Euler Numbers and Polynomials Associated with Zeta Functions

Abstract

For 𝑠∈ℂ, the Euler zeta function and the Hurwitz-type Euler zeta function are defined by 𝜁_𝐸(𝑠)=2∑^∞_𝑛=1((−1)^𝑛/𝑛^𝑠), and 𝜁_𝐸(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, 𝜁_𝐸(−𝑘)=𝐸^∗_𝑘, and 𝜁_𝐸(−𝑘,𝑥)=𝐸^∗_𝑘(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.

1. Introduction

Throughout this paper, ℤ, ℚ, ℂ, ℤ_𝑝, ℚ_𝑝, and ℂ_𝑝 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 ℚ_𝑝. Let 𝑣_𝑝 be the normalized exponential valuation of ℂ_𝑝 such that |𝑝|_𝑝=𝑝^{−𝑣𝑝(𝑝)}=𝑝⁻¹. If 𝑞∈ℂ_𝑝, we normally assume that |𝑞−1|_𝑝<1. We use the notation[𝑥]_𝑞=1−𝑞^𝑥1−𝑞,[𝑥]_−𝑞=1−(−𝑞)^𝑥1+𝑞.(1.1)Hence, lim_𝑞→1[𝑥]_𝑞=1, for any 𝑥 with |𝑥|_𝑝≤1 in the present 𝑝-adic case.

Let 𝑝 be a fixed odd prime. For 𝑑(=𝑜𝑑𝑑), a fixed positive integer with (𝑝,𝑑)=1, let𝑋=𝑋_𝑑=lim_←𝑁ℤ𝑑𝑝^𝑁ℤ,𝑋₁=ℤ_𝑝,𝑋^∗=⋃_{0<𝑎<𝑑𝑝(𝑎,𝑝)=1}(𝑎+𝑑𝑝ℤ_𝑝)𝑎+𝑑𝑝^𝑁ℤ_𝑝={𝑥∈𝑋∣𝑥≡𝑎(mod𝑑𝑝^𝑛)},(1.2)where 𝑎∈ℤ lies in 0≤𝑎<𝑑𝑝^𝑁.

In [1], we note that𝜇_−𝑞(𝑎+𝑑𝑝^𝑁ℤ_𝑝)=(1+𝑞)(−1)^𝑎𝑞^𝑎1+𝑞^𝑑𝑝𝑁=(−𝑞)^𝑎[𝑑𝑝^𝑁]_−𝑞(1.3)is distribution on 𝑋 for 𝑞∈ℂ_𝑝 with |1−𝑞|_𝑝<1. This distribution yields an integral as follows:𝐼_−𝑞(𝑓)=_ℤ𝑝𝑓(𝑥)𝑑𝜇_−𝑞(𝑥)=lim_𝑁→∞1[𝑝^𝑁]^𝑝_−𝑞^𝑁−1_𝑥=0𝑓(𝑥)(−𝑞)^𝑥,for𝑓∈𝑈𝐷(𝑍_𝑝),(1.4) 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 ℤ_𝑝 as follows:𝐼₋₁(𝑓)=_ℤ𝑝𝑓(𝑥)𝑑𝜇₋₁(𝑥)=lim^𝑝_𝑁→∞^𝑁−1_𝑥=0𝑓(𝑥)(−1)^𝑥,(1.5) (cf. [1–5]). For any positive integer 𝑁, we set𝜇_𝑞(𝑎+𝑙𝑝^𝑁ℤ_𝑝𝑞)=^𝑎[𝑙𝑝^𝑁]_𝑞(1.6) (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]):𝐼_𝑞(𝑓)=_ℤ𝑝𝑓(𝑥)𝑑𝜇_𝑞(𝑥)=_𝑋𝑓(𝑥)𝑑𝜇_𝑞(𝑥),(1.7)where 𝑓∈𝑈𝐷(ℤ_𝑝)=thespaceofuniformlydifferentiablefunctiononℤ_𝑝 with values in ℂ_𝑝, (cf. [2, 11, 16–20]). In view of notation, 𝐼₋₁ can be written symbolically as 𝐼₋₁(𝑓)=lim_𝑞→−1𝐼_𝑞(𝑓). If we take 𝑓(𝑥)=𝑞^−𝑥[𝑥]^𝑛_𝑞, then we can derive the 𝑞-extension of Bernoulli numbers and polynomials from 𝑝-adic 𝑞-integrals on ℤ_𝑝 as follows:𝛽_𝑛,𝑞=_ℤ𝑝𝑞^−𝑥[𝑥]^𝑛_𝑞𝑑𝜇_𝑞(𝑥),𝛽_𝑛,𝑞(𝑥)=_ℤ𝑝𝑞^−𝑦[𝑦+𝑥]_𝑞𝑑𝜇_𝑞(𝑦)(1.8) (cf. [11, 20]). Thus, we note that𝛽_0,𝑞=𝑞−1log𝑞,𝛽_𝑚,𝑞=1(𝑞−1)^𝑚𝑚_𝑖=0𝑚𝑖𝑖[𝑖]_𝑞,(1.9) (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

