Infinite order linear differential equation satisfied by p-adic Hurwitz-type Euler zeta functions

Abstract

In 1900, at the international congress of mathematicians, Hilbert claimed that the Riemann zeta function $\zeta (s)$ is not the solution of any algebraic ordinary differential equations on its region of analyticity. In 2015, Van Gorder (J Number Theory 147:778–788, 2015) considered the question of whether $\zeta (s)$ satisfies a non-algebraic differential equation and showed that it formally satisfies an infinite order linear differential equation. Recently, Prado and Klinger-Logan (J Number Theory 217:422–442, 2020) extended Van Gorder’s result to show that the Hurwitz zeta function $\zeta (s,a)$ is also formally satisfies a similar differential equation

$$\begin{aligned} T\left[ \zeta (s,a) - \frac{1}{a^s}\right] = \frac{1}{(s-1)a^{s-1}}. \end{aligned}$$

But unfortunately in the same paper they proved that the operator T applied to Hurwitz zeta function $\zeta (s,a)$ does not converge at any point in the complex plane ${\mathbb {C}}$. In this paper, by defining $T_{p}^{a}$, a p-adic analogue of Van Gorder’s operator T,  we establish an analogue of Prado and Klinger-Logan’s differential equation satisfied by $\zeta _{p,E}(s,a)$ which is the p-adic analogue of the Hurwitz-type Euler zeta functions

$$\begin{aligned} \zeta _E(s,a)=\sum _{n=0}^\infty \frac{(-1)^n}{(n+a)^s}. \end{aligned}$$

In contrast with the complex case, due to the non-archimedean property, the operator $T_{p}^{a}$ applied to the p-adic Hurwitz-type Euler zeta function $\zeta _{p,E}(s,a)$ is convergent p-adically in the area of $s\in {\mathbb {Z}}_{p}$ with $s\ne 1$ and $a\in K$ with $|a|_{p}>1,$ where K is any finite extension of ${\mathbb {Q}}_{p}$ with ramification index over ${\mathbb {Q}}_{p}$ less than $p-1.$

1 Introduction

Throughout this paper we shall use the following notations.

$$\begin{aligned} \qquad {\mathbb {C}} ~~~&- ~\text {the field of complex numbers}.\\ \qquad p ~~~&- ~\text {an odd rational prime number}. \\ \qquad {\mathbb {Z}}_p ~~~&- ~\text {the ring of { p}-adic integers}. \\ \qquad {\mathbb {Q}}_p~~~&- ~\text {the field of fractions of}~{\mathbb {Z}}_p.\\ \qquad {\mathbb {C}}_p ~~~&- ~\text {the completion of a fixed algebraic closure}~\overline{{\mathbb {Q}}}_p~ \text {of}~{\mathbb {Q}}_{p}. \end{aligned}$$

The Riemann zeta function $\zeta (s)$ is defined as

$$\begin{aligned} \zeta (s)=\sum _{n=1}^{\infty }\frac{1}{n^{s}},\quad \text {Re}(s) > 1, \end{aligned}$$

(1.1)

it can be analytically continued to the whole complex plane except for a single pole at $s=1$ with residue 1. In 1900, at the international congress of mathematicians, David Hilbert [5] claimed that $\zeta (s)$ is not the solution of any algebraic ordinary differential equations on its region of analyticity. In 2015, Van Gorder [19] considered the question of whether $\zeta (s)$ satisfies a non-algebraic differential equation and showed that it formally satisfies an infinite order linear differential equation. In fact, he established the differential equation

$$\begin{aligned} T[\zeta (s)-1]=\frac{1}{s-1} \end{aligned}$$

(1.2)

formally, where

$$\begin{aligned} T = \sum _{n=0}^\infty L_n \end{aligned}$$

(1.3)

and

$$\begin{aligned} L_n&:= p_n(s) \exp (nD), \\ p_n(s)&:= {\left\{ \begin{array}{ll} 1&{} \text { if } n = 0 \\ \frac{1}{(n+1)!}\prod _{j = 0}^{n-1} (s+j) &{}\text { if } n >0, \end{array}\right. }\\ \exp (nD)&:= id + \sum _{k=1}^\infty \frac{n^k}{k!} D_s^k \end{aligned}$$

for $D_s^k := \frac{\partial ^k}{\partial s^k}.$

For $0 < a \le 1$, Re$(s) >1$, in 1882 Hurwitz [4] defined the partial zeta functions

$$\begin{aligned} ~ \zeta (s,a)=\sum _{n=0}^{\infty }\frac{1}{(n+a)^{s}}, \end{aligned}$$

(1.4)

which generalized (1.1). As (1.1), this function can also be analytically continued to a meromorphic function in the complex plane with a simple pole at $s=1$. Recently, Prado and Klinger-Logan [15] extended Van Gorder’s result to show that the Hurwitz zeta function $\zeta (s,a)$ also formally satisfies a similar differential equation

$$\begin{aligned} T\left[ \zeta (s,a) - \frac{1}{a^s}\right] = \frac{1}{(s-1)a^{s-1}} \end{aligned}$$

(1.5)

for $s \in {\mathbb {C}}$ satisfying $s + n \ne 1$ for all $n \in {\mathbb {Z}}_{\ge 0},$ where T is the Van Gorder’s operator defined as in (1.3) (see [15, Corollary 4]). But unfortunately, in the same paper they proved that

$$\begin{aligned} T\left[ \zeta (s, a) - \frac{1}{a^s} \right] =\sum _{n=0}^{\infty } p_n(s)\exp (nD)\left[ \zeta (s,a)- \frac{1}{a^s}\right] , \end{aligned}$$

the operator T applied to Hurwitz zeta function, does not converge at any point in the complex plane ${\mathbb {C}}$ (see [15, Theorem 8]). Then they defined a generalized operator G instead of T. That is, let ${\mathcal {M}}$ be the collection of meromorphic functions on ${\mathbb {C}}$ and $f\in {\mathcal {M}}$, define $G:{\mathcal {M}}\rightarrow {\mathcal {M}}$ by

$$\begin{aligned} G[f](s) = \sum _{n=0}^\infty p_n(s)f(s+n). \end{aligned}$$

(1.6)

Under this linear operator, we have a convergent difference equation

$$\begin{aligned} G\left[ \zeta (s,a) - \frac{1}{a^s}\right] = \frac{1}{(s-1)a^{s-1}}. \end{aligned}$$

(1.7)

But it needs to mention that G is not a differential operator.

