} \\ &\quad=\Biggl(\prod_{i=1}^{r}a_{i} \Biggr)\sum_{l_{1},\ldots,l_{r}=0}^{n-1}\sum _{m=0}^{\infty}B_{m}\biggl(\frac{a_{1}l_{1}+\cdots+a_{r}l_{r}}{n}\Big|a_{1}, \ldots ,a_{r}\biggr)n^{m}\frac{t^{m}}{m!} \\ &\quad=\sum_{m=0}^{\infty}\Biggl(\prod _{i=1}^{r}a_{i}\Biggr)n^{m}\sum _{l_{1},\ldots,l_{r}=0}^{n-1}B_{m}\biggl( \frac{a_{1}l_{1}+\cdots+a_{r}l_{r}}{n}\Big|a_{1},\ldots ,a_{r}\biggr) \frac{t^{m}}{m!}. \end{aligned}
(17)

Thus, by (17), we get

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} (a_{1}x_{1}+\cdots +a_{r}x_{r})^{m} \,d\mu_{0}(x_{1}) \cdots \,d\mu_{0}(x_{r}) \\ &\quad=\Biggl(\prod_{i=1}^{r}a_{i} \Biggr)n^{m}\sum_{l_{1},\ldots ,l_{r}=0}^{n-1}B_{m} \biggl(\frac{a_{1}l_{1}+\cdots+a_{r}l_{r}}{n}\Big|a_{1},\ldots,a_{r}\biggr), \end{aligned}
(18)

where $$n\in\mathbb{N}$$ and $$m\in\mathbb{Z}\geq0$$.

Therefore, by Theorem 1 and (17), we obtain the following theorem.

### Theorem 5

For $$n\in\mathbb{N}$$ and $$m\geq0$$, we have

$$B_{m}(a_{1},\ldots, a_{r})=n^{m}\sum _{l_{1},\ldots,l_{r}=0}^{n-1}B_{m}\biggl( \frac {a_{1}l_{1}+\cdots+a_{r}l_{r}}{n}\Big|a_{1},\ldots,a_{r}\biggr).$$

Moreover,

$$B_{m}(x|a_{1},\ldots, a_{r})=n^{m}\sum _{l_{1},\ldots ,l_{r}=0}^{n-1}B_{m}\biggl( \frac{a_{1}l_{1}+\cdots+a_{r}l_{r}+x}{n}\Big|a_{1},\ldots,a_{r}\biggr).$$

### Remark

Let $$a_{1}=1$$ and $$r=1$$. Then we have

$$\int_{\mathbb{Z}_{p}}e^{(x+n)t}\,d\mu_{0}(x)-\int _{\mathbb{Z}_{p}}e^{xt}\,d\mu_{0}(x)=t\sum _{l=0}^{n-1}e^{lt}.$$

Thus, we have

$$\sum_{m=0}^{\infty}\biggl\{ \int _{\mathbb{Z}_{p}}(x+n)^{m}\,d\mu _{0}(x)-\int _{\mathbb{Z}_{p}}x^{m}\,d\mu_{0}(x)\biggr\} \frac{t^{m}}{m!}(x)=t\sum_{l=0}^{n-1}\sum _{m=0}^{\infty}l^{m}\frac{t^{m}}{m!}.$$
(19)

By (19), we get

$$\frac{1}{m+1}\biggl\{ \int_{\mathbb{Z}_{p}}(x+n)^{m+1}\,d\mu_{0}(x)-\int_{\mathbb {Z}_{p}}x^{m+1}\,d\mu_{0}(x)\biggr\} =\sum_{l=0}^{n-1}l^{m},$$
(20)

where $$n\in\mathbb{N}$$ and $$m\in\mathbb{Z}\geq0$$.

It is easy to show that

$$\int_{\mathbb{Z}_{p}}e^{(x+y)t}\,d\mu_{0}(y) =\frac{t}{e^{t}-1}e^{xt}=\sum _{n=0}^{\infty}B_{n}(x)\frac{t^{n}}{n!},$$
(21)

where $$B_{n}(x)$$ is the nth Bernoulli polynomial.

