Special Values and Integral Representations for the Hurwitz-Type

Home , Dirichlet beta function

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Bernoulli polynomials have many interesting properties and arise in various areas of mathematics (see [9, 34, 35]). Setting x = 0 in (1.4), we get the Bernoulli numbers B = B (0), m N = N 0 . m m ∈ 0 ∪{ } From an easy manipulations of the generating function (1.4), we see that Bm(1) = Bm (m = 1) and B1(1) = B1, where m N0 (see [13, p. 1480, (2.3)] and [34, p.6 529, (17)]). − ∈ The Hurwitz zeta function is a fundamental tool for studying Stark’s conjectures of algebraic number fields, since it represents the partial zeta function of cyclotomic field (see e.g., [18, p. 993, (4.2)] or [23, p. 4248]). It also plays a role in the evaluation of fundamental determinants that appears in mathematical physics (see [14] or [15, p. 161, (1.14) and (1.15)]). In this paper, we consider the Hurwitz-type Euler zeta function, which is defined as a deformation of the Hurwitz zeta function ∞ ( 1)n (1.5) ζ (s, x)= − , E (n + x)s n=0 X for s C and x = 0, 1, 2,.... This series converges for Re(s) > 0, and it can∈ be analytically6 − continued− to the complex plane without any pole. It also represents a partial zeta function of cyclotomic fields in one version of Stark’s conjectures in algebraic number theory (see [23, p. 4249, (6.13)]), and its special case, the alternative series, ∞ ( 1)n−1 (1.6) ζ (s)= − , E ns n=1 X is also a particular case of Witten’s zeta functions in mathematical physics. (See [31, p. 248, (3.14)]). We here mention that several properties of ζE(s) can be found in [2, 6, 12]. For example, in the form on [2, p. 811], the left hand side is the special values of the Riemann zeta functions at positive integers, and the right hand side is the special values of Euler zeta functions at positive integers. Using the Fourier expansion formula 2Γ(1 s) ∞ sin 2nπx + πs (1.7) ζ(s, x)= − 2 , (2π)1−s n1−s n=1 X Espinosa and Moll [15] evaluated several fundamental integrals with integrands involving Hurwitz zeta functions, where Γ is the Euler gamma- function (see, e.g., [30] or [35, Chapter 43]), who studied “the Hurwitz transform” meaning an integral over [0, 1] of a Fourier series multiplied by the Hurwitz zeta function ζ(s, x), and obtained numerous results for those which aries from the Hurwitz formula. Mez˝o[29] determined the Fourier series expansion of the log-Barnes function, which is the analogue of the classical result of Kummer and Malmsten. Applying this expansion, he also got some integrals similar to the Espinosa–Moll log–Gamma integrals with respect to log G in [15, 16], where G is Barnes G function. 3

Recently, Shpot, Chaudhary and Paris [37] evaluated two integrals over −a x [0, 1] involving products of the function ζ1(s, x) = ζ(s, x) x for Re(∈s) > 1. As an application, they also calculated the O(g) weak-coupling− expansion coefficient c1(ǫ) of the Casimir energy for a film with Dirichlet- Neumann boundary conditions, first stated by Symanzik [38] under the 4 framework of gφ4−ǫ theory in quantum physics. Here we will follow the approach of Espinosa and Moll in [15] to evaluate several integrals with integrands involving Hurwitz-type Euler zeta functions ζE(s, x). Our main tool is the following Fourier expansion of ζE(s, x)

2Γ(1 s) ∞ sin (2n + 1)πx + πs (1.8) ζ (s, x)= − 2 E π1−s (2n + 1)1−s n=0 X (comparing with (1.7) above). This expression, valid for Re(s) < 1 and 0 < x 1, is derived by Williams and Zhang [40, p. 36, (1.7)]. The≤ paper is organized as follows. In Section 3, we determine the Fourier coeﬃcients of ζE(s, x). In Section 4, we evaluate integrals with integrands consisting of products of two Hurwitz-type Euler zeta functions. These will imply some classical relations for the integral of products of two Euler polynomials (see Proposi- tion 4.8 and 4.10). We also evaluate the moments of the Hurwitz-type Euler zeta functions and the Euler polynomials (see Proposition 4.9 and 4.10). In Section 5, we prove several further consequences of integrals with integrands consisting of products of the Hurwitz-type Euler zeta functions, the exponential function, the power of sine and cosine. In Section 6, we obtain an Euler-type formula for the Dirichlet beta function β(2m) and consider the Catalan’s constant G compiled by Adamchik in [1, Entry 19]. From this, we get a ﬁnite closed-form expression for any β(2m) in terms of the elementary functions (see [24]). In Section 7, we prove the relations between values of a class of the Hurwitz-type (or Lerch-type) Euler zeta functions at rational arguments.