For Re$(s)>0$, the Euler zeta function (also called alternative series or Dirichlet eta function) is defined by

$$\begin{aligned} ~ \zeta _{E}(s)=\sum _{n=1}^{\infty }\frac{(-1)^{n-1}}{n^{s}}. \end{aligned}$$

(1.8)

This function can be analytically continued to the complex plane without any pole. For Re$(s)>0$, (1.1) and (1.8) are connected by the following equation

$$\begin{aligned} ~ \zeta _{E}(s)=(1-2^{1-s})\zeta (s). \end{aligned}$$

(1.9)

By Weil’s history [21, p. 273–276] (also see a survey by Goss [3, Sect. 2]), Euler used (1.8) to “prove”

$$\begin{aligned} ~ \frac{\zeta _{E}(1-s)}{\zeta _{E}(s)}=\frac{-\Gamma (s)(2^{s}-1)\text {cos}(\pi s/2)}{(2^{s-1}-1)\pi ^{s}}, \end{aligned}$$

(1.10)

which leads to the functional equation of $\zeta (s)$.

For $s\in {\mathbb {C}}$ and $a\ne 0,-1,-2,\ldots ,$ the Hurwitz-type Euler zeta function is defined as the Hurwitz zeta function (1.4) twisted by $(-1)^{n}$

$$\begin{aligned} \zeta _E(s,a)=\sum _{n=0}^\infty \frac{(-1)^n}{(n+a)^s}. \end{aligned}$$

(1.11)

This function can also be analytically continued to the complex plane without any pole. It represents a partial zeta function of cyclotomic fields in one version of Stark’s conjectures in algebraic number theory (see [11, p. 4249, (6.13)]). Recently, several interesting properties for the function $\zeta _{E}(s,a)$ have been studied, including its Fourier expansion and several integral representations [7], special values and power series expansions [6], convexity properties [2], etc.

In [10], using the fermionic p-adic integral (see (2.6) below), we defined $\zeta _{p,E}(s,a),$ the p-adic analogue of Hurwitz-type Euler zeta functions (1.11), which interpolates (1.11) at nonpositive integers (see Theorem 2.4 below), so called the p-adic Hurwitz-type Euler zeta functions. In the same paper, we also proved several properties of $\zeta _{p,E}(s,a),$ including the analyticity, the convergent Laurent series expansion, the distribution formula, the difference equation, the reflection functional equation, the derivative formula and the p-adic Raabe formula.

In this note, we define a p-adic analogue of the operator T,  denoted by $T_{p}^{a}$ (see (2.10) below). Under this operator, the p-adic Hurwitz-type Euler zeta function $\zeta _{p,E}(s,a)$ satisfies an infinite order linear differential equation

$$\begin{aligned} T_{p}^{a}\left[ \zeta _{p,E}(s,a)-\langle a\rangle ^{1-s}\right] =\frac{1}{s-1}\left( \langle a-1 \rangle ^{1-s}-\langle a\rangle ^{1-s}\right) \end{aligned}$$

(1.12)

(see Theorem 3.5). In contrast with the complex case, the left hand side of the above equation is convergent everywhere for $s\in {\mathbb {Z}}_{p}$ with $s\ne 1$ and $a\in K$ with $|a|_{p}>1,$ where K is any finite extension of ${\mathbb {Q}}_{p}$ with ramification index over ${\mathbb {Q}}_{p}$ less than $p-1$ (see Corollary 3.8 and Remarks 3.7 and 3.9 below).

2 Preliminaries

2.1 p-adic Teichmüller character

To our purpose, in this subsection, we recall some notions from p-adic analysis, including the p-adic Teichmüller character $\omega _v(a)$ and the projection function $\langle a\rangle$ for $a\in {\mathbb {C}}_{p}^{\times }$. Our approach follows Tangedal and Young in [18] closely.

Given $a\in {\mathbb {Z}}_{p}, p\not \mid a$ and $p>2,$ there exists a unique $(p-1)$th root of unity $\omega (a)\in {\mathbb {Z}}_p$ such that

$$\begin{aligned} a\equiv \omega (a)\pmod {p}, \end{aligned}$$

where $\omega$ is the Teichmüller character. Let $\langle a\rangle =\omega ^{-1}(a)a,$ so $\langle a\rangle \equiv 1\pmod p.$

In what follows we extend the definition domain of the projection function $\langle a\rangle$ from ${\mathbb {Z}}_{p}$ to ${\mathbb {C}}_{p}$. Fixed an embedding of $\overline{{\mathbb {Q}}}$ into ${\mathbb {C}}_{p},$ denote the image of the set of positive real rational powers of p under this embedding in ${\mathbb {C}}_{p}^{\times }$ by $p^{{\mathbb {Q}}},$ and the group of roots of unity with order not divisible by p in ${\mathbb {C}}_{p}^{\times }$ by $\mu$. Given $a\in {\mathbb {C}}_{p}$ with $|a|_{p}=1$, there exists a unique element ${\hat{a}}\in \mu$ such that

$$\begin{aligned} |a-{\hat{a}}|_{p} < 1, \end{aligned}$$

(2.1)

which is also named the Teichmüller representative of a; it can also be defined from ${\hat{a}}=\lim _{n\rightarrow \infty }a^{p^{n!}}$. Then we extend this definition to $a\in {\mathbb {C}}_{p}^{\times }$ by

$$\begin{aligned} {\hat{a}}:=(\widehat{a/p^{v_{p}(a)}}), \end{aligned}$$

(2.2)

that is, we define ${\hat{a}}={\hat{u}}$ if $a=p^{r}u$ with $p^{r}\in p^{{\mathbb {Q}}}$ and $|u|_{p}=1$, then we define the function $\langle \cdot \rangle$ on ${\mathbb {C}}_{p}^{\times }$ by

$$\begin{aligned} \langle a\rangle =p^{-v_{p}(a)}a/{\hat{a}}. \end{aligned}$$

Now we define $\omega _v(\cdot )$ on ${\mathbb {C}}_{p}^{\times }$ by

$$\begin{aligned} \omega _v(a)=\frac{a}{\langle a\rangle }=p^{v_{p}(a)}{\hat{a}}. \end{aligned}$$

(2.3)

From this we get an internal product decomposition of multiplicative groups

$$\begin{aligned} {\mathbb {C}}_{p}^{\times }\simeq p^{{\mathbb {Q}}}\times \mu \times D, \end{aligned}$$