𝑡𝑒^𝑡=−1^∞_𝑛=0𝐵_𝑛𝑡^𝑛𝑛!,|𝑡|<2𝜋(1.10) (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 that 𝐵₀=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 𝐵_2𝑘+1=0 for 𝑘∈ℕ. It is well known that Riemann zeta function is defined by𝜁(𝑠)=^∞_𝑛=11𝑛^𝑠,for𝑠∈𝐶.(1.11) 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]):𝐵𝜁(−𝑛)=−^𝑛+1𝑛+1,𝑛∈𝑁={1,2,3,…}.(1.12) 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 function𝜁(𝑠)=_𝑝∶prime11−𝑝^−𝑠,(1.13)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 thatsin𝑧𝑧𝑧=(1−²𝜋²𝑧)(1−²(2𝜋)²𝑧)(1−³(3𝜋)²)….(1.14)Thus,1−𝑧cot𝑧=2^∞_𝑚=1𝜁(2𝑚)𝜋^2𝑚𝑧^2𝑚(1.15) (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. For 𝑛∈ℕ,𝜁(2𝑛)=^∞_𝑘=11𝑘^2𝑛=(−1)^𝑛−1(2𝜋)^2𝑛𝐵2(2𝑛)!^2𝑛,for𝑛∈𝑁.(1.16)

It is easy to see that𝑧cot𝑧=2𝑖𝑧𝑒^2𝑖𝑧−1+𝑖𝑧=1+^∞_𝑘=1(−1)^𝑘2^2𝑘𝐵_2𝑘𝑧(2𝑘)!^2𝑘.(1.17)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 𝐸^∗_𝑘 in the Taylor series expansion2𝑒^𝑡=+1^∞_𝑛=0𝐸^∗_𝑛𝑡^𝑛𝑛!,where|𝑡|<𝜋(1.18) (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 that𝐸^∗₀=1,𝐸^∗_𝑛=−^𝑛_𝑙=0𝑛𝑙𝐸^∗_𝑙,for𝑛∈𝑁.(1.19) The first few are 1,−1/2,0,1/4,…, and 𝐸^∗_2𝑘=0 for 𝑘=1,2,…. The Euler polynomials are also defined by2𝑒^𝑡𝑒+1^𝑥𝑡=^∞_𝑛=0𝐸^∗_𝑛𝑡(𝑥)^𝑛=𝑛!^∞_𝑛=0(^𝑛_𝑘=0𝑛𝑘𝐸^∗_𝑘𝑥^𝑛−𝑘)𝑡^𝑛𝑛!.(1.20)For 𝑠∈ℂ, the Euler zeta function and Hurwitz's type Euler zeta function are defined by𝜁_𝐸(𝑠)=2^∞_𝑛=1(−1)^𝑛𝑛^𝑠,𝜁_𝐸(𝑠,𝑥)=2^∞_𝑛=0(−1)^𝑛(𝑛+𝑥)^𝑠(1.21) (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,𝜁_𝐸(−𝑘)=𝐸^∗_𝑘,𝜁_𝐸(−𝑘,𝑥)=𝐸^∗_𝑘(𝑥)(1.22) (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 with 𝑝-adic Fermionic Integrals