Thus, by (21), we get

$$\int_{\mathbb{Z}_{p}}(x+y)^{n} d \mu_{0}(y)=B_{n}(x)\quad (n\geq0).$$
(22)

From (20) and (21), we note that

$$\frac{1}{m+1}\bigl\{ B_{m+1}(n)-B_{m+1}\bigr\} =\sum _{l=0}^{n-1}l^{m},$$

where $$m\in\mathbb{Z}\geq0$$ and $$n\in\mathbb{N}$$.

From (5) and (6), we can derive the following equation:

$$\int_{\mathbb{Z}_{p}}e^{(x+1)t}\,d\mu_{-1}(x)+\int _{\mathbb{Z}_{p}}e^{xt}\,d\mu_{-1}(x)=2.$$

Thus, we have

$$\int_{\mathbb{Z}_{p}}e^{(x+y)t}\,d\mu_{-1}(y)= \frac{2}{e^{t}+1}e^{xt}=\sum_{n=0}^{\infty}E_{n}(x) \frac{t^{n}}{n!},$$

where $$E_{n}(x)$$ is the nth Euler polynomial.

Witt’s formula for the Euler polynomials is given by

$$\int_{\mathbb{Z}_{p}}(x+y)^{n} d \mu_{-1}(y)=E_{n}(x)\quad (n\geq0)\ (\mbox{see [26, 27]}).$$
(23)

When $$x=0$$, $$E_{n}=E_{n}(0)$$ are called the Euler numbers.

For $$r\in\mathbb{N}$$, the generating function of higher-order Euler polynomials can be derived from the multivariate p-adic fermionic integral on $$\mathbb{Z}_{p}$$ as follows:

\begin{aligned}[b] \int_{\mathbb{Z}_{p}}\cdots\int _{\mathbb{Z}_{p}}e^{(x_{1}+\cdots+x_{r}+x)t}\,d\mu_{-1}(x_{1}) \cdots \,d\mu_{-1}(x_{r})&=\biggl(\frac{2}{e^{t}+1} \biggr)^{r}e^{xt} \\ &=\sum_{n=0}^{\infty}E^{(r)}_{n}(x) \frac{t^{n}}{n!}. \end{aligned}

Thus we get

$$\int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}}(x_{1}+ \cdots+x_{r}+x)^{n} \,d\mu _{-1}(x_{1}) \cdots \,d\mu_{-1}(x_{r})=E^{(r)}_{n}(x)\quad (n \in\mathbb{Z}\geq0).$$
(24)

It is easy to show that

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}}(x_{1}+\cdots+x_{r}+x)^{n}\,d\mu _{-1}(x_{1})\cdots \,d\mu_{-1}(x_{r}) \\ &\quad=\sum_{l_{1}+\cdots+l_{r}=n}\binom{n}{l_{1},\ldots,l_{r}}\int _{\mathbb {Z}_{p}}x^{l_{1}}_{1} \,d\mu_{-1}(x_{1}) \cdots\int_{\mathbb {Z}_{p}}x^{l_{r-1}}_{r-1}\,d\mu_{-1}(x_{r-1}) \\ &\qquad{}\times\int_{\mathbb{Z}_{p}}(x_{r}+x)^{l_{r}}\,d\mu_{-1}(x_{r}) \\ &\quad=\sum_{l_{1}+\cdots+l_{r}=n}\binom{n}{l_{1},\ldots ,l_{r}}E_{l_{1}}E_{l_{2}} \cdots E_{l_{r-1}}E_{l_{r}}(x). \end{aligned}
(25)

From (24) and (25), we have

$$E^{(r)}_{n}(x)=\sum _{l_{1}+\cdots+l_{r}=n}\binom{n}{l_{1},\ldots ,l_{r}}E_{l_{1}}\cdots E_{l_{r-1}}E_{l_{r}}(x).$$
(26)

When $$x=0$$, $$E^{(r)}_{n}=E^{(r)}_{n}(0)$$ are called the higher-order Euler numbers.