(2.4)

where $D=\{a\in {\mathbb {C}}_{p}: |a-1|_{p} < 1\},$ given by

$$\begin{aligned} a=p^{v_{p}(a)}\cdot {\hat{a}}\cdot \langle a\rangle \mapsto (p^{v_{p}(a)},{\hat{a}},\langle a\rangle ). \end{aligned}$$

(2.5)

As remarked by Tangedal and Young in [18], this decomposition of ${\mathbb {C}}_{p}^{\times }$ depends on the choice of embedding of $\overline{{\mathbb {Q}}}$ into ${\mathbb {C}}_{p}$; the projections $p^{v_{p}(a)},{\hat{a}},\langle a\rangle$ are uniquely determined up to roots of unity. However for $a\in {\mathbb {Q}}_{p}^{\times }$ the projections $p^{v_{p}(a)},{\hat{a}},\langle a\rangle$ are uniquely determined and do not depend on the choice of the embedding. Notice that the projections $a\mapsto p^{v_{p}(a)}$ and $a\mapsto {\hat{a}}$ are constant on discs of the form $\{a\in {\mathbb {C}}_{p}:|a-y|_{p} < |y|_{p}\}$ and therefore have derivative zero whereas the projections $a\mapsto \langle a\rangle$ has derivative $\frac{d}{da}\langle a\rangle =\langle a\rangle /a$.

2.2 The fermionic p-adic integral and the p-adic Hurwitz-type Euler zeta functions

In this subsection, we recall the definition of the p-adic Hurwitz-type Euler zeta functions $\zeta _{p,E}(s,a)$ from the fermionic p-adic integral. For details, we refer to [10].

Let $UD({\mathbb {Z}}_p)$ be the space of all uniformly (or strictly) differentiable ${\mathbb {C}}_p$-valued functions on ${\mathbb {Z}}_p$ (see [1, §11.1.2]). The fermionic p-adic integral $I_{-1}(f)$ on ${\mathbb {Z}}_p$ of a function $f\in UD({\mathbb {Z}}_p)$ is defined by

$$\begin{aligned} I_{-1}(f)=\int _{{\mathbb {Z}}_p}f(t)d\mu _{-1}(t) =\lim _{r\rightarrow \infty }\sum _{k=0}^{p^r-1}f(k)(-1)^k. \end{aligned}$$

(2.6)

The fermionic p-adic integral (2.6) was independently found by Katz [8, p. 486] (in Katz’s notation, the $\mu ^{(2)}$-measure), Shiratani and Yamamoto [17], Osipov [14], Lang [12] (in Lang’s notation, the $E_{1,2}$-measure), Kim [9] from very different viewpoints.

For $a\in {\mathbb {C}}_{p}^{\times }$ and $s\in {\mathbb {C}}_{p},$ the two-variable function $\langle a\rangle ^{s}$ ([16, p. 141]) is defined by

$$\begin{aligned} \langle a\rangle ^{s}=\sum _{n=0}^{\infty }\left( {\begin{array}{c}s\\ n\end{array}}\right) (\langle a\rangle -1)^{n}, \end{aligned}$$

(2.7)

when this sum is convergence. The analytic property of $\langle a\rangle ^{s}$ is stated in the following proposition.

Proposition 2.1

(see Tangedal and Young [18]) For any $a\in {\mathbb {C}}_{p}^{\times }$ the function $s\mapsto \langle a\rangle ^{s}$ is a $C^{\infty }$ function of s on ${\mathbb {Z}}_{p}$ and is analytic on a disc of positive radius about $s=0$; on this disc it is locally analytic as a function of a and independent of the choice made to define the $\langle \cdot \rangle$ function. If a lies in a finite extension K of ${\mathbb {Q}}_{p}$ whose ramification index over ${\mathbb {Q}}_{p}$ is less than $p-1$ then $s\mapsto \langle a\rangle ^{s}$ is analytic for $|s|_{p} < |\pi |_{p}^{-1}p^{-1/(p-1)}$, where $(\pi )$ is the maximal ideal of the ring of integers $O_{K}$ of K. If $s\in {\mathbb {Z}}_{p},$ the function $a\mapsto \langle a\rangle ^{s}$ is an analytic function of a on any disc of the form $\{a\in {\mathbb {C}}_{p}:|a-y|_{p} <|y|_{p}\}$.

Now we are at the position to recall the definition for the p-adic Hurwitz-type Euler zeta function $\zeta _{p,E}(s,a)$.

Definition 2.2

(see [10, Definition 3.3]) For $a\in {\mathbb {C}}_{p}\backslash {\mathbb {Z}}_{p}$, we define the p-adic Hurwitz-type Euler zeta function $\zeta _{p,E}(s,a)$ by the formula

$$\begin{aligned} \zeta _{p,E}(s,a)=\int _{{\mathbb {Z}}_{p}}\langle a+t\rangle ^{1-s}d\mu _{-1}(t). \end{aligned}$$

(2.8)

The following theorem summarize the analytic property of $\zeta _{p,E}(s,x)$ and Tangedal and Young proved a similar result for p-adic multiple zeta functions (see [18, Theorem 3.1]).

Theorem 2.3

(see [10, Theorem 3.4]) For any choice of $a\in {\mathbb {C}}_{p}\backslash {\mathbb {Z}}_{p}$ the function $\zeta _{p,E}(s,a)$ is a $C^{\infty }$ function of s on ${\mathbb {Z}}_{p}$, and is an analytic function of s on a disc of positive radius about $s=0$; on this disc it is locally analytic as a function of a and independent of the choice made to define the $\langle \cdot \rangle$ function. If a is so chosen to lie in a finite extension K of ${\mathbb {Q}}_{p}$ whose ramification index over ${\mathbb {Q}}_{p}$ is less than $p-1$ then $\zeta _{p,E}(s,a)$ is analytic for $|s|_{p} < |\pi |_{p}^{-1}p^{-1/(p-1)}$. If $s\in {\mathbb {Z}}_{p}$, the function $\zeta _{p,E}(s,a)$ is locally analytic as a function of a on ${\mathbb {C}}_{p}\backslash {\mathbb {Z}}_{p}$.

