# Euler Numbers and Polynomials Associated  with Zeta Functions

Authors: Taekyun Kim
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

Fors∈ℂ, the Euler zeta function and the Hurwitz-type Euler zeta
function are defined by ζE(s)=2∑n=1∞((−1)n/ns), and ζE(s,x)=2∑n=0∞((−1)n/(n+x)s). Thus, we note that the Euler zeta functions are entire functions in whole complex s-plane, and these zeta functions have the values of the Euler numbers or the Euler
polynomials at negative integers. That is, ζE(−k)=Ek∗, and ζE(−k,x)=Ek∗(x). We give some interesting identities between the Euler numbers and the zeta functions. Finally, we will give the new values of the Euler zeta function at positive even integers.


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## Body

## 1. Introduction

Throughout this paper,ℤ, ℚ, ℂ, ℤp, ℚp, and ℂp will, respectively, denote the ring of rational
integers, the field of rational numbers, the field of complex numbers, the ring p-adic rational integers, the field of p-adic rational numbers, and the completion of
the algebraic closure of ℚp.
Let vp be the normalized exponential valuation of ℂp such that |p|p=p−vp(p)=p−1.
If q∈ℂp,
we normally assume that |q−1|p<1.
We use the notation(1.1)[x]q=1−qx1−q,[x]−q=1−(−q)x1+q.Hence, lim⁡q→1[x]q=1, for any x with |x|p≤1 in the present p-adic case.Letp be a fixed odd prime. For d(=odd), a fixed positive integer with (p,d)=1,
let(1.2)X=Xd=lim⁡N←ℤdpNℤ,X1=ℤp,X*=⋃0<a<dp(a,p)=1(a+dpℤp)a+dpNℤp={x∈X∣x≡a⁢(mod⁡dpn)},where a∈ℤ  lies in 0≤a<dpN.In [1], we note that(1.3)μ−q(a+dpNℤp)=(1+q)(−1)aqa1+qdpN=(−q)a[dpN]−qis distribution on X for q∈ℂp with |1−q|p<1. This distribution yields an integral as
follows:(1.4)I−q(f)=∫ℤpf(x)dμ−q(x)=lim⁡N→∞1[pN]−q∑x=0pN−1f(x)(−q)x,forf∈UD(Zp), which has a sense as we
see readily that the limit is convergent (see [1]). Let q=1.
Then, we have the fermionic p-adic integral on ℤp as follows:(1.5)I−1(f)=∫ℤpf(x)dμ−1(x)=lim⁡N→∞∑x=0pN−1f(x)(−1)x, (cf. [1–5]). For any positive integer N, we set(1.6)μq(a+lpNℤp)=qa[lpN]q  (cf. [1–3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]) and this can be extended to a
distribution on X.
This distribution yields p-adic bosonic q-integral as follows (see [11, 20]):(1.7)Iq(f)=∫ℤpf(x)dμq(x)=∫Xf(x)dμq(x),where f∈UD(ℤp)=the space of uniformly differentiable function onℤp with values in ℂp,
(cf. [2, 11, 16–20]). In view of notation, I−1 can be written symbolically as I−1(f)=lim⁡q→−1Iq(f). If we take f(x)=q−x[x]qn,
then we can derive the q-extension of Bernoulli numbers and
polynomials from p-adic q-integrals on ℤp as follows:(1.8)βn,q=∫ℤpq−x[x]qndμq(x),βn,q(x)=∫ℤpq−y[y+x]qdμq(y)
(cf. [11, 20]). Thus, we note that(1.9)β0,q=q−1log⁡q,βm,q=1(q−1)m∑i=0m(mi)i[i]q,
(cf. [11, 14, 20]). In the complex plane, the
ordinary Bernoulli numbers are a sequence of signed rational numbers that can