From (6), we note that

$$\int_{\mathbb{Z}_{p}}e^{(x+n)t}\,d\mu_{-1}(x)+(-1)^{n-1}\int_{\mathbb {Z}_{p}}e^{xt}\,d\mu_{-1}(x)=2\sum_{l=0}^{n-1}(-1)^{n-1-l}e^{lt}\quad (n\in\mathbb{N}).$$
(27)

Thus, by (27), we get

$$\int_{\mathbb{Z}_{p}}e^{xt}\,d\mu_{-1}(x)=\frac{2}{e^{nt}+(-1)^{n-1}} \sum_{l=0}^{n-1}(-1)^{n-1-l}e^{lt},$$
(28)

and

\begin{aligned}[b] &\sum_{m=0}^{\infty} \biggl\{ \int_{\mathbb{Z}_{p}}(x+n)^{m} \,d\mu _{-1}(x)+(-1)^{n-1} \int_{\mathbb{Z}_{p}}x^{m} \,d\mu_{-1}(x)\biggr\} \frac{t^{m}}{m!}\\ &\quad=\sum_{m=0}^{\infty}\Biggl\{ 2\sum _{l=0}^{n-1}(-1)^{n-1-l}l^{m}\Biggr\} \frac{t^{m}}{m!}\quad (n\in\mathbb{N}). \end{aligned}
(29)

By comparing the coefficients on the both sides of (29), we get

$$\int_{\mathbb{Z}_{p}}(x+n)^{m} \,d\mu_{-1}(x)+(-1)^{n-1}\int_{\mathbb {Z}_{p}}x^{m} \,d\mu_{-1}(x) =2\sum_{l=0}^{n-1}(-1)^{n-1-l}l^{m},$$
(30)

where $$n\in\mathbb{N}$$ and $$m\in\mathbb{Z}\geq0$$.

Therefore, by (23) and (30), we obtain the following lemma.

### Lemma 6

For $$m\geq0$$, $$n\in\mathbb{N}$$, we have

$$E_{m}(n)+(-1)^{n-1}E_{m} =2\sum _{l=0}^{n-1}(-1)^{n-1-l}l^{m}.$$

Let us assume that $$n\in\mathbb{N}$$ with $$n\equiv1\ (\operatorname{mod}2)$$.

Then, by (28), we get

$$\int_{\mathbb{Z}_{p}}e^{xt}\,d\mu_{-1}(x)=\frac{2}{e^{nt}+1}\sum_{l=0}^{n-1}(-1)^{l}e^{lt}.$$
(31)

Now, we consider the multivariate p-adic fermionic integral on $$\mathbb{Z}_{p}$$ related to the higher-order Euler numbers as follows:

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} e^{(x_{1}+x_{2}+\cdots +x_{r})t}\,d\mu_{-1}(x_{1}) \cdots \,d\mu_{-1}(x_{r}) \\ &\quad=\biggl(\frac{2}{e^{nt}+1}\biggr)^{r}\sum _{l_{1}=0}^{n-1}\cdots\sum_{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}e^{(l_{1}+\cdots+l_{r})t} \\ &\quad=\Biggl(\sum_{l=0}^{\infty}E^{(r)}_{l}n^{l} \frac{t^{l}}{l!}\Biggr)\sum_{l_{1}=0}^{n-1} \cdots\sum_{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}\sum _{k=0}^{\infty}(l_{1}+ \cdots+l_{r})^{k}\frac{t^{k}}{k!} \\ &\quad=\sum_{l_{1}=0}^{n-1}\cdots\sum _{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}\sum _{m=0}^{\infty}\Biggl\{ \sum_{k=0}^{m} \binom{m}{k}(l_{1}+\cdots+l_{r})^{k}n^{m-k}E^{(r)}_{m-k} \Biggr\} \frac{t^{m}}{m!} \\ &\quad=\sum_{m=0}^{\infty}\Biggl\{ \sum _{l_{1}=0}^{n-1}\cdots\sum_{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}} \sum_{k=0}^{m}\binom {m}{k}n^{m-k}E^{(r)}_{m-k}(l_{1}+ \cdots+l_{r})^{k}\Biggr\} \frac{t^{m}}{m!}. \end{aligned}
(32)