It needs to mention that the p-adic Hurwitz-type Euler zeta function $\zeta _{p,E}(s,a)$ interpolates its complex counterpart $\zeta _{E}(s,a)$ (1.11) p-adically, that is,

Theorem 2.4

(see [10, Theorem 3.8]) Suppose that $a\in {\mathbb {C}}_{p}$ and $|a|_p>1.$ For $m\in {\mathbb {N}},$

$$\begin{aligned} \zeta _{p,E}(1-m,a)=\frac{1}{\omega _v^{m}(a)}E_m(a)=\frac{1}{\omega _v^{m}(a)}\zeta _E(-m,a), \end{aligned}$$

where the Euler polynomials $E_{m}(x)$ is defined by the generating function

$$\begin{aligned} \frac{2e^{xz}}{e^z+1}=\sum _{m=0}^\infty E_m(x)\frac{z^m}{n!}, \quad |z|<\pi . \end{aligned}$$

(2.9)

2.3 The p-adic operator $T_{p}^{a}$

In this subsection, we give a definition of $T_{p}^{a}$, the p-adic analogue of the operator T (see (1.3)). Let $E=\{x\in {\mathbb {C}}_{p}: |x|_{p}<p^{-\frac{1}{p-1}}\}$ be the region of convergence of the power series $\sum _{k=0}^{\infty }\frac{x^{k}}{k!}$. The p-adic exponential function is given by

$$\begin{aligned} \text {exp}_{p}(x)=\sum _{k=0}^{\infty }\frac{x^{k}}{k!},\quad (x\in E) \end{aligned}$$

(see [16, p. 70]) and the p-adic Van Gorder’s operator is defined as follows

$$\begin{aligned} T _{p}^{a}= \sum _{n=0}^\infty L_{p,n}^{a}, \end{aligned}$$

(2.10)

where

$$\begin{aligned} L_{p,n}^{a}:= & {} P_{p,n}^{a}(s) \exp _{p}(nD), \nonumber \\ P_{p,n}^{a}(s) := & {} {\left\{ \begin{array}{ll} \frac{2}{s-1}&{} \text { if } n = 0\\ \frac{1}{\omega _{v}(a)}&{} \text { if } n = 1\\ \frac{1}{n!\omega _{v}^{n}(a)}\prod _{j = 1}^{n-1} (s-1+j) &{}\text { if } n\ge 2, \end{array}\right. } \nonumber \\ \exp _{p}(nD) := & {} id + \sum _{k=1}^\infty \frac{n^k}{k!} D_s^k \end{aligned}$$

(2.11)

for $D_s^k := \frac{\partial ^k}{\partial s^k}.$

3 Main results

In this section, we shall prove (1.12). First we need to establish the following identity for $\zeta _{p,E}(s,a)$.

Lemma 3.1

Let $\zeta _{p,E}(s,a)$ be the p-adic Hurwitz-type Euler zeta function. Then, for $s\in {\mathbb {Z}}_{p}$ with $s\ne 1$ and $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$, we have that

$$\begin{aligned}&\frac{2}{s-1}\left( \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right) +\frac{1}{\omega _{v}(a)}\left( \zeta _{p,E}(s+1,a)-\langle a \rangle ^{1-(s+1)}\right) \nonumber \\&\qquad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \nonumber \\&\quad =\frac{1}{s-1}\left( \langle a-1 \rangle ^{1-s}-\langle a \rangle ^{1-s}\right) . \end{aligned}$$

(3.1)

Remark 3.2

This is a p-adic analogue of complex identities for the Hurwitz zeta function $\zeta (s,a)$ (see [15, Lemma 1]) and for the Riemann zeta function $\zeta (s)$ (see [15, (3.3)]).

Proof of Lemma 3.1

Fix $s\in {\mathbb {Z}}_{p}$ and $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1.$ For any $r\in {\mathbb {N}}$ we have

$$\begin{aligned} -\frac{1}{(s-1)\langle a\rangle ^{s-1}}= & {} \frac{1}{s-1}\left( \sum _{k=0}^{p^{r}-1}\frac{(-1)^{k+1}}{\langle k+a\rangle ^{s-1}}+\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a\rangle ^{s-1}}\right) \nonumber \\= & {} \frac{1}{s-1}\left( \sum _{k=1}^{p^{r}}\frac{(-1)^{k}}{\langle k-1+a\rangle ^{s-1}}+\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a\rangle ^{s-1}}\right) \nonumber \\= & {} \frac{1}{s-1}\left( \sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k-1+a\rangle ^{s-1}}+\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a\rangle ^{s-1}}+\frac{(-1)^{p^{r}}}{\langle p^{r}-1+a\rangle ^{s-1}}\right) \nonumber \\= & {} \frac{1}{s-1}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}}\left( \left\langle \frac{k+a}{k-1+a}\right\rangle ^{s-1}+1\right) \nonumber \\&\quad -\frac{1}{s-1}\frac{1}{\langle p^{r}-1+a\rangle ^{s-1}}\nonumber \\&\quad (\text {since}\, p \text { is an odd prime}). \end{aligned}$$

(3.2)

Since $|a|_{p}>1$, for $k\in {\mathbb {N}}$, we have $|k+a|_{p}>1$, thus

$$\begin{aligned} \left| 1-\frac{1}{k+a}\right| _{p}=1 \end{aligned}$$

and

$$\begin{aligned} \left| \frac{1}{1-\frac{1}{k+a}}-1\right| _{p}=\left| \frac{\frac{1}{k+a}}{1-\frac{1}{k+a}}\right| _{p}<1. \end{aligned}$$

Then from (2.1) we see that

$$\begin{aligned} \widehat{\frac{1}{1-\frac{1}{k+a}}}=1 \end{aligned}$$

and by (2.3)

$$\begin{aligned} \omega _{v}\left( \frac{1}{1-\frac{1}{k+a}}\right) =1. \end{aligned}$$

Again by (2.3), we have

$$\begin{aligned} \left\langle \frac{k+a}{k-1+a}\right\rangle= & {} \left\langle \frac{1}{1-\frac{1}{k+a}}\right\rangle \nonumber \\= & {} \omega _{v}^{-1}\left( \frac{1}{1-\frac{1}{k+a}}\right) \left( \frac{1}{1-\frac{1}{k+a}}\right) \nonumber \\= & {} \frac{1}{1-\frac{1}{k+a}}\nonumber \\= & {} \left( 1-\frac{1}{k+a}\right) ^{-1}. \end{aligned}$$

(3.3)

From [16, p.140, Lemma 47.6], for $s\in {\mathbb {Z}}_{p}$ we have the expansion

$$\begin{aligned} (1+x)^{s}=\sum _{n=0}^{\infty }\left( {\begin{array}{c}s\\ n\end{array}}\right) x^{n},\quad |x|_{p}<1. \end{aligned}$$