be defined by the identity(1.10)tet−1=∑n=0∞Bntnn!,|t|<2π
(cf. [1–33]).These numbers arise in the series expansions of
trigonometric functions, and are extremely important in number theory and
analysis. From the generating function of Bernoulli numbers, we note thatB0=1,B1=−1/2,⁢B2=1/6,⁢B4=−1/30,⁢B6=1/42,⁢B8=−1/30,⁢B10=5/66,⁢B12=−691/2730,⁢B14=7/6,⁢B16=−3617/510,⁢B18=43867/798,⁢B20=−174611/330,…, and B2k+1=0 for k∈ℕ.
It is well known that Riemann zeta function is defined by(1.11)ζ(s)=∑n=1∞1ns,fors∈C. We also note that the
Riemann zeta function is closely related to Bernoulli numbers at positive
integer or negative integer in the complex plane. Riemann did develop the
theory of analytic continuation needed to rigorously define ζ(s) for all s∈ℂ−{0}.
From this zeta function, he derived the following formula (cf. [1–33]):(1.12)ζ(−n)=−Bn+1n+1,n∈N={1,2,3,…}. Thus, we note that ζ(−n)=0 if n is an even integer and greater than 0. These
are called the trivial zeros of the zeta function. In 1859, starting with
Euler's factorization of the zeta function(1.13)ζ(s)=∏p:prime11−p−s,he derived an explicit formula
for the prime numbers in terms of zeros of the zeta function. He also posed the
Riemann hypothesis: if ζ(z)=0, then either z is a trivial zero or z lies on the critical line Re⁡(z)=1/2 (cf. [4, 5, 16–20, 27, 28, 29, 30, 31, 32, 33]). It is well known that(1.14)sinzz=(1−z2π2)(1−z2(2π)2)(1−z3(3π)2)….Thus,(1.15)1−zcot⁡z=2∑m=1∞ζ(2m)π2mz2m
(cf. [4, 5, 10, 16–20, 27, 28, 29, 30, 31, 32, 33]).
From this, we can derive the
following famous formula.Lemma 1.1.
Forn∈ℕ,(1.16)ζ(2n)=∑k=1∞1k2n=(−1)n−1(2π)2n2(2n)!B2n,forn∈N.It is easy to see that(1.17)zcot⁡z=2ize2iz−1+iz=1+∑k=1∞(−1)k22kB2k(2k)!z2k.However, it is not known the
values of ζ(2k+1) for k∈ℕ.
In the case of k=1,
Apery proved that ζ(3) is irrational number (see [34]). The constants Ek* in the Taylor series expansion(1.18)2et+1=∑n=0∞En*tnn!,where|t|<π (cf. [3–5, 10, 27]) are known as the first-kind Euler numbers. From the generating function of the first-kind Euler
numbers, we note that(1.19)E0*=1,En*=−∑l=0n(nl)El*,forn∈N. The first few are 1,⁢−1/2,⁢0,⁢1/4,…, and E2k*=0 for k=1,2,…. The Euler polynomials are also defined by(1.20)2et+1ext=∑n=0∞En*(x)tnn!=∑n=0∞(∑k=0n(nk)Ek*xn−k)tnn!.For s∈ℂ,
the Euler zeta function and Hurwitz's type Euler zeta function are defined by(1.21)ζE(s)=2∑n=1∞(−1)nns,ζE(s,x)=2∑n=0∞(−1)n(n+x)s (cf. [2, 4, 5, 9, 10, 27]). Thus, we note that Euler zeta
functions are entire functions in the whole complex s-plane and these zeta functions have the
values of the Euler numbers or Euler polynomials at negative integers. That is,(1.22)ζE(−k)=Ek*,ζE(−k,x)=Ek*(x) (cf. [2, 4, 5, 9, 10, 27]).In this paper, we give some interesting identities
between Euler numbers and zeta functions. Finally, we will give the values of
the Euler zeta function at positive even integers.