2. Preliminaries The methods employed in this section may also be used to derive several results of the Hurwitz-type Euler zeta functions, which are analogous of the known results for Hurwitz zeta functions. The series expansion (1.5) for ζE(s, x) has a meaning if Re(s) > 0. In the following, we recall some basic results of this functions given in [40]. From (1.5), we can easily derive the following basic properties for ζE(s, x)

−s (2.1) ζE(s, x)+ ζE(s, x +1)= x ,

k−1 n (2.2) ζ (s,kx)= k−s ( 1)nζ s, + x for k odd, E − E k n=0 X 4 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

∞ s (2.3) ζ (s, x) x−s = − ζ (s + n)xn E − − n E n=0 X (see [4, (5), (6) and (7)]). For fixed s = 1, the series in (2.3) converges absolutely for x < 1. For Re(s) > 1, (1.2)6 and (1.6) are connected by the following equation| | (2.4) ζ (s)=(1 21−s)ζ(s). E − In [40, 3], Williams and Zhang established the following integral representation § e−πisΓ(1 s) e(1−x)zzs−1 (2.5) ζ (s, x)= − dz, E 2πi ez +1 ZC where C is the contour consisting of the real axis from + to ǫ, the circle z = ǫ and the real axis from ǫ to + . By (2.5) and the generating∞ function |of| the Euler polynomials ∞ 2exz ∞ zn (2.6) = E (x) , z < π, ez +1 n n! | | n=0 X we have ( 1)mm! ∞ E (1 x) zn ζ ( m, x)= − n − dz, E − 4πi n! zm+1 |z|=ǫ n=0 Z X that is, ( 1)m 1 (2.7) ζ ( m, x)= − E (1 x)= E (x), m N E − 2 m − 2 m ∈ 0 (comparing with (1.3) above). Setting s = m in (2.4), from (1.3) and (2.7) it is easy to see that − 2 (2.8) E (0) = E (1) = 1 2m+1 B m − m m +1 − m+1 (see [34, p. 529, (16)]). The above expansion (2.7) has also been derived in [40, p. 41, (3.8)]. The integers 1 E =2mE , m N , m m 2 ∈ 0 are called m-th Euler numbers. For example, E0 =1, E2 = 1, E4 =5, and E = 61. Notice that the Euler numbers with odd subscripts− vanish, that 6 − is, E2m+1 = 0 for all m N0. The Euler polynomials can be expressed in terms of the Euler numbers∈ in the following way: m m E 1 m−i (2.9) E (x)= i x m i 2i − 2 i=0 X (see [34, p. 531, (29)]). Some properties of Euler polynomials can be easily derived from their generating functions, for example, from (2.6), we have (2.10) E (x +1)+ E (x)=2xm, m N , m m ∈ 0 5 which have numerous applications (see [13, p. 1489, (5.4)] and [34, p. 530, (23) and (24)]). For further properties of the Euler polynomials and numbers including their applications, we refer to [2, 8, 17, 22, 34]. It may be interesting to point out that there is also a connection between the generalized Euler numbers and the ideal class group of the pn+1-th cyclotomic field when p is a prime number. For details, we refer to a recent paper [19], especially [19, Proposition 3.4]. The result 1 4Γ(1 s) πs (2.11) ζE(s, x)dx = 2−−s cos λ(2 s), 0 π 2 − Z valid for Re(s) < 1, follows directly from the representation (1.8). Here λ(s) is the Dirichlet lambda function ∞ 1 (2.12) λ(s)= = (1 2−s)ζ(s), Re(s) > 1, (2n + 1)s − n=0 X where ζ(s) is the Riemann zeta function (see [2, p. 807, 23.2.20] and [35, p. 32, Eq. 3:6:2]). If we put s = 2m with m N0 in (2.11) and use (2.7), then we obtain − ∈ 1 8(2m)!( 1)m (2.13) E (x)dx = − λ(2m + 2). 2m π2m+2 Z0 If we replace m by 2m and let n = 0 in (4.15) below, then the left hand side of (2.13) becomes 1 2(2m)! (2.14) E (x)dx = E (0), 2m −(2m + 1)! 2m+1 Z0 where m N0. From (2.13) and (2.14), the values of the Dirichlet lambda function at∈ even positive integers is given as π2m (2.15) λ(2m)=( 1)m E (0), − 4(2m 1)! 2m−1 − where m N. In particular, if we put m = 1, m = 2 and m = 3 in (2.15), ∈ π2 π4 π6 then we have the first few values λ(2) = 8 ,λ(4) = 96 and λ(6) = 960 (see [2, p. 808]).