Thus by (3.3) we get

$$\begin{aligned} \left\langle \frac{k+a}{k-1+a}\right\rangle ^{s-1}= & {} \left( 1-\frac{1}{k+a}\right) ^{1-s}=\sum _{n=0}^{\infty }\left( {\begin{array}{c}1-s\\ n\end{array}}\right) \frac{(-1)^{n}}{(k+a)^{n}}\nonumber \\= & {} 1+\sum _{n=1}^{\infty } \frac{\prod _{j = 0}^{n - 1}(s-1+j)}{n!}\frac{1}{(k+a)^{n}}. \end{aligned}$$

(3.4)

Substituting the above expansion into (3.2), we have

$$\begin{aligned}&-\frac{1}{(s-1)\langle a\rangle ^{s-1}}\nonumber \\&\quad =\frac{1}{s-1}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}}\left( \left( 1+\sum _{n=1}^{\infty }\frac{\prod _{j = 0}^{n - 1}(s-1+j)}{n!}\frac{1}{(k+a)^{n}}\right) +1\right) \nonumber \\&\qquad -\frac{1}{s-1}\frac{1}{\langle p^{r}-1+a\rangle ^{s-1}}\nonumber \\&\quad =\frac{2}{s-1}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}} \nonumber \\&\qquad +\frac{1}{s-1}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}}\sum _{n=1}^\infty \frac{\prod _{j = 0}^{n - 1}(s-1+j)}{n!}\frac{1}{(k+a)^n} \nonumber \\&\qquad -\frac{1}{s-1}\frac{1}{\langle p^{r}-1+a\rangle ^{s-1}}\nonumber \\&\quad =\frac{2}{s-1}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}} \nonumber \\&\qquad +\frac{1}{s-1}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}}\left( \frac{s-1}{k+a}+\sum _{n=2}^\infty \frac{\prod _{j = 0}^{n - 1}(s-1+j)}{n!}\frac{1}{(k+a)^n}\right) \nonumber \\&\qquad -\frac{1}{s-1}\frac{1}{\langle p^{r}-1+a\rangle ^{s-1}}. \end{aligned}$$

(3.5)

Since $|a|_{p}>1$ and $k\in {\mathbb {N}}$, by (2.3) we have

$$\begin{aligned} \omega _{v}(k+a)=\omega _{v}(a) \end{aligned}$$

and

$$\begin{aligned} k+a=\omega _{v}(k+a)\langle k+a \rangle =\omega _{v}(a)\langle k+a \rangle . \end{aligned}$$

Substituting the above identity into (3.5), we get

$$\begin{aligned} -\frac{1}{(s-1)\langle a\rangle ^{s-1}}= & {} \frac{2}{s-1}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}}+\frac{1}{\omega _{v}(a)}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s}}\nonumber \\&\quad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}} \nonumber \\&\quad -\frac{1}{s-1}\frac{1}{\langle p^{r}-1+a\rangle ^{s-1}}. \end{aligned}$$

(3.6)

Taking $r\rightarrow \infty$ in the above equality, by the continuity of the p-adic function $\langle a \rangle ^{s}$ in a (see the last sentence of Proposition 2.1), we have

$$\begin{aligned} \lim _{r\rightarrow \infty }\langle p^{r}-1+a\rangle ^{s-1} =\langle a-1\rangle ^{s-1} \end{aligned}$$

and

$$\begin{aligned} -\frac{1}{(s-1)\langle a\rangle ^{s-1}}= & {} \frac{2}{s-1}\lim _{r\rightarrow \infty }\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}}+\frac{1}{\omega _{v}(a)}\lim _{r\rightarrow \infty }\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s}} \nonumber \\&\quad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\lim _{r\rightarrow \infty }\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}}\nonumber \\&\quad (\text {see Proposition 3.3})\nonumber \\&\quad -\frac{1}{s-1}\frac{1}{\langle a-1\rangle ^{s-1}}\nonumber \\= & {} \frac{2}{s-1}\left( \lim _{r\rightarrow \infty }\sum _{k=0}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s-1}}-\frac{1}{\langle a \rangle ^{s-1}}\right) \nonumber \\&\quad +\frac{1}{\omega _{v}(a)}\left( \lim _{r\rightarrow \infty }\sum _{k=0}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s}}-\frac{1}{\langle a \rangle ^{s}}\right) \nonumber \\&\quad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \lim _{r\rightarrow \infty }\sum _{k=0}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}}-\frac{1}{\langle a \rangle ^{s+n-1}}\right) \nonumber \\&\quad -\frac{1}{s-1}\frac{1}{\langle a-1\rangle ^{s-1}}. \end{aligned}$$

(3.7)

Then by the definitions of the fermionic p-adic integral (2.7) and the p-adic Hurwitz-type zeta function $\zeta _{p,E}(s,a)$ (2.8), we have

$$\begin{aligned} -\frac{1}{(s-1)\langle a\rangle ^{s-1}}= & {} \frac{2}{s-1}\left( \int _{{\mathbb {Z}}_{p}}\langle k+a \rangle ^{1-s}d\mu _{-1}(a)-\frac{1}{\langle a \rangle ^{s-1}}\right) \nonumber \\&\quad +\frac{1}{\omega _{v}(a)}\left( \int _{{\mathbb {Z}}_{p}}\langle k+a \rangle ^{-s}d\mu _{-1}(a)-\frac{1}{\langle a \rangle ^{s}}\right) \nonumber \\&\quad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \int _{{\mathbb {Z}}_{p}}\langle k+a \rangle ^{1-(s+n)}d\mu _{-1}(a)-\frac{1}{\langle a \rangle ^{s+n-1}}\right) \nonumber \\&\quad -\frac{1}{s-1}\frac{1}{\langle a-1\rangle ^{s-1}}\nonumber \\= & {} \frac{2}{s-1}\left( \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right) \nonumber \\&\quad +\frac{1}{\omega _{v}(a)}\left( \zeta _{p,E}(s+1,a)-\langle a \rangle ^{1-(s+1)}\right) \nonumber \\&\quad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \nonumber \\&\quad -\frac{1}{s-1}\frac{1}{\langle a-1\rangle ^{s-1}}, \end{aligned}$$

(3.8)

which is equivalent to

$$\begin{aligned}&\frac{2}{s-1}\left( \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right) +\frac{1}{\omega _{v}(a)}\left( \zeta _{p,E}(s+1,a)-\langle a \rangle ^{1-(s+1)}\right) \nonumber \\&\qquad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \nonumber \\&\quad =\frac{1}{s-1}\left( \langle a-1 \rangle ^{1-s}-\langle a \rangle ^{1-s}\right) . \end{aligned}$$

(3.9)

This completes the proof. $\square$

As pointed out by the referee, in order to move the limit to the inside of the summation $\sum _{n=2}^{\infty }$ in (3.7) of the above lemma, we need to show that the convergence of the inner limit is uniform for $r\in {\mathbb {N}}$. To this end, we add the following proposition.