## 2. Preliminaries/Euler Numbers Associated withp-adic Fermionic Integrals

Letf1(x) be the translation defined by f1(x)=f(x+1).
Then we have(2.1)I−1(f1)=−I−1(f)+2f(0).If we take f(x)=e(x+y)t,
then we can derive the first-kind Euler polynomials from the integral equation
of I−1(f) as follows:(2.2)∫ℤpe(x+y)tdμ−1(y)=ext2et+1=∑n=0∞En*(x)tnn!.That is,(2.3)∫ℤpyndμ−1(y)=En*,∫ℤp(x+y)ndμ−1(y)=En*(x).For n∈ℕ,
we have the following integral equation:(2.4)∫ℤpf(x+n)dμ−1(x)+(−1)n−1∫ℤpf(x)dμ−1(x)=2∑l=0n−1(−1)n−1+lf(l).From this we note
that(2.5)Ek*(n)−Ek*=2∑l=0n−1(−1)l−1lk,ifn≡0⁢(mod⁡2),Ek*(n)+Ek*=2∑l=0n−1(−1)llk,ifn≡1⁢(mod⁡2).Letf(x)=sinax (or f(x)=cosax). By using the fermionic p-adic q-integral on ℤp,
we see that(2.6)0=∫ℤpsinax⁢dμ−1(x)+∫ℤpsinax⁢dμ−1(x)=(cosa+1)∫ℤpsinax⁢dμ−1(x)+sina∫ℤpcosax⁢dμ−1(x), see [12],(2.7)2=(cosa+1)∫ℤpcosax⁢dμ−1(x)−sina∫ℤpsinax⁢dμ−1(x).Thus, we obtain(2.8)∫ℤpcosax⁢dμ−1(x)=1,∫ℤpsinax⁢dμ−1(x)=−sinacosa+1, see [12].
From this we note that(2.9)tan⁡a2=−∫ℤpsinax⁢dμ−1(x)=∑n=0∞(−1)n+1a2n+1(2n+1)!E2n+1*.By the same motivation, we can
also observe that(2.10)a2cot⁡a2=∫ℤpcosax⁢dμ1(x)=∑n=0∞(−1)nB2n(2n)!a2n, see [12]. These formulae are also treated
in Section 3.Letf(x)=et(2x+1).
Then we can derive the generating function of the second-kind Euler numbers
from fermionic p-adic integral equation as follows:(2.11)∫ℤpet(2x+1)dμ−1(x)=2et+e−t=1cosht=∑n=0∞Entnn!.Thus, we have(2.12)(E+1)n+(E−1)n=2δ0,n,where we have used the symbolic
notation En for En.
The first few are E0=1,E1=0,E2=−1,⁢E3=0,E4=5,…, E2k+1=0 for k∈ℕ.
In particular,(2.13)E2n=−∑k=0n−1(2n2k)E2k.Recently, Simsek, Ozden, Cangül,
Cenkci, Kurt, and others have studied the various extensions of the first kind Euler
numbers by using fernionic p-adic invariant q-integral on ℤp,
see [2–5, 16, 20, 27]. It seems to be also interesting to study the q-extensions of the second-kind Euler numbers
due to Simsek et al. (see [4, 5, 16]).