3. The Fourier expansion of ζE(s, x)

In this section, by using the Fourier expansion (1.8) for ζE(s, x), we shall evaluate definite integrals of the Hurwitz-type transform 1 H(f,s)= f(x)ζE(s, x)dx. Z0 The expansion is valid for Re(s) < 1. For the basics we refer the reader to definite integrals of the Hurwitz transform, e.g., [15] and [25]. The following is the Fourier coefficients of ζE(s, x), as a direct conse- quence of (1.8) (see e.g., [15, p. 164, Proposition 2.1]). 6 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

Proposition 3.1. The Fourier coeﬃcients of ζE(s, x) are given by 1 πs(2k + 1)s−1 πs (3.1) sin((2k + 1)πx)ζE(s, x)dx = csc 0 2Γ(s) 2 Z and 1 πs(2k + 1)s−1 πs (3.2) cos((2k + 1)πx)ζE(s, x)dx = sec . 0 2Γ(s) 2 Z Remark 3.2. Recently, Luo [27], Bayad [7], Navas, Francisco and Varona [33] investigated Fourier expansions of the Apostol-Bernoulli polynomials, which are special values of Hurwitz-Lerch zeta function at non-positive integers [21, Lemma 2.1]. Proof of Proposition 3.1. The proof follows from the similar argument with [15, Proposition 2.1], by applying the orthogonality of the trigonometric functions and the expansion (1.8) above. In the following proposition, we shall compute the Hurwitz-type transform of powers of sine and cosine, including some related integrals (see [15, 11]). § R− N Proposition 3.3. Let s 0 =( , 0] and n . Then we have (3.3) ∈ −∞ ∈ 1 n−1 k 2n−1 Γ(1 s) πs ( 1) 2n 1 sin (πx)ζE(s, x)dx = 1−s −2n−2 cos − 1−s − 0 π 2 2 (2k + 1) n k 1 Z Xk=0 − − and (3.4) 1 n−1 2n−1 Γ(1 s) πs 1 2n 1 cos (πx)ζE(s, x)dx = 1−s −2n−2 sin 1−s − . 0 π 2 2 (2k + 1) n k 1 Z Xk=0 − − Proof. The proofs follow from the similar arguments with [15, Examples 11.1 and 11.2], which is based on (3.1) and (3.2), and the integral representation 1 n−1 2n 1 (3.5) sin2n−1(πx)= ( 1)k − sin((2k + 1)πx) 22n−2 − n k 1 Xk=0 − − by Kogan in [17, p. 31, 1.320]. As in Espinosa and Moll [15], we may evaluate the integrals using the Fourier expansions of functions. The following result about this idea will become the basic tool for the proofs of the results in this paper. Theorem 3.4. Let f(w, x) be deﬁned for x [0, 1] and a parameter w. Let ∈ ∞

(3.6) f(w, x)= an(w) cos((2n + 1)πx)+ bn(w) sin((2n + 1)πx) n=0 X be its Fourier expansion, so that 1 (3.7) an(w)=2 f(w, x) cos((2n + 1)πx)dx, Z0 7