Thus, from (32), we can derive the following equation:

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} {(x_{1}\cdots+x_{r})^{m}}\,d\mu _{-1}(x_{1})\cdots \,d\mu_{-1}(x_{r})\\ &\quad=\sum_{l_{1}=0}^{n-1}\cdots\sum _{l_{r}=0}^{n-1}\sum_{k=0}^{m}(-1)^{l_{1}+\cdots+l_{r}} \binom {m}{k}E^{(r)}_{m-k}n^{m-k}(l_{1}+ \cdots+l_{r})^{k}, \end{aligned}
(33)

where $$m\geq0$$ and $$n\in\mathbb{N}$$ with $$n\equiv1\ (\operatorname{mod}2)$$.

Therefore, by (24) and (33), we obtain the following theorem.

### Theorem 7

For $$m\geq0$$ and $$n\in\mathbb{N}$$ with $$n\equiv1\ (\operatorname{mod}2)$$, we have

$$E^{(r)}_{m}=\sum_{l_{1}=0}^{n-1} \cdots\sum_{l_{r}=0}^{n-1}\sum _{k=0}^{m}(-1)^{l_{1}+\cdots+l_{r}}\binom {m}{k}E^{(r)}_{m-k}n^{m-k}(l_{1}+ \cdots+l_{r})^{k},$$

Moreover,

$$E^{(r)}_{m}(x)=\sum_{l_{1}=0}^{n-1} \cdots\sum_{l_{r}=0}^{n-1}\sum _{k=0}^{m}(-1)^{l_{1}+\cdots+l_{r}}\binom {m}{k}E^{(r)}_{m-k}n^{m-k}(l_{1}+ \cdots+l_{r}+x)^{k}.$$

From (32), we note that

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} e^{(x_{1}+\cdots +x_{r})t}\,d\mu_{-1}(x_{1}) \cdots \,d\mu_{-1}(x_{r}) \\ &\quad=\biggl(\frac{2}{e^{nt}+1}\biggr)^{r}\sum _{l_{1}=0}^{n-1}\cdots\sum_{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}e^{(l_{1}+\cdots+l_{r})t} \\ &\quad=\sum_{l_{1}=0}^{n-1}\cdots\sum _{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}\biggl( \frac{2}{e^{nt}+1}\biggr)^{r}e^{(\frac {l_{1}+\cdots+l_{r}}{n})nt} \\ &\quad=\sum_{m=0}^{\infty}\sum _{l_{1}=0}^{n-1}\cdots\sum_{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}E^{(r)}_{m} \biggl(\frac{l_{1}+\cdots +l_{r}}{n}\biggr)n^{m}\frac{t^{m}}{m!}, \end{aligned}
(34)

where $$n\in\mathbb{N}$$ with $$n\equiv1\ (\operatorname{mod}2)$$.

Thus, by (34), we get

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} (x_{1}+\cdots+x_{r})^{m} \,d\mu _{-1}(x_{1})\cdots \,d\mu_{-1}(x_{r}) \\ &\quad=\sum_{l_{1}=0}^{n-1}\cdots\sum _{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}E^{(r)}_{m} \biggl(\frac{l_{1}+\cdots+l_{r}}{n}\biggr)n^{m}, \end{aligned}
(35)

where $$m\in\mathbb{Z}\geq0$$, $$n\in\mathbb{N}$$ with $$n\equiv1 \ (\operatorname{mod}2)$$.

Therefore by (24) and (35), we obtain the following theorem.