Let 𝑓₁(𝑥) be the translation defined by 𝑓₁(𝑥)=𝑓(𝑥+1). Then we have𝐼₋₁(𝑓₁)=−𝐼₋₁(𝑓)+2𝑓(0).(2.1)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:_ℤ𝑝𝑒^{(𝑥+𝑦)𝑡}𝑑𝜇₋₁(𝑦)=𝑒^𝑥𝑡2𝑒^𝑡=+1^∞_𝑛=0𝐸^∗_𝑛(𝑥)𝑡^𝑛𝑛!.(2.2)That is,_ℤ𝑝𝑦^𝑛𝑑𝜇₋₁(𝑦)=𝐸^∗_𝑛,_ℤ𝑝(𝑥+𝑦)^𝑛𝑑𝜇₋₁(𝑦)=𝐸^∗_𝑛(𝑥).(2.3)For 𝑛∈ℕ, we have the following integral equation:_ℤ𝑝𝑓(𝑥+𝑛)𝑑𝜇₋₁(𝑥)+(−1)^𝑛−1_ℤ𝑝𝑓(𝑥)𝑑𝜇₋₁(𝑥)=2^𝑛−1_𝑙=0(−1)^{𝑛−1+𝑙}𝑓(𝑙).(2.4)From this we note that𝐸^∗_𝑘(𝑛)−𝐸^∗_𝑘=2^𝑛−1_𝑙=0(−1)^𝑙−1𝑙^𝑘𝐸,if𝑛≡0(mod2),_∗𝑘(𝑛)+𝐸^∗_𝑘=2^𝑛−1_𝑙=0(−1)^𝑙𝑙^𝑘,if𝑛≡1(mod2).(2.5)

Let 𝑓(𝑥)=sin𝑎𝑥 (or 𝑓(𝑥)=cos𝑎𝑥). By using the fermionic 𝑝-adic 𝑞-integral on ℤ_𝑝, we see that0=_ℤ𝑝sin𝑎𝑥𝑑𝜇₋₁(𝑥)+_ℤ𝑝sin𝑎𝑥𝑑𝜇₋₁(𝑥)=(cos𝑎+1)_ℤ𝑝sin𝑎𝑥𝑑𝜇₋₁(𝑥)+sin𝑎_ℤ𝑝cos𝑎𝑥𝑑𝜇₋₁(𝑥),(2.6) see [12],2=(cos𝑎+1)_ℤ𝑝cos𝑎𝑥𝑑𝜇₋₁(𝑥)−sin𝑎_ℤ𝑝sin𝑎𝑥𝑑𝜇₋₁(𝑥).(2.7)Thus, we obtain_ℤ𝑝cos𝑎𝑥𝑑𝜇₋₁(𝑥)=1,_ℤ𝑝sin𝑎𝑥𝑑𝜇₋₁(𝑥)=−sin𝑎cos𝑎+1,(2.8) see [12]. From this we note that𝑎tan2=−_ℤ𝑝sin𝑎𝑥𝑑𝜇₋₁(𝑥)=^∞_𝑛=0(−1)^𝑛+1𝑎^2𝑛+1𝐸(2𝑛+1)!^∗2𝑛+1.(2.9)By the same motivation, we can also observe that𝑎2𝑎cot2=_ℤ𝑝cos𝑎𝑥𝑑𝜇₁(𝑥)=^∞_𝑛=0(−1)^𝑛𝐵_2𝑛𝑎(2𝑛)!^2𝑛,(2.10) see [12]. These formulae are also treated in Section 3.