1 (3.8) bn(w)=2 f(w, x) sin((2n + 1)πx)dx. Z0 R− Then, for s 0 , we have (3.9) ∈ 1 Γ(1 s) πs πs f(w, x)ζE(s, x)dx = 1−s sin C(s,w) + cos S(s,w) 0 π 2 2 Z and (3.10) 1 Γ(1 s) πs πs f(w, x)ζE(s, 1 x)dx = 1−s cos S(s,w) sin C(s,w) , 0 − π 2 − 2 Z where ∞ a (w) ∞ b (w) C(s,w)= n and S(s,w)= n . (2n + 1)1−s (2n + 1)1−s n=0 n=0 X X Proof. The proof follows from the similar argument with [15, Theorem 2.1], by multiplying the formula (3.6) by ζE(s, x) and integatationg both sides from 0 to 1, then applying (3.1) and (3.2).

4. Integral representations of Hurwitz-type Euler zeta functions In this section, we shall evaluate integrals with integrands consisting of products of Hurwitz-type Euler zeta functions. These will imply some integral representations for the function and classical relations for the integral of products of two Euler polynomials.

′ R− Theorem 4.1. Let s,s 0 . Then we have (4.1) ∈ 1 2Γ(1 s)Γ(1 s′) π(s s′) ζ (s′, x)ζ (s, x)dx = − − λ(2 s s′) cos − . E E π2−s−s′ − − 2 Z0 Similarly, we have (4.2) 1 2Γ(1 s)Γ(1 s′) π(s + s′) ζ (s′, x)ζ (s, 1 x)dx = − − λ(2 s s′) cos . E E − π2−s−s′ − − 2 Z0 Proof. In view of (3.4), (3.7) and (3.8), the expansion (1.8) shows that the ′ coeﬃcients of ζE(s , x) are given by

′ 2Γ(1 s′) sin πs 1 ′ − 2 an(s )= 1−s′ 1−s′ , π (2n + 1) (4.3) ′ 2Γ(1 s′) cos πs 1 ′ − 2 bn(s )= 1−s′ 1−s′ . π (2n + 1) 8 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

From (2.12), (3.9) and (4.3), we obtain (4.4) 1 ′ ′ ′ 2Γ(1 s)Γ(1 s ) πs πs ζE(s , x)ζE(s, x)dx = −2−s−s′ − sin sin 0 π 2 2 Z πs πs′ ∞ 1 + cos cos 2 2 (2n + 1)2−s−s′ n=0 X 2Γ(1 s)Γ(1 s′) π(s s′) = − − cos − λ(2 s s′), π2−s−s′ 2 − − which gives (4.1). The proof of (4.2) is similar. Corollary 4.2. Let s,s′ R− and ∈ 0 1 2s+1 δ (s)= − , s =0. 2 1 2s 6 − Then we have π(s−s′) 1 cos ′ ′ 2 ′ ζE(s , x)ζE(s, x)dx = δ2(1 s s ) B(1 s, 1 s ) π(s+s′) (4.5) 0 − − − − Z cos 2 λ(s + s′ 1). × − Similarly, we have 1 ζ (s′, x)ζ (s, 1 x)dx = δ (1 s s′)B(1 s, 1 s′) (4.6) E E − 2 − − − − Z0 λ(s + s′ 1). × − Here B(s,s′) is the beta function, Γ(s)Γ(s′) (4.7) B(s,s′)= , Γ(s + s′) deﬁned for s,s′ C with Re(s), Re(s′) > 1, see [40, (1.36)]. ∈ Proof. The functional equation of Riemann’s zeta function is well-known, and it is given by (see [15, (1.35)]) πs Γ(s) (4.8) ζ(1 s) = 2 cos ζ(s). − 2 (2π)s From (2.12) and (4.8), we have ′ 2−s−s′ ′ δ2(1 s s )π ′ (4.9) λ(2 s s )= − − ′ λ(s + s 1). − − −2Γ(2 s s′) cos π(2−s−s ) − − − 2 Finally, using (4.1) and (4.9), we obtain (4.5). The proof of (4.6) is similar.

