# A Newq-Analogue of Bernoulli Polynomials Associated with p-Adic q-Integrals

Authors: Lee-Chae Jang
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

We will study a newq-analogue of Bernoulli polynomials associated with p-adic q-integrals. Furthermore, we examine the Hurwitz-type q-zeta functions, replacing p-adic rational integers x with a q-analogue [x]q for a p-adic number q with |q−1|p<1, which interpolate q-analogue of Bernoulli polynomials.


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## Body

## 1. Introduction

Letp be a fixed odd prime number. Throughout this
paper ℤp, ℚp, ℂ,
and ℂp will, respectively,
represent the ring of p-adic rational integers, the field of p-adic rational numbers, the complex number
field, and the p-adic completion of the algebraic closure of ℚp.
The p-adic absolute value in ℂp is normalized so that |p|p=1/p and q is a p-adic number in ℂp with |q−1|p<1.
We use the notation(1.1)[x]q=1−qx1−q(cf. [1–13]) for all x∈ℤp.
Hence, limq→1|x|q=x.
For a fixed odd positive integer d with (p,d)=1,
let(1.2)X=Xd=limn←ℤdpnℤ,X1=ℤp,X∗=→0<a<dp(a,p)=1∪(a+dpℤp),a+dpnℤp={x∈X∣x≡a(moddpn)},where a∈ℤ lies in 0≤a<dpn.
For any n∈ℕ,(1.3)μq(a+dpnℤp)=qa[dpn]qis known to be a distribution on X (cf. [1–13]).We say thatf is uniformly differentiable function at a
point a∈ℤp and denote this property by f∈UD(ℤp),
if the difference quotients(1.4)Ff(x,y)=f(x)−f(y)x−yhave a limit l=f′(a) as (x,y)→(a,a) (cf. [2, 6, 7]). The p-adic q-integral of a function f∈UD(ℤp) was defined as(1.5)Iq(f)=∫ℤpf(x)dμq(x)=limN→∞1[pN]q∑x=0pN−1f(x)qx.By usingp-adic q-integrals on ℤp,
it is well known that(1.6)tet−1ext=∑n=0∞∫ℤp(x+s)ndμ1(s)tnn!,where μ1(x+pnℤp)=1/pn.
Then we note that the Bernoulli polynomials Bn(x) were defined as(1.7)tet−1ext=∑n=0∞Bn(x)tnn!.From (1.6) and (1.7), we
have(1.8)Bn(x)=∫ℤp(x+s)ndμ1(s)for all n∈ℕ∪{0}.
We note that [0]q=(1−q0)/(1−q)=0.In Section2, we study a q-analogue of Bernoulli polynomials associated
with p-adic q-integrals—simply, we say q-Bernoulli polynomials. In Section 3, we
examine Hurwitz-type  q-zeta functions, replacing p-adic rational integers x with a q-analogue [x]q for a p-adic number q with |q−1|p<1,
which interpolate q-analogue of Bernoulli polynomials.