Proposition 3.3

For $s\in {\mathbb {Z}}_{p}$ and $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$, the series

$$\begin{aligned} \sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}} \end{aligned}$$

converges uniformly for $r\in {\mathbb {N}}$ and

$$\begin{aligned}&\lim _{r\rightarrow \infty }\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}}\nonumber \\&\quad =\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\lim _{r\rightarrow \infty }\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}}. \end{aligned}$$

(3.10)

Proof

For $n\ge 2$ we have

$$\begin{aligned} \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!}=\left( {\begin{array}{c}s+n-2\\ n-1\end{array}}\right) \frac{1}{n}. \end{aligned}$$

By [16, p. 138, Proposition 47.2(v)], for $s\in {\mathbb {Z}}_{p},$

$$\begin{aligned} \left| \left( {\begin{array}{c}s+n-2\\ n-1\end{array}}\right) \right| _{p}\le 1. \end{aligned}$$

Since $|n|_{p}=\left( \frac{1}{p}\right) ^{v_{p}(n)}$, we have

$$\begin{aligned} \left| \frac{1}{n}\right| _{p}=p^{v_{p}(n)}\le n, \end{aligned}$$

thus for $n\ge 2$,

$$\begin{aligned} \left| \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!}\right| _{p}=\left| \left( {\begin{array}{c}s+n-2\\ n-1\end{array}}\right) \frac{1}{n}\right| _{p}\le n. \end{aligned}$$

(3.11)

Since for any $a\in {\mathbb {C}}_{p}^{\times }$, ${\hat{a}}\in \mu$ is a root of unity, we have $|{\hat{a}}|_{p}=1$ and by (2.3)

$$\begin{aligned} |\omega _{v}(a)|_{p}=|p^{v_{p}(a)}{\hat{a}}|_{p}=|p|_{p}^{v_{p}(a)}=\left( \frac{1}{p}\right) ^{v_{p}(a)}, \end{aligned}$$

thus

$$\begin{aligned} \left| \frac{1}{\omega _{v}^{n}(a)}\right| _{p}=p^{nv_{p}(a)}. \end{aligned}$$

(3.12)

Combining (3.11) and (3.12), for any fixed $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$ we have

$$\begin{aligned} \left| \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)} \right| _{p}\le p^{nv_{p}(a)}\cdot n. \end{aligned}$$

(3.13)

Now fix $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1,$ we know that $\log _{p}\langle y+a \rangle$ is a continuous function in $y\in {\mathbb {Z}}_{p}.$ Since ${\mathbb {Z}}_{p}$ is compact in the p-adic topology, by the Weierstrass maximum value theorem ([13, p. 61, Theorem 4.3]) there exists a $y_{0}\in {\mathbb {Z}}_{p}$ such that

$$\begin{aligned} |\log _{p}\langle y+a \rangle |_{p} \le |\log _{p}\langle y_{0}+a \rangle |_{p} \end{aligned}$$

(3.14)

for all $y\in {\mathbb {Z}}_{p}$. Since $\langle y_{0}+a \rangle -1\in (p)$, we have

$$\begin{aligned} |\langle y_{0}+a \rangle -1|_{p} \le p^{-1} < p^{-1/(p-1)} \end{aligned}$$

(3.15)

and by [20, p. 51, Lemma 5.5],

$$\begin{aligned} |\log _{p}\langle y_{0}+a \rangle |_{p}=|\langle y_{0}+a \rangle -1|_{p}\le p^{-1}. \end{aligned}$$

(3.16)

Combining (3.14), (3.15) and (3.16), we see that

$$\begin{aligned} |\log _{p}\langle y+a \rangle |_{p}\le p^{-1} \end{aligned}$$

(3.17)

for all $y\in {\mathbb {Z}}_{p}$. By [18, p. 1245, (2.22)], for $(x,s)\in {\mathbb {C}}_{p}^{\times }\times {\mathbb {C}}_{p}$ satisfying $|s|_{p} < p^{-1/(p-1)}|\log _{p}\langle x \rangle |_{p}^{-1},$ we have

$$\begin{aligned} \langle x \rangle ^{s}=\exp _{p}(s\log _{p} \langle x \rangle ). \end{aligned}$$

(3.18)

Let $D={\mathbb {Z}}_{p}\times {\mathbb {Z}}_{p}$. For $(y,s)\in D$, at first we have $|s|_{p}\le 1$ and by (3.17), we see that $p^{-1/(p-1)} |\log _{p}\langle y+a \rangle |_{p}^{-1}\ge p^{-1/(p-1)}\cdot p=p^{\frac{p-2}{p-1}}>1,$ thus $|s|_{p} <p^{-1/(p-1)}|\log _{p}\langle y+a \rangle |_{p}^{-1}.$ Then by (3.18) we have

$$\begin{aligned} \langle y+a \rangle ^{s}=\exp _{p}(s\log _{p} \langle y+a \rangle ) \end{aligned}$$

(3.19)

for $(y,s)\in D$. Hence the two variable function $f(y,s)=\langle y+a\rangle ^{s}$ is continuous on the domain D. Since $D={\mathbb {Z}}_{p}\times {\mathbb {Z}}_{p}$ is compact in the p-adic topology, for any fixed $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$ it is bounded as a function for $(y,s)\in D$, so there exists a positive constant $N_{a}$ such that for any $k\in {\mathbb {N}}$ and $n\in {\mathbb {N}},$

$$\begin{aligned} \left| \frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}}\right| _{p}= & {} \left| (-1)^{k} \langle k+a \rangle ^{1-s-n}\right| _{p}\nonumber \\= & {} \left| (-1)^{k} f(k,1-s-n)\right| _{p}\nonumber \\\le & {} N_{a} \end{aligned}$$

(3.20)

and by the non-archimedean property, for any $r\in {\mathbb {N}}$,

$$\begin{aligned} \left| \sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}}\right| _{p} \le N_{a}. \end{aligned}$$

(3.21)