Corollary 4.3. Let s R−. Then we have ∈ 0 1 (4.10) ζ2 (s, x)dx = 2Γ2(1 s)π2s−2λ(2 2s) E − − Z0 9 and 1 (4.11) ζ (s, x)ζ (s, 1 x)dx = 2Γ2(1 s)π2s−2λ(2 2s) cos(πs). E E − − − Z0 Proof. Put s = s′ in (4.1) and (4.2), we get our results. Corollary 4.4. Let m N. Then we have ∈ 1 1 (2m)! 2 λ(2m + 1) (4.12) ζ2 m + , x dx =2 . E − 2 22mm! π2m Z0 Proof. Setting s = m + 1 in (4.10), then by − 2 1 √π(2m)! Γ m + = (see [15, p. 167]), 2 22mm! we get our result. Corollary 4.5. Let s R− and m N. Then we have ∈ 0 ∈ 1 (m 1)!λ(s m) (4.13) E (x)ζ (s, x)dx =( 1)m+12δ (m s) − − , m−1 E − 2 − (1 s) Z0 − m where (s) = s(s + 1) (s + k 1) is the Pochhammer symbol. k ··· − Remark 4.6. The case m = 1 in (4.13) implies (2.11) from (2.12) and (4.8). Proof of the Corollary 4.5. Let s′ =1 m in (4.5), we have − π(s−1+m) 1 cos 2 ζE(1 m, x)ζE(s, x)dx = δ2(m s) B(1 s, m) π(s+1−m) 0 − − − Z cos 2 λ(s m). × − The result then follows from (2.7), since π(s−1+m) cos 2 =( 1)m+1 π(s+1−m) − cos 2 and B(1 s, m)=(m 1)!/(1 s) . − − − m Lemma 4.7. For m N, we have ∈ 1 (4.14) λ(1 m)=( 1)m+1 E (0). − − 2δ (m 1) m−1 2 − Proof. This follows from (1.3) with x =1, (2.8) and (2.12). The following formula on the integral of two Euler polynomials has a long history, it has already appeared in a book by N¨orlund [32, p. 36]. See e.g. [34, p. 530, (25)] and [36, p. 64, (52)]. Corollary 4.8. Let m, n N . Then we have ∈ 0 1 m!n! (4.15) E (x)E (x)dx = 2( 1)n+1 E (0). m n − (m + n + 1)! m+n+1 Z0 10 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

Proof of Corollary 4.8. Set s =1 n, where n N in (4.13), then by (2.7) and (4.14), we obtain − ∈ 1 (m 1)!λ(1 (m + n)) E (x)E (x)dx = 4( 1)m+1δ (m + n 1) − − m−1 n−1 − 2 − (n) Z0 m n (m 1)! = 2( 1) − Em+n−1(0) − (n)m (m 1)!(n 1)! = 2( 1)n − − E (0), − (m + n 1)! m+n−1 − where m, n N. This completes our proof. ∈ Corollary 4.9. For n N0, the moments of the Hurwitz-type Euler zeta function are given by ∈ 1 n n j!λ(s j 1) xnζ (s, x)dx = ( 1)jδ (j s + 1) − − E j − 2 − (1 s) 0 j=0 j+1 (4.16) Z X − n!λ(s n 1) +( 1)nδ (n s + 1) − − . − 2 − (1 s) − n+1 Proof. The proof of (4.16) is obtained by using the expansion of xn in terms of Euler polynomials (2.10) and the evaluation (4.13). Corollary 4.10. Let m N and n N . Then we have ∈ ∈ 0 n 1 ( 1)m n E (0) (4.17) xnE (x)dx = − j E (0) + m+n . m−1 m m+j m+j m+n 0 j=0 j n ! Z X Proof. By (4.14), we have

j+m 1 (4.18) λ(1 (j + m +1))= ( 1) Ej+m(0). − − 2δ2(j + m) Letting s = 1 m in (4.16), then using (2.7) and (4.18), we easily deduce that − 1 n n j! n! xnE (x)dx =( 1)m E (0) + E (0) m−1 − j (m) m+j (m) m+n 0 j=0 j+1 n+1 ! Z X and this implies our result.