## 2. A Newq-Analogue of Bernoulli Polynomials

In this section, from the view of (1.8), we can define a
new q-analogue of Bernoulli polynomials as
follows:(2.1)βnq(x)=∫ℤp([x]q+[s]q)ndμq(s).We note that βnq=βnq(0) are called the q-Bernoulli numbers. Then we find some properties
of q-Bernoulli numbers and polynomials as
follows.Theorem 2.1.
Forn∈ℕ∪{0},
one has(2.2)βnq=1(1−q)n∑l=0n(nl)(−1)l1[l+1]q.Proof .
From (1.5) with x=0,
we can find the following:(2.3)βnq=∫ℤp[s]qndμq(s)=limN→∞∑j=0pN−1[j]qnqj[pN]q=∑l=0n(nl)(−1)l1(1−q)n−1limN→∞∑j=0pN−1qj(l+1)11−qpN=1(1−q)n∑l=0n(nl)(−1)l1[l+1]q.Theorem 2.2.
Forn∈ℕ∪{0} and d being an odd
positive integer with (p,d)=1, one has(2.4)βnq(x)=[d]qn−1∑l=0n(nl)βlqd∑i=0d−1qi(l+1)([xd]qd+[id]qd)n−l.Proof .
From (1.5), we can
derive (2.4) as follows:
(2.5)βnq(x)=∫ℤp([x]q+[s]q)ndμq(s)=limN→∞1[dpN]q∑a=0dpN−1([x]q+[a]q)nqa=limN→∞1−q1−qdpN∑i=0d−1∑k=0pN−1([x]q+[i+dk]q)nqi+dk=limN→∞1[d]q1[pN]qd∑i=0d−1qi∑k=0pN−1[d]qn([xd]qd+[id]qd+qi[k]qd)n(qd)k=limN→∞1[d]q1[pN]qd∑i=0d−1qi∑k=0pN−1∑l=0n(nl)([xd]qd+[id]qd)n−l(qi[k]qd)l(qd)k=[d]qn−1∑i=0d−1∑l=0n(nl)qi(l+1)([xd]qd+[id]qd)n−llimN→∞1[pN]qd∑k=0pN−1[k]qdl(qd)k=[d]qn−1∑i=0d−1∑l=0n(nl)qi(l+1)([xd]qd+[id]qd)n−l∫ℤp[s]qdldμqd(s)=[d]qn−1∑l=0n(nl)βlqd∑i=0d−1qi(l+1)([xd]qd+[id]qd)n−l,since a=i+dk and(2.6)([x]q+[i+dk]q)n=[d]qn([xd]qd+[id]qd+qi[k]qd)nfor a=0,1,…,dpN−1, i=0,1,…,d−1, and k=0,1,…,pN−1.
LetGq(x,t) be the generating function of q-Bernoulli polynomials as
follows:(2.7)Gq(x,t)=∑n=0∞βnq(x)tnn!.From (2.2) and (2.7), we can obtain
the following theorem.Theorem 2.3.
LetGq(x,t) be as in the above generating function. Then,
one has(2.8)Gq(x,t)=(1−q)∑m=0∞qme([x]q+[m]q)t.Proof .
By using (2.2) and
(2.7), we can derive (2.8) as follows:(2.9)Gq(x,t)=e([x]q+βq)t=e[x]qteβqt=e[x]qt∑n=0∞βnqtnn!=e[x]qt∑n=0∞1(1−q)n∑l=0n(nl)(−1)l1[l+1]qtnn!=e[x]qt∑n=0∞1(1−q)n−1∑l=0n(nl)(−1)l11−ql+1tnn!=e[x]qt∑n=0∞1(1−q)n−1∑l=0n(nl)(−1)l∑m=0∞q(l+1)mtnn!=e[x]qt(1−q)∑m=0∞qm∑n=0∞1(1−q)n∑l=0n(nl)(−1)lqlmtnn!=e[x]qt(1−q)∑m=0∞qm∑n=0∞[m]qntnn!=(1−q)∑m=0∞qme([x]q+[m]q)t.


## 3. A New Formula for Hurwitz-Typeq-Zeta Functions

In this section, we consider the generating functionsF(t,x) which interpolate the q-Bernoulli polynomials βn*q(x) as follows:(3.1)F(t,x)=∑m=0∞qme([x]q+[m]q)t=∑m=0∞βn*q(x)tmm!.From (3.1), we directly obtain
the following theorem.Theorem 3.1.
For eachk∈ℕ∪{0},
one has(3.2)βk*q(x)=∑m=0∞qm([x]q+[m]q)k.Proof .
By thekth differentiation on both sides of (3.1), we
can derive (3.2) as follows:(3.3)βk*q(x)=dkdtkF(x,t)|t=0=∑m=0∞qm([x]q+[m]q)k.
We remark that(3.4)−βk*q(x)k=1k∑m=0∞qm([x]q+[m]q)kfor k∈ℕ.
From (3.2), we derive a q-extension of
Hurwitz-type zeta function as follows: for s∈ℂ with ℜ(s)>1 and ℜ(x)>0,
we define(3.5)ζq(s,x)=11−s∑m=0∞qm([x]q+[m]q)s.Note that the functions ζq(s,x) are analytic on ℜ(s)>1 and they have simple pole at s=1.
From (3.2), (3.4), and (3.5), we can see that
Hurwitz-type  q-zeta functions interpolate q-Bernoulli polynomials as follows.Theorem 3.2.
For eachk∈ℕ,
one has(3.6)ζq(1−k,x)=−βk*q(x)k.



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*Source: 295307-2008-10-12.xml*