Open Access

Some identities of Barnes-type special polynomials

Advances in Difference Equations20152015:42

https://doi.org/10.1186/s13662-015-0385-y

Received: 30 December 2014

Accepted: 21 January 2015

Published: 11 February 2015

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 \(\mathbb{Z}_{p}\).

Keywords

Barnes-type Bernoulli polynomialhigher-order Euler polynomialmultivariate p-adic fermionic integralBarnes-type Euler polynomial

MSC

11B6811S40

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 [1433]).

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

Declarations

Acknowledgements

Authors wish to express their sincere gratitude to the referees for their valuable suggestions and comments.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Mathematics, Kyungpook National University

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© Lim and Do; licensee Springer. 2015