Remark 4.11. Corollary 4.10 gives some identities of Em(0) for each value of m N. For instance, when m = 1 and 2, we have ∈ n n E (0) 1 j+1 = (E (0) + 1), j j +1 −n +1 n+1 j=0 X n n E (0) 1 j+2 = (2E (0) n), j (j + 1)(j + 2) −2(n + 1)(n + 2) n+2 − j=0 X where n N . ∈ 0 11

5. The exponential function In this section, we shall evaluate the Hurwitz-type transform of the exponential function (see [15, 4]). As [15, 4], the result is expressed in terms of the transcendental function§ §

∞ (5.1) F (x, s)= λ(n +2 s)xn for x < 1, − | | n=0 X where λ(s) denotes the Dirichlet lambda function (deﬁne 00 = 1 if x = 0).

Theorem 5.1. Let s R− and t < 1. Then we have ∈ 0 | | 1 2(e2πt + 1)Γ(1 s) (5.2) e2πtxζ (s, x)dx = − Re eπis/2F (2it, s) , E π2−s Z0 where F (x, s) is given in (5.1).

Remark 5.2. Setting t = 0 in (5.2), we get (2.11).

Proof of Theorem 5.1. The generating function for the Euler polynomials (2.6) gives et +1 ∞ tn ext = E (x) , 2 n n! n=0 X so that

1 et +1 ∞ tn 1 (5.3) extζ (s, x)dx = E (x)ζ (s, x)dx. E 2 n! n E 0 n=0 0 Z X Z By (4.7), we have

n! Γ(1 s)Γ(n + 1) (5.4) = B(1 s, n +1) = − . (1 s) − Γ(n s + 2) − n+1 − Therefore, from (4.13) and (5.4), (5.3) gives (5.5) 1 ∞ tn extζ (s, x)dx =(et + 1) ( 1)nδ (n s + 1)B(1 s, n + 1) E n! − 2 − − 0 n=0 Z X λ(s n 1). × − − By (4.9), we have

π(n−s+2) 2 Γ(n s + 2) cos 2 (5.6) λ(s n 1) = − λ(n s+2), − − −δ (n s + 1) πn−s+2 − 2 − 12 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM thus in viewing of (5.4) and (5.6), we my also write (5.5) in the form (5.7) 1 ∞ tn Γ(n + 1) extζ (s, x)dx =(et + 1)Γ(1 s) ( 1)nδ (n s + 1) E − n! − 2 − Γ(n s + 2) 0 n=0 Z X − λ(s n 1) × − − 2(et + 1)Γ(1 s) ∞ t n π(n s) = − ( 1)n cos − π2−s π − 2 n=0 X λ(n s + 2). × − Finally, by replacing t with 2πt and using the identity π(n s) cos − = Re(eiπ(s−n)/2) = Re(( i)neiπs/2), 2 − we get 1 2πtx e ζE(s, x)dx Z0 2(e2πt + 1)Γ(1 s) ∞ π(n s) = − ( 2t)n λ(n s + 2) cos − π2−s − − 2 n=0 X 2(e2πt + 1)Γ(1 s) ∞ = − ( 2t)n λ(n s + 2)Re(( i)neiπs/2) π2−s − − − n=0 X 2(e2πt + 1)Γ(1 s) ∞ = − Re eπis/2 λ(n s + 2)(2it)n π2−s − n=0 ! X 2(e2πt + 1)Γ(1 s) = − Re eπis/2F (2it, s) . π2−s This completes our proof.

Corollary 5.3. Let m N0 and t < 1, t =0. Then we have (5.8) ∈ | | 6 1 ( 1)m4(e2πt + 1)m! e2πtxE (x)dx = − m (2πt)m+2 Z0 m−1 πt ⌊ 2 ⌋ tanh(πt) ( 1)rλ(2r + 2)(2t)2r+2 , ×  2 − −  r=0  X  where x is the ﬂoor (greatest integer) function.  ⌊ ⌋ Proof. Suppose that k N . Let s = 2k in (5.2). From (5.1), we have ∈ 0 − Re eπis/2F (2it, s) = Re eπi(−2k)/2F (2it, 2k) − ∞ =( 1)kRe λ(r +2+2k)(2it)r − (5.9) r=0 ! ∞ X =( 1)k λ(2r +2+2k)( 1)r(2t)2r. − − r=0 X 13