### Theorem 8

For $$m\in\mathbb{Z}\geq0$$, $$n\in\mathbb{N}$$ with $$n\equiv1 \ (\operatorname{mod}2)$$, we have

$$E^{(r)}_{m}=n^{m}\sum_{l_{1}=0}^{n-1} \cdots\sum_{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}E^{(r)}_{m} \biggl(\frac{l_{1}+\cdots+l_{r}}{n}\biggr),$$

Moreover,

$$E^{(r)}_{m}(x)=n^{m}\sum _{l_{1}=0}^{n-1}\cdots\sum_{l_{r}=0}^{n-1}(-1)^{l_{1}+\cdots+l_{r}}E^{(r)}_{m} \biggl(\frac{l_{1}+\cdots+l_{r}+x}{n}\biggr).$$

For $$a_{1}, a_{2},\ldots, a_{r}\in\mathbb{C}_{p}\backslash\{0\}$$, let us consider the Barnes-type multiple Euler polynomials as follows:

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} e^{(a_{1}x_{1}+\cdots +a_{r}x_{r}+x)t}\,d\mu_{-1}(x_{1}) \cdots \,d\mu_{-1}(x_{r}) \\ &\quad=\biggl(\frac{2}{e^{a_{1}t}+1}\biggr)\times\cdots\times\biggl(\frac {2}{e^{a_{r}t}+1} \biggr)e^{xt} \\ &\quad=\sum_{n=0}^{\infty}E_{n}(x|a_{1}, \ldots,a_{r})\frac{t^{n}}{n!}. \end{aligned}
(36)

When $$x=0$$, $$E_{n}(a_{1},\ldots,a_{r})=E_{n}(0|a_{1},\ldots,a_{r})$$ is called the nth Barnes-type Euler number.

For $$d\in\mathbb{N}$$ with $$d\equiv1\ (\operatorname{mod}2)$$, we observe that

$$\int_{\mathbb{Z}_{p}}f(x)\,d\mu_{-1}(x)=\sum _{a=0}^{d-1}(-1)^{a}\int _{\mathbb{Z}_{p}}f(a+dx)\,d\mu_{-1}(x).$$
(37)

From (37), we can derive the following equation:

\begin{aligned} &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} e^{(a_{1}x_{1}+\cdots +a_{r}x_{r})t}\,d\mu_{-1}(x_{1}) \cdots \,d\mu_{-1}(x_{r}) \\ &\quad=\sum_{l_{1}=0}^{d-1}\cdots\sum _{l_{r}=0}^{d-1}(-1)^{l_{1}+\cdots+l_{r}}\int _{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} e^{\{a_{1}l_{1}+\cdots+a_{r}l_{r}+(a_{1}x_{1}+\cdots+a_{r}x_{r})d\} t}\,d\mu_{-1}(x_{1})\cdots \,d\mu_{-1}(x_{r}) \\ &\quad=\sum_{n=0}^{\infty}d^{n}\sum _{l_{1}=0}^{d-1}\cdots\sum _{l_{r}=0}^{d-1}(-1)^{l_{1}+\cdots+l_{r}}\int _{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} \biggl( \frac{a_{1}l_{1}+\cdots+a_{r}l_{r}}{d} \\ &\qquad{}+a_{1}x_{1}+\cdots +a_{r}x_{r} \biggr)^{n} \,d\mu_{-1}(x_{1})\cdots \,d\mu_{-1}(x_{r}) \frac{t^{n}}{n!}. \end{aligned}
(38)

By (38), we get

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}} (a_{1}x_{1}+\cdots +a_{r}x_{r})^{n}\,d\mu_{-1}(x_{1}) \cdots \,d\mu_{-1}(x_{r}) \\ &\quad=d^{n}\sum_{l_{1}=0}^{d-1}\cdots\sum _{l_{r}=0}^{d-1}(-1)^{l_{1}+\cdots+l_{r}}\int _{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}} \biggl( \frac{a_{1}l_{1}+\cdots+a_{r}l_{r}}{d}\\ &\qquad{}+a_{1}x_{1}+\cdots +a_{r}x_{r} \biggr)^{n} \,d\mu_{-1}(x_{1})\cdots \,d\mu_{-1}(x_{r}). \end{aligned}
(39)

Therefore, by (36) and (39), we obtain the following theorem.