Then combining (3.13) and (3.21), for any fixed $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$ and for any $r\in {\mathbb {N}}$ we have

$$\begin{aligned} \left| \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}}\right| _{p}\le N_{a}\cdot p^{nv_{p}(a)}\cdot n. \end{aligned}$$

(3.22)

Since $|a|_{p}>1$, i.e., $v_{p}(a)<0$, we have $\lim _{n\rightarrow \infty } N_{a}\cdot p^{nv_{p}(a)}\cdot n=0,$ which implies the series

$$\begin{aligned} N_{a} \sum _{n=2}^{\infty }p^{nv_{p}(a)} n \end{aligned}$$

(3.23)

is convergent. Finally by (3.22), (3.23) and the Weierstrass test (see [13, p. 230, Theorem 5.1]), we see that the series

$$\begin{aligned} \sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\sum _{k=1}^{p^{r}-1}\frac{(-1)^{k}}{\langle k+a \rangle ^{s+n-1}} \end{aligned}$$

converges uniformly for $r\in {\mathbb {N}}$. Then applying [13, p. 185, Theorem 3.5] we conclude that the limit $r\rightarrow \infty$ can be moved to the inside of the above series, which is the desired result. $\square$

The following result ensures the convergence of (3.1), which is a p-adic analogue of [15, Lemma 2].

Lemma 3.4

The left hand side of (3.1) in Lemma 3.1 converges p-adically for $s\in {\mathbb {Z}}_{p}$ with $s\ne 1$ and $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$.

Proof

By Proposition 2.1 and Theorem 2.3, for $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$, $\zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}$ is a $C^{\infty }$ function of s on ${\mathbb {Z}}_{p}$. Since ${\mathbb {Z}}_{p}$ is compact in the p-adic topology, for any fixed $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$ it is bounded as a function for $s\in {\mathbb {Z}}_{p}$, i.e., there exists a positive constant $M_{a}$ such that

$$\begin{aligned} \left| \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right| _{p}\le M_{a}. \end{aligned}$$

(3.24)

Then combining (3.13) and (3.24), for any fixed $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$ we have

$$\begin{aligned} \left| \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)} \left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \right| _{p}\le M_{a}\cdot p^{nv_{p}(a)}\cdot n. \end{aligned}$$

(3.25)

Since $|a|_{p}>1$, i.e., $v_{p}(a)<0$, we have $\lim _{n\rightarrow \infty } M_{a}\cdot p^{nv_{p}(a)}\cdot n=0,$ which implies

$$\begin{aligned} \lim _{n\rightarrow \infty }\left| \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \right| _{p}=0, \end{aligned}$$

thus the series

$$\begin{aligned} \sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \end{aligned}$$

is convergent under the p-adic topology. $\square$

The above result implies the following theorem.

Theorem 3.5

Let $T_{p}^{a}$ be as defined in (2.10). Then $\zeta _{p,E}(s,a)$ formally satisfies the following differential equation

$$\begin{aligned} T_{p}^{a}\left[ \zeta _{p,E}(s,a)-\langle a\rangle ^{1-s}\right] =\frac{1}{s-1}\left( \langle a-1\rangle ^{1-s}-\langle a\rangle ^{1-s}\right) \end{aligned}$$

(3.26)

for $s\in {\mathbb {Z}}_{p}$ with $s\ne 1$ and $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$.

Proof

Denote by $D_s^k := \frac{\partial ^k}{\partial s^k}$. For any analytic function f(s) on ${\mathbb {Z}}_{p}$ and $n\in {\mathbb {N}}$ we have

$$\begin{aligned} \exp _{p}(nD)f(s)= & {} \left( id + \sum _{k=1}^\infty \frac{n^k}{k!} D_s^k\right) f(s)\nonumber \\= & {} f(s)+ \sum _{k=1}^\infty \frac{n^k}{k!}\frac{\partial ^k f(s)}{\partial s^k}\nonumber \\= & {} f(s+n), \end{aligned}$$

(3.27)

which maybe interpreted operationally through its formal Taylor expansion in n. By Proposition 2.1 and Theorem 2.3, for $a\in {\mathbb {C}}_{p}$ with $|a|_{p}>1$, the function $\zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}$ is analytic for $s\in {\mathbb {Z}}_{p}.$ Thus from (3.27) we get

$$\begin{aligned} L_{p,n}^{a}\left[ \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right]= & {} P_{p,n}^{a}(s) \exp _{p}(nD)\left[ \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right] \nonumber \\= & {} P_{p,n}^{a}(s)\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \end{aligned}$$

(3.28)

for $n\ge 0$ and by the definition of $T_{p}^{a}$ (2.10) and Lemma 3.1

$$\begin{aligned}&T_{p}^{a}\left[ \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right] \nonumber \\&\quad =\sum _{n=0}^\infty L_{p,n}^{a}\left[ \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right] \nonumber \\&\quad =\sum _{n=0}^\infty P_{p,n}^{a}(s)\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \nonumber \\&\quad =\frac{2}{s-1}\left( \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right) +\frac{1}{\omega _{v}(a)}\left( \zeta _{p,E}(s+1,a)-\langle a \rangle ^{1-(s+1)}\right) \nonumber \\&\qquad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) \nonumber \\&\quad =\frac{1}{s-1}\left( \langle a-1 \rangle ^{1-s}-\langle a\rangle ^{1-s}\right) , \end{aligned}$$

(3.29)

which is the desired result. $\square$

In what follows, we shall investigate the area of convergence for Theorem 3.5 and show that the operator $T_{p}^{a}$ applied to the p-adic Hurwitz-type Euler zeta function $\zeta _{p,E}(s,a)$ is convergent in certain area of the p-adic plane. First we need to prove the following proposition.

Proposition 3.6

Let K be a finite extension of ${\mathbb {Q}}_{p}$ with ramification index e over ${\mathbb {Q}}_{p}$ less than $p-1.$ Let $s\in {\mathbb {C}}_{p}$ with $|s|_{p}<r_{p}:=p^{\frac{1}{e}-\frac{1}{p-1}},$ and $a\in K\backslash {\mathbb {Z}}_{p}.$ For any $n \ge 2$ the series

$$\begin{aligned} \exp _{p}(nD)\left[ \zeta _{p,E}(s,a) -\langle a \rangle ^{1-s}\right] =\sum _{k = 0}^\infty \frac{D_s^k \left( \zeta _{p,E}(s,a) -\langle a \rangle ^{1-s}\right) }{k!}n^k \end{aligned}$$

converges.