From [35, p. 37, Eq. 3:14:7], we obtain πx 4 ∞ tanh = ( 1)rλ(2r)x2r, 2 −πx − r=1 X so (5.10) πx πx ∞ tanh = ( 1)rλ(2r + 2)x2r+2 4 2 − r=0 X ∞ k−1 = ( 1)rλ(2r + 2)x2r+2 + ( 1)rλ(2r + 2)x2r+2. − − r=0 Xr=k X Replacing x by 2t in (5.10), we have (5.11) ∞ πt k−1 ( 1)rλ(2r + 2)(2t)2r+2 = tanh(πt) ( 1)rλ(2r + 2)(2t)2r+2. − 2 − − r=0 Xr=k X On the other hand, by (5.9), we have ∞ (2t)−2k−2 ( 1)rλ(2r + 2)(2t)2r+2 − Xr=k (5.12) ∞ =( 1)k λ(2r +2+2k)( 1)r(2t)2r − − r=0 X = Re eπi(−2k)/2F (2it, 2k) − for t < 1, t =0. By (5.11) and (5.12), we get | | 6 Re eπi(−2k)/2F (2it, 2k) − (5.13) ( 1)2k πt k−1 = − tanh(πt) ( 1)rλ(2r + 2)(2t)2r+2 . (2t)2k+2 2 − − r=0 ! X Let s = 2k 1 in (5.2), similar with the above, we obtain − − Re eπis/2F (2it, s) πi(−2k−1)/2 = Re e F (2it, 2k 1) (5.14) − − ( 1)2k+1 πt k = − tanh(πt) ( 1)rλ(2r + 2)(2t)2r+2 . (2t)2k+3 2 − − r=0 ! X The proof now follows directly from (2.7), (5.2), (5.13) and (5.14).

6. An expression for Catalan’s constant and an Euler-type formula for the Dirichlet beta function Following Berndt [9, p. 200, Entry 17(v)], Espinosa and Moll [15, p. 177, (8.1)], for x> 0, we deﬁne (6.1) G (s, x)= ζ (s, x) ζ (s, 1 x). E E − E − 14 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

From (1.8), the Fourier expansion for ζE(s, x), we have the following Fourier expansion of GE(s, x) 4Γ(1 s) πs ∞ cos((2n + 1)πx) (6.2) G (s, x)= − sin . E π1−s 2 (2n + 1)1−s n=0 X We introduce the anti-symmetrized Hurwitz transform for GE(s, x) as follows (also see [15]) 1 1 (6.3) H (f)= f(w, x)G (s, x)dx, A 2 E Z0 where f(w, x) is given in (3.6). For a function f(w, x) with Fourier expansion as in (3.6), we have 1 1 Γ(1 s) πs ∞ a (w) (6.4) f(w, x)G (s, x)dx = − sin n . 2 E π1−s 2 (2n + 1)1−s 0 n=0 Z X From [17, p. 408, 3.612], we get 1 cos((2n + 1)πx) (6.5) dx =( 1)n. cos(πx) − Z0 By (6.2) and (6.5), we see that 1 1 GE(s, x) 2Γ(1 s) πs (6.6) dx = 1−−s sin β(1 s), 2 0 cos(πx) π 2 − Z where ∞ ( 1)n (6.7) β(s)= − (2n + 1)s n=0 X is the Dirichlet beta function (see [2, p. 807, 23.2.21]). It is necessary to remark that by the procedure of analytic continuation, the function β(s) can extend to all points in the complex plane, without any singularities. In particular, when s = 1, (6.6) becomes − 1 1 G ( 1, x) 2 (6.8) E − dx = β(2). 2 cos(πx) −π2 Z0 Espinosa and Moll [15, p. 178, (8.8)] established the integral representation of the Catalan constant G (see [1]) as 1 1 x 4G (6.9) 2 − dx = . cos(πx) π2 Z0 Using (2.7), (6.1) with s = 1 and (6.9), the left hand side of (6.8) becomes − 1 1 G ( 1, x) 1 1 x 1 E − dx = − 2 dx 2 cos(πx) 2 cos(πx) (6.10) Z0 Z0 2G = . − π2 15