Let 𝑓(𝑥)=𝑒^{𝑡(2𝑥+1)}. Then we can derive the generating function of the second-kind Euler numbers from fermionic 𝑝-adic integral equation as follows:_ℤ𝑝𝑒^{𝑡(2𝑥+1)}𝑑𝜇₋₁2(𝑥)=𝑒^𝑡+𝑒^−𝑡=1=cosh𝑡^∞_𝑛=0𝐸_𝑛𝑡^𝑛𝑛!.(2.11)Thus, we have(𝐸+1)^𝑛+(𝐸−1)^𝑛=2𝛿_0,𝑛,(2.12)where we have used the symbolic notation 𝐸_𝑛 for 𝐸^𝑛. The first few are 𝐸₀=1,𝐸₁=0,𝐸₂=−1,𝐸₃=0,𝐸₄=5,…, 𝐸_2𝑘+1=0 for 𝑘∈ℕ. In particular,𝐸_2𝑛=−^𝑛−1_𝑘=0𝐸2𝑛2𝑘^2𝑘.(2.13)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 ℤ_𝑝, 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 numbers 𝐸_𝑘 are defined by the following expansion:1𝑠𝑒𝑐ℎ𝑥==cosh𝑥2𝑒^𝑥𝑒^2𝑥=+1^∞_𝑘=0𝐸_𝑘𝑥^𝑘𝜋𝑘!,for|𝑥|<2(3.1) (cf. [10]). From (1.18) and (3.1), we can derive the following equation:𝐸_𝑘=^𝑘_𝑙=0𝑘𝑙2^𝑙𝐸^∗_𝑙𝑘𝑙,whereisbinomialcoefficient.(3.2) By (3.2) and (1.18), we easily see that 𝐸₀=1,𝐸₁=0,𝐸₂=−1,𝐸₃=0,𝐸₄=5,𝐸₆=61,…, and 𝐸_2𝑘+1=0 for 𝑘=1,2,3,…. As Euler formula, it is well known that𝑒^𝑖𝑥=cos𝑥+𝑖sin𝑥,where𝑖=(−1)^1/2.(3.3) From (3.3), we note that cos𝑥=(𝑒^𝑖𝑥+𝑒^−𝑖𝑥)/2. Thus, we have2sec𝑥=𝑒^𝑖𝑥+𝑒^−𝑖𝑥=𝑠𝑒𝑐ℎ(𝑖𝑥)=^∞_𝑛=0𝑖^𝑛𝐸_𝑛𝑥𝑛!^𝑛=^∞_𝑛=0(−1)^𝑛𝐸_2𝑛𝑥(2𝑛)!^2𝑛+𝑖^∞_𝑛=0(−1)^𝑛𝐸_2𝑛+1𝑥(2𝑛+1)!^2𝑛+1=^∞_𝑛=0(−1)^𝑛𝐸_2𝑛𝑥(2𝑛)!^2𝑛.(3.4)From (3.4), we derive𝑥sec𝑥=^∞_𝑛=0(−1)^𝑛𝐸_2𝑛𝑥(2𝑛)!^2𝑛+1𝜋,for|𝑥|<2.(3.5) The Fourier series of an odd function on the interval (−𝑝,𝑝) is the sine series:𝑓(𝑥)=^∞_𝑛=1𝑏_𝑛sin(𝑛𝜋𝑥𝑝),(3.6)where𝑏_𝑛=2𝑝^𝑝₀𝑓(𝑥)sin(𝑛𝜋𝑥𝑝)𝑑𝑥.(3.7)Let us consider 𝑓(𝑥)=sin𝑎𝑥 on [−𝜋,𝜋]. From (3.6) and (3.7), we note thatsin𝑎𝑥=^∞_𝑛=1𝑏_𝑛sin𝑛𝑥,(3.8)where𝑏_𝑛=2𝜋^𝜋₀2sin𝑎𝑥sin𝑛𝑥𝑑𝑥=𝜋^𝜋₀[cos(𝑛−𝑎)𝑥−cos(𝑛+𝑎)𝑥2=1]𝑑𝑥𝜋[sin(𝑛−𝑎)𝑥−𝑛−𝑎sin(𝑛+𝑎)𝑥]𝑛+𝑎^𝜋0=(−1)^𝑛−12𝜋𝑛sin𝑎𝜋(𝑛²−𝑎²).(3.9)In (3.8), if we take 𝑥=𝜋/2, then we havesin𝜋𝑎2=^∞_𝑛=1𝑏_2𝑛−1(−1)^𝑛−1=2𝜋sin𝑎𝜋^∞_𝑛=1(−1)^𝑛−12𝑛−1(2𝑛−1)²−𝑎²=2𝜋sin𝑎𝜋^∞_𝑛=1(2𝑛−1)(−1)^𝑛−1(2𝑛−1)²(1−(𝑎/(2𝑛−1))²)=2𝜋sin𝑎𝜋^∞_𝑛=1(−1)^𝑛−12𝑛−1^∞_𝑘=0𝑎^2𝑘(2𝑛−1)^2𝑘=2𝜋sin𝑎𝜋^∞_𝑘=0(^∞_𝑛=1(−1)^𝑛−1(2𝑛−1)^2𝑘+1)𝑎^2𝑘(3.10)From (3.10), we note that𝜋𝑎2sec(𝜋𝑎2)=^∞_𝑘=0(2^∞_𝑛=1(−1)^𝑛−1(2𝑛−1)^2𝑘+1)𝑎^2𝑘+1.(3.11)In (3.5), it is easy to see that𝜋𝑎2sec(𝜋𝑎2)=^∞_𝑛=0(−1)^𝑛𝐸_2𝑛(𝜋(2𝑛)!2)^2𝑛+1𝑎^2𝑛+1.(3.12)By (3.11) and (3.12), we obtain the following.