Remark 3.7

This is mainly because the non-archimedean property of the p-adic metric and it is quite different from the complex situation for Hurwitz zeta functions. In that case, by [15, Proposition 5], we have “for any $s \in {\mathbb {C}}$, we can find some $N \ge 0$ so that the series

$$\begin{aligned} \exp (ND)\left[ \zeta (s,a) - \frac{1}{a^s}\right] =\sum _{k = 0}^\infty \frac{D_s^k \left( \zeta (s,a) - \frac{1}{a^s}\right) }{k!}N^k \end{aligned}$$

diverges.”

Proof of Propsoition 3.6

Let $(\pi )$ be the maximal ideal of the ring of integers $O_{K}$ of K. Then

$$\begin{aligned} |\pi |_{p}=|p|_{p}^{\frac{1}{e}}=\left( \frac{1}{p}\right) ^{\frac{1}{e}}. \end{aligned}$$

By Proposition 2.1 and Theorem 2.3, given $a\in K\backslash {\mathbb {Z}}_{p},$ the function $\zeta _{p,E}(s,a) -\langle a \rangle ^{1-s}$ is analytic for

$$\begin{aligned} |s|_{p}<r_{p}:= |\pi |_{p}^{-1}p^{-1/(p-1)}=p^{\frac{1}{e}-\frac{1}{p-1}}. \end{aligned}$$

Fix $s_{0}\in {\mathbb {C}}_{p}$ with $|s_{0}|_{p}<r_{p}.$ For any $s\in {\mathbb {C}}_{p}$ with $|s-s_{0}|_{p}<r_{p}$, we have

$$\begin{aligned} |s|_{p}\le \max \{|s-s_{0}|_{p},|s_{0}|_{p}\}<r_{p}, \end{aligned}$$

so the disc $\{s: |s-s_{0}|< r_{p}\}$ is contained in the disc $\{s:|s|< r_{p}\}.$ In fact,

$$\begin{aligned} \{s: |s-s_{0}|< r_{p}\}=\{s:|s|< r_{p}\}. \end{aligned}$$

Thus $\zeta _{p,E}(s,a)$ can be expanded as a power series around $s_{0}$ with the radius of convergence equal to $r_{p}.$

Since $e < p-1$ as the assumption, we have $r_{p}>1$ and for any $n\in {\mathbb {N}},$ we have $|(s_{0}+n)-s_{0}|_{p}=|n|_{p}\le 1<r_{p}.$ From the discussion above, we have the following convergent power series expansion of $\zeta _{p,E}(s,a)$ at $s_{0}$

$$\begin{aligned} \zeta _{p,E}(s_{0}+n,a) -\langle a \rangle ^{1-(s_{0}+n)} =\sum _{k = 0}^\infty \frac{D_s^k\big |_{s=s_{0}} \left( \zeta _{p,E}(s,a) -\langle a \rangle ^{1-s}\right) }{k!}n^k. \end{aligned}$$

Then by the definition of exp$_{p}(nD)$ (2.10), we see that

$$\begin{aligned} \zeta _{p,E}(s_{0}+n,a) -\langle a \rangle ^{1-(s_{0}+n)}= & {} \sum _{k = 0}^\infty \frac{D_s^k\big |_{s=s_{0}} \left( \zeta _{p,E}(s,a) -\langle a \rangle ^{1-s}\right) }{k!}n^k\nonumber \\= & {} \exp _{p}(nD)\big |_{s=s_{0}} \left[ \zeta _{p,E}(s,a) -\langle a \rangle ^{1-s}\right] , \end{aligned}$$

(3.30)

which is the desired result. $\square$

From the above proposition we have the following result which asserts that the operator $T_{p}^{a}$ applied to the p-adic Hurwitz-type Euler zeta function $\zeta _{p,E}(s,a)$ is convergent in the p-adic topology.

Corollary 3.8

Let K be stated as in the Proposition 3.6. Then

$$\begin{aligned} T_{p}^{a}\left[ \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right] =\sum _{n=1}^\infty P_{p,n}^{a}(s)\exp _{p}(nD)\left[ \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right] \end{aligned}$$

(3.31)

converges for $s\in {\mathbb {Z}}_{p}$ with $s\ne 1$ and $a\in K$ with $|a|_{p}>1.$

Remark 3.9

Notice that in the complex situation, we have

$$\begin{aligned} T\left[ \zeta (s, a) - \frac{1}{a^s} \right] =\sum _{n=0}^{\infty } p_n(s)\exp (nD)\left[ \zeta (s,a)- \frac{1}{a^s}\right] \end{aligned}$$

diverges for all complex numbers $s\in {\mathbb {C}}$ (see [15, Theorem 8]).

Proof of Corollary 3.8

By (3.29) we have

$$\begin{aligned}&T_{p}^{a}\left[ \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right] \nonumber \\&\quad =\sum _{n=0}^\infty P_{p,n}^{a}(s)\exp _{p}(nD)\left[ \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right] \nonumber \\&\quad =\frac{2}{s-1}\left( \zeta _{p,E}(s,a)-\langle a \rangle ^{1-s}\right) +\frac{1}{\omega _{v}(a)}\left( \zeta _{p,E}(s+1,a)-\langle a \rangle ^{1-(s+1)}\right) \nonumber \\&\qquad +\sum _{n=2}^\infty \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)}\left( \zeta _{p,E}(s+n,a)-\langle a \rangle ^{1-(s+n)}\right) . \end{aligned}$$

(3.32)

Suppose that $s\in {\mathbb {Z}}_{p}$ with $s\ne 1$ and $a\in K$ with $|a|_{p}>1.$ By (2.11) and (3.13), for $n\ge 2$ we have

$$\begin{aligned} |P_{p,n}^{a}(s)|_{p}= \left| \frac{\prod _{j = 1}^{n - 1}(s-1+j)}{n!\omega _{v}^{n}(a)} \right| _{p}\le p^{nv_{p}(a)}\cdot n \end{aligned}$$

and $\lim _{n\rightarrow \infty } P_{p,n}^{a}(s)=0.$ Then combining the conclusions of Proposition 3.6 and Lemma 3.4, for each $n \ge 2,$ both the series

$$\begin{aligned} \exp _{p}(nD)\left[ \zeta _{p,E}(s,a) -\langle a \rangle ^{1-s}\right] =\sum _{k = 0}^\infty \frac{D_s^k \left( \zeta _{p,E}(s,a) -\langle a \rangle ^{1-s}\right) }{k!}n^k \end{aligned}$$

and the right hand side of (3.32) converge, which have established our result. $\square$