## 3. Some Relationships between Euler Numbers and Zeta Functions

In this section, we also consider Bernoulli and the
second Euler numbers in the complex plane. The second-kind Euler numbersEk are defined by the following expansion:(3.1)sech⁢x=1coshx=2exe2x+1=∑k=0∞Ekxkk!,for|x|<π2 (cf. [10]). From (1.18) and (3.1), we can
derive the following equation:(3.2)Ek=∑l=0k(kl)2lEl*,where(kl)⁢is binomial coefficient. By (3.2) and (1.18), we
easily see that E0=1,E1=0,E2=−1,⁢E3=0,E4=5,E6=61,…, and E2k+1=0 for k=1,2,3,…. As Euler formula, it is well known that(3.3)eix=cosx+i⁢sinx,wherei=(−1)1/2. From (3.3), we note that cosx=(eix+e−ix)/2. Thus, we have(3.4)sec⁡x=2eix+e−ix=sech(ix)=∑n=0∞inEnn!xn=∑n=0∞(−1)nE2n(2n)!x2n+i∑n=0∞(−1)nE2n+1(2n+1)!x2n+1=∑n=0∞(−1)nE2n(2n)!x2n.From (3.4), we derive(3.5)x⁢secx=∑n=0∞(−1)nE2n(2n)!x2n+1,for|x|<π2. The Fourier series of an
odd function on the interval (−p,p) is the sine series:(3.6)f(x)=∑n=1∞bn⁢sin⁡(nπxp),where(3.7)bn=2p∫0pf(x)sin⁡(nπxp)dx.Let us consider f(x)=sinax on [−π,π].
From (3.6) and (3.7), we note that(3.8)sinax=∑n=1∞bn⁢sinnx,where(3.9)bn=2π∫0πsinax⁢sinnx⁢dx=2π∫0π[cos⁡(n−a)x−cos⁡(n+a)x2]dx=1π[sin⁡(n−a)xn−a−sin⁡(n+a)xn+a]0π=(−1)n−12π⁢sinaπ(nn2−a2).In (3.8), if we take x=π/2,
then we have(3.10)sin⁡πa2=∑n=1∞b2n−1(−1)n−1=2π⁢sinaπ∑n=1∞(−1)n−12n−1(2n−1)2−a2=2π⁢sinaπ∑n=1∞(2n−1)(−1)n−1(2n−1)2(1−(a/(2n−1))2)=2π⁢sinaπ∑n=1∞(−1)n−12n−1∑k=0∞a2k(2n−1)2k=2π⁢sinaπ∑k=0∞(∑n=1∞(−1)n−1(2n−1)2k+1)a2kFrom (3.10), we note that(3.11)πa2sec⁡(πa2)=∑k=0∞(2∑n=1∞(−1)n−1(2n−1)2k+1)a2k+1.In (3.5), it is easy to see that(3.12)πa2sec⁡(πa2)=∑n=0∞(−1)nE2n(2n)!(π2)2n+1a2n+1.By (3.11) and (3.12), we obtain the
following.Theorem 3.1.
Forn∈ℕ,
(3.13)∑k=1∞(−1)k−1(2k−1)2n+1=∑k=0∞(−1)k(2k+1)2n+1=(−1)nE2n2(2n)!(π2)2n+1.It is easy to see that(3.14)∑n=1∞(−1)n(2n+1)2k+1=2∑n=1∞1(4n−3)2k+1+∑n=1∞1(2n)2k+1−∑n=1∞1n2k+1−1=124k+1ζ(2k+1,14)−22k+1−122k+1ζ(2k+1)−1.By (3.13) and (3.14), we obtain the
following.Corollary 3.2.
Forn∈ℕ,(3.15)ζ(2n+1,14)+22n(1−22n+1)ζ(2n+1)=(−1)nE2n2(2n)!π2n+122n.By simple calculation, we easily see that(3.16)i⁢tanx=eix−e−ixeix+e−ix=1−2e2ix−1+4e4ix−1.Thus, we have(3.17)x⁢tanx=−xi+2xie2xi−1−4xie4xi−1=∑n=1∞(−1)nB2n4n(1−4n)(2n)!x2n.From (3.17), we can easily derive(3.18)tanx=∑n=0∞(−1)n+14n+1(1−4n+1)B2n+2(2n+2)!x2n+1.By (3.3), we also see that(3.19)i⁢tanx=1−2e2ix+1=i∑n=0∞E2n+1*(2n+1)!22n+1(−1)n+1x2n+1.Thus, we have(3.20)tanx=∑n=0∞E2n+1*(2n+1)!22n+1(−1)n+1x2n+1.By (3.18) and (3.20), we obtain the