Theorem 3.1. For 𝑛∈ℕ,
^∞_𝑘=1(−1)^𝑘−1(2𝑘−1)^2𝑛+1=^∞_𝑘=0(−1)^𝑘(2𝑘+1)^2𝑛+1=(−1)^𝑛𝐸_2𝑛(𝜋2(2𝑛)!2)^2𝑛+1.(3.13)

It is easy to see that^∞_𝑛=1(−1)^𝑛(2𝑛+1)^2𝑘+1=2^∞_𝑛=11(4𝑛−3)^2𝑘+1+^∞_𝑛=11(2𝑛)^2𝑘+1−^∞_𝑛=11𝑛^2𝑘+1=1−12^4𝑘+11𝜁(2𝑘+1,42)−^2𝑘+1−12^2𝑘+1𝜁(2𝑘+1)−1.(3.14)By (3.13) and (3.14), we obtain the following.

Corollary 3.2. For 𝑛∈ℕ,1𝜁(2𝑛+1,4)+2^2𝑛(1−2^2𝑛+1)𝜁(2𝑛+1)=(−1)^𝑛𝐸_2𝑛𝜋2(2𝑛)!^2𝑛+12^2𝑛.(3.15)

By simple calculation, we easily see that𝑒𝑖tan𝑥=^𝑖𝑥−𝑒^−𝑖𝑥𝑒^𝑖𝑥+𝑒^−𝑖𝑥2=1−𝑒^2𝑖𝑥+4−1𝑒^4𝑖𝑥−1.(3.16)Thus, we have𝑥tan𝑥=−𝑥𝑖+2𝑥𝑖𝑒^2𝑥𝑖−−14𝑥𝑖𝑒^4𝑥𝑖=−1^∞_𝑛=1(−1)^𝑛𝐵_2𝑛4^𝑛(1−4^𝑛)𝑥(2𝑛)!^2𝑛.(3.17)From (3.17), we can easily derivetan𝑥=^∞_𝑛=0(−1)^𝑛+14^𝑛+1(1−4^𝑛+1)𝐵_2𝑛+2𝑥(2𝑛+2)!^2𝑛+1.(3.18)By (3.3), we also see that2𝑖tan𝑥=1−𝑒^2𝑖𝑥+1=𝑖^∞_𝑛=0𝐸^∗_2𝑛+12(2𝑛+1)!^2𝑛+1(−1)^𝑛+1𝑥^2𝑛+1.(3.19)Thus, we havetan𝑥=^∞_𝑛=0𝐸^∗_2𝑛+12(2𝑛+1)!^2𝑛+1(−1)^𝑛+1𝑥^2𝑛+1.(3.20)By (3.18) and (3.20), we obtain the following.

Theorem 3.3. For 𝑛∈ℕ,
𝜁(2𝑛)=(−1)^𝑛−1(2𝜋)^2𝑛𝐸^∗_2𝑛−14(2𝑛−1)!(1−4^𝑛),(3.21)where 𝐸^∗_𝑛 are the first-kind Euler numbers.

It is easy to see that^∞_𝑘=11(2𝑘+1)^2𝑛1=(1−4^𝑛)𝜁(2𝑛)=(−1)^𝑛(2𝜋)^2𝑛4^𝑛+1𝐸(2𝑛−1)!^∗2𝑛−1.(3.22)Therefore, we obtain the following corollary.

Corollary 3.4. For 𝑛∈ℕ,^∞_𝑘=11(2𝑘+1)^2𝑛=(−1)^𝑛(2𝜋)^2𝑛4^𝑛+1𝐸(2𝑛−1)!^∗2𝑛−1.(3.23)

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 that𝜁_𝐸(𝑠)=2^∞_𝑛=1(−1)^𝑛𝑛^𝑠=−2^∞_𝑛=01(2𝑛+1)^𝑠+12^𝑠−1𝜁(𝑠),𝑠∈𝐶.(3.24) By (3.24), Theorem 3.3, and Corollary 3.4, we obtain the following theorem.