### Theorem 9

For $$d\in\mathbb{N}$$ with $$d\equiv1\ (\operatorname{mod}2)$$, $$n\geq0$$, we have

$$E_{n}(a_{1},\ldots,a_{r})=d^{n}\sum _{l_{1}=0}^{d-1}\cdots\sum _{l_{r}=0}^{d-1}(-1)^{l_{1}+\cdots+l_{r}}E_{n}\biggl( \frac{a_{1}l_{1}+\cdots +a_{r}l_{r}}{d}\Big|a_{1},\ldots,a_{r}\biggr).$$

Moreover,

$$E_{n}(x|a_{1},\ldots,a_{r})=d^{n}\sum _{l_{1}=0}^{d-1}\cdots\sum _{l_{r}=0}^{d-1}(-1)^{l_{1}+\cdots+l_{r}}E_{n}\biggl( \frac{x+a_{1}l_{1}+\cdots +a_{r}l_{r}}{d}\Big|a_{1},\ldots,a_{r}\biggr).$$

### Remark

Note that

\begin{aligned}[b] E_{n}(x|a_{1}, \ldots,a_{r})&=\sum_{l=0}^{n} \binom {n}{l}x^{l}E_{n-l}(a_{1}, \ldots,a_{r}) \\ &=\sum_{l=0}^{n}\binom{n}{l}x^{n-l}E_{l}(a_{1}, \ldots,a_{r}). \end{aligned}

Thus, we have

\begin{aligned}[b] &\int_{\mathbb{Z}_{p}}\cdots \int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb {Z}_{p}} (a_{1}x_{1}+\cdots+a_{r}x_{r}\\ &\qquad{}+b_{1}y_{1}+ \cdots+b_{s}y_{s})^{n} \,d\mu _{-1}(y_{1}) \cdots \,d\mu_{-1}(y_{s}) \,d\mu_{0}(x_{1})\cdots \,d\mu_{0}(x_{r}) \\ &\quad=\int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}}E_{n}(a_{1}x_{1}+ \cdots +a_{r}x_{r}|b_{1},\ldots,b_{s})\,d\mu_{0}(x_{1})\cdots \,d\mu_{0}(x_{r}) \\ &\quad=\sum_{l=0}^{n}\binom{n}{l}E_{n-l}(b_{1}, \ldots,b_{s})\int_{\mathbb {Z}_{p}}\cdots\int _{\mathbb{Z}_{p}}(a_{1}x_{1}+\cdots+a_{r}x_{r})^{l} \,d\mu _{0}(x_{1})\cdots \,d\mu_{0}(x_{r}) \\ &\quad=\sum_{l=0}^{n}\binom{n}{l}E_{n-l}(b_{1}, \ldots ,b_{s})B_{l}(a_{1},\ldots,a_{r}). \end{aligned}
(40)

Now, we define mixed-type Barnes-type Euler and Bernoulli numbers as follows:

\begin{aligned}[b] & EB_{n}(b_{1}, \ldots,b_{s};a_{1},\ldots,a_{r}) \\ &\quad=\int_{\mathbb{Z}_{p}}\cdots\int_{\mathbb{Z}_{p}}\cdots\int _{\mathbb {Z}_{p}} (a_{1}x_{1}+\cdots+a_{r}x_{r}\\ &\qquad{}+b_{1}y_{1}+ \cdots+b_{s}y_{s})^{n} \,d\mu _{-1}(y_{1}) \cdots \,d\mu_{-1}(y_{s}) \,d\mu_{0}(x_{1})\cdots \,d\mu_{0}(x_{r}), \end{aligned}
(41)

where $$a_{1},\ldots,a_{r},b_{1},\ldots,b_{s}\neq0$$.

By (40) and (41), we get

$$EB_{n}(b_{1},\ldots,b_{s};a_{1}, \ldots,a_{r})=\sum_{l=0}^{n} \binom {n}{l}E_{n-l}(b_{1},\ldots,b_{s})B_{l}(a_{1}, \ldots,a_{r}).$$

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