following.Theorem 3.3.
Forn∈ℕ,
(3.21)ζ(2n)=(−1)n−1(2π)2nE2n−1*4(2n−1)!(1−4n),where En* are the first-kind Euler numbers.It is easy to see that(3.22)∑k=1∞1(2k+1)2n=(1−14n)ζ(2n)=(−1)n(2π)2n4n+1(2n−1)!E2n−1*.Therefore, we obtain the
following corollary.Corollary 3.4.
Forn∈ℕ,(3.23)∑k=1∞1(2k+1)2n=(−1)n(2π)2n4n+1(2n−1)!E2n−1*.Now we try to give the new value of the Euler zeta
function at positive integers. From the definition of the Euler zeta function,
we note that(3.24)ζE(s)=2∑n=1∞(−1)nns=−2∑n=0∞1(2n+1)s+12s−1ζ(s),s∈C. By (3.24), Theorem 3.3, and
Corollary 3.4, we obtain the following theorem.Theorem 3.5.
Forn∈ℕ,(3.25)ζE(2n)=(−1)n−1π2n(2−4n)2(2n−1)!(1−4n)E2n−1*.Remark 3.6.
We note thatζ(2)=π2/6, ζE(2)=−π2/6, ζ(4)=π4/90 and ζE(4)=−7π4/360….
For q∈ℂ with |q|<1,  s∈ℂ, q-ζ-function is defined by(3.26)ζq(s)=∑n=1∞qn[n]qs−1s−1(1−q)slogq (cf. [10, 14]). Note that ζq(s) is analytic continuation in ℂ with only one simple pole at s=1,
and(3.27)ζq(1−k)=−βk,qk,wherek⁢is a positive integer (cf. [14]). By simple calculation,
we easily see that(3.28)∑n=1∞(−1)nqn[n]q2k+1∑j=0∞θ2j+1[n]q2j+1(2j+1)!+1logq∑j=0k−1(1−q)2k−2j(2k−2j−1)(2j+1)!−1logq∑j=0k−1(1−q)2k−2j(2k−2j−1)(2j+1)!=∑j=0k−1(−1)jθ2j+1(2j+1)!(−ζq(2k−2j)+ζq2(2k−2j)2[2]q2k−2j)−q1+qθ2k+1(2k+1)!(−1)k+∑j=k+1∞θ2j+1(−1)j(2j+1)!H2j−2k,q(−q−1)1+q,where Hn,q(−q) are Carlitz's q-Euler numbers with lim⁡q→1Hn,q(−q)=En* (cf. [6, 21, 22]). If q→1,
then we have(3.29)∑n=1∞(−1)nn2k+1sin⁡(nθ)=∑j=0k−1(−1)j(2j+1)!θ2j+1(222k−2j−1)×(−1)k−j+1(2π)2k−2j2⁢⋅(2k−2j)!B2k−2j−12θ2k+1(2k+1)!(−1)k.For k∈ℕ,
and θ=π/2,
it is easy to see that(3.30)∑n=1∞(−1)n(2n−1)2k+1=∑j=0k−1(−1)kπ2k+1(22k−2j−2)B2k−2j(2j+1)!(2k−2j)!22j+2−π2k+1(−1)k(2k+1)!22k+2.From (3.30) and Theorem 3.1, we can
also derive the following equation:(3.31)∑j=0k−1(−1)k−1π2k+1(22k−2j−2)B2k−2j(2j+1)!(2k−2j)!22j+2+π2k+1(−1)k(2k+1)!22k+2=(−1)kE2k2(2k)!(π2)2k+1.Thus, we have(3.32)∑j=0k−1(22k−2j−2)B2k−2j(2j+1)!(2k−2j)!22j+2=1(2k+1)!22k+2−E2k22k+2(2k)!.



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