Theorem 3.5. For 𝑛∈ℕ,𝜁_𝐸(2𝑛)=(−1)^𝑛−1𝜋^2𝑛(2−4^𝑛)2(2𝑛−1)!(1−4^𝑛)𝐸^∗_2𝑛−1.(3.25)

Remark 3.6. We note that 𝜁(2)=𝜋²/6, 𝜁_𝐸(2)=−𝜋²/6, 𝜁(4)=𝜋⁴/90 and 𝜁_𝐸(4)=−7𝜋⁴/360…. For 𝑞∈ℂ with |𝑞|<1, 𝑠∈ℂ, 𝑞-𝜁-function is defined by𝜁_𝑞(𝑠)=^∞_𝑛=1𝑞^𝑛[𝑛]^𝑠_𝑞−1𝑠−1(1−𝑞)^𝑠log𝑞(3.26) (cf. [10, 14]). Note that 𝜁_𝑞(𝑠) is analytic continuation in ℂ with only one simple pole at 𝑠=1, and𝜁_𝑞𝛽(1−𝑘)=−^𝑘,𝑞𝑘,where𝑘isapositiveinteger(3.27) (cf. [14]). By simple calculation, we easily see that^∞_𝑛=1(−1)^𝑛𝑞^𝑛[𝑛]_𝑞^∞2𝑘+1_𝑗=0𝜃^2𝑗+1[𝑛]_𝑞^2𝑗+1+1(2𝑗+1)!log𝑞^𝑘−1_𝑗=0(1−𝑞)^{2𝑘−2𝑗}−1(2𝑘−2𝑗−1)(2𝑗+1)!log𝑞^𝑘−1_𝑗=0(1−𝑞)^{2𝑘−2𝑗}=(2𝑘−2𝑗−1)(2𝑗+1)!_{𝑘−1𝑗=0}(−1)^𝑗𝜃^2𝑗+1(2𝑗+1)!(−𝜁_𝑞(2𝑘−2𝑗)+𝜁_𝑞22(2𝑘−2𝑗)[2]_𝑞^{2𝑘−2𝑗}𝑞)−𝜃1+𝑞(2𝑘+^2𝑘+11)!(−1)^𝑘+^∞_𝑗=𝑘+1𝜃^2𝑗+1(−1)^𝑗𝐻(2𝑗+1)!^{2𝑗−2𝑘,𝑞}(−𝑞⁻¹),1+𝑞(3.28)where 𝐻_𝑛,𝑞(−𝑞) are Carlitz's 𝑞-Euler numbers with lim_𝑞→1𝐻_𝑛,𝑞(−𝑞)=𝐸^∗_𝑛 (cf. [6, 21, 22]). If 𝑞→1, then we have^∞_𝑛=1(−1)^𝑛𝑛^2𝑘+1sin(𝑛𝜃)=^𝑘−1_𝑗=0(−1)^𝑗𝜃(2𝑗+1)!^2𝑗+1(22^{2𝑘−2𝑗}−1)×(−1)^{𝑘−𝑗+1}(2𝜋)^{2𝑘−2𝑗}𝐵2⋅(2𝑘−2𝑗)!^{2𝑘−2𝑗}−12𝜃^2𝑘+1(2𝑘+1)!(−1)^𝑘.(3.29)For 𝑘∈ℕ, and 𝜃=𝜋/2, it is easy to see that^∞_𝑛=1(−1)^𝑛(2𝑛−1)^2𝑘+1=^𝑘−1_𝑗=0(−1)^𝑘𝜋^2𝑘+1(2^{2𝑘−2𝑗}−2)𝐵_{2𝑘−2𝑗}(2𝑗+1)!(2𝑘−2𝑗)!2^2𝑗+2−𝜋^2𝑘+1(−1)^𝑘(2𝑘+1)!2^2𝑘+2.(3.30)From (3.30) and Theorem 3.1, we can also derive the following equation:^𝑘−1_𝑗=0(−1)^𝑘−1𝜋^2𝑘+1(2^{2𝑘−2𝑗}−2)𝐵_{2𝑘−2𝑗}(2𝑗+1)!(2𝑘−2𝑗)!2^2𝑗+2+𝜋^2𝑘+1(−1)^𝑘(2𝑘+1)!2^2𝑘+2=(−1)^𝑘𝐸_2𝑘(𝜋2(2𝑘)!2)^2𝑘+1.(3.31)Thus, we have^𝑘−1_𝑗=0(2^{2𝑘−2𝑗}−2)𝐵_{2𝑘−2𝑗}(2𝑗+1)!(2𝑘−2𝑗)!2^2𝑗+2=1(2𝑘+1)!2^2𝑘+2−𝐸_2𝑘2^2𝑘+2(2𝑘)!.(3.32)

Acknowledgments

The present research has been conducted by the research grant of Kwangwoon University in 2008. The author wishes to express his sincere gratitude to the referees and Professor Lance L. Littlejohn for their valuable suggestions and comments.