From (6.8) and (6.10), we then have π2 1 G ( 1, x) (6.11) G = β(2) = E − dx. − 4 cos(πx) Z0 This form of Catalan’s constant is Entry 14 and 19 in the list of expressions for G compiled by Adamchik in [1]. From (6.6), we have the following lemma, which were given in [2, p. 807, 23.2.23], [35, p. 36, Eq. 3:13:4] and [40, (4.13)]. Lemma 6.1. Let m N. Then we have ∈ 1 4(2m 1)! (6.12) sec(πx)E (x)dx =( 1)m − β(2m). 2m−1 − π2m Z0 Proof. For m N, letting s = (2m 1) in (6.6), we have ∈ − − 1 1 G ( (2m 1), x) 2(2m 1)! (6.13) E − − dx = − ( 1)mβ(2m). 2 cos(πx) π2m − Z0 It is easy to see that by (2.6), E (1 x)=( 1)mE (x), m N . m − − m ∈ 0 Hence, by (2.7) and (6.1), the left hand side of (6.13) becomes 1 1 G ( (2m 1), x) 1 1 E (x) E (1 x) E − − dx = 2m−1 − 2m−1 − dx 2 cos(πx) 4 cos(πx) (6.14) Z0 Z0 1 1 E (x) = 2m−1 dx 2 cos(πx) Z0 for m N. The proof now follows directly from (6.13) and (6.14). ∈ We now evaluate β(2m) using (4.17). This new formula can be regarded as an analogue of Theorem 1 of [24]. Theorem 6.2 (Euler-type formula for β(2m)). Let m N. Then we have (6.15) ∈ ∞ ( 1)n+mπ2m+2nE n E (0) β(2m)= − 2n 2m+2j−1 . 4 (2n 2j + 1)!(2m +2j 1)! n=1 j=1 ! X X − − Proof. From the Taylor expansion for sec(πx), namely ∞ ( 1)nE sec(πx)= − 2n (πx)2n, (2n)! n=0 X it is clear that 1 ∞ ( 1)nπ2nE 1 sec(πx)E (x)dx = − 2n x2nE (x)dx, 2m−1 (2n)! 2m−1 0 n=0 0 Z X Z where m N. Note that E2m(0) = 0, m N. Therefore, by (4.17) and (6.12), we∈ get our result. ∈ For instance, if we set m = 1 in (6.15), then we obtain: 16 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

Corollary 6.3. ∞ ( 1)n+1nπ2n+2E β(2) = G = − 2n . 4(2n + 2)! n=1 X Remark 6.4. For m N0 we write s = 2m in (1.8) and let x =1/2. We obtain easily ∈ − 1 ( 1)m2(2m)! ∞ ( 1)n (6.16) ζ 2m, = − − , E − 2 π2m+1 (2n + 1)2m+1 n=0 X so (6.16) becomes, after using (2.7), 1 1 ( 1)m2(2m)! ∞ ( 1)n (6.17) E = − − , 2 2m 2 π2m+1 (2n + 1)2m+1 n=0 X 2m By using E2m = 2 E2m (1/2) , m N0, we are easy to show that, for the Dirichlet beta function (6.7), ∈ E (6.18) β(2m +1)=( 1)m 2m π2m+1, − 22m+2(2m)! where Em are the Euler number, that is, the integers obtained as the coefficients of zm/m! in the Taylor expansion of 1/ cosh z, z < π/2 (see [24, (3)] and [40, (4.10)]). The formula (6.12) for β(2m) gives| us| π2m 1 (6.19) β(2m)=( 1)m sec(πx)E (x)dx, − 4(2m 1)! 2m−1 − Z0 which is also Equation 23.2.23 of [2], Equation 3:13:4 of [35] and (4.13) of [40]. (6.19) may give some hint for the question “why it is so difficult to find closed-form expressions for even beta values?” However, Lima [24] has derived the exact closed-form expression for a certain class of zeta series related to β(2m) and a finite number of odd zeta values.

7. Integral representations of Lerch-type Euler zeta functions and special values for Lerch-type Euler zeta functions at rational arguments The counterpart of the Hurwitz-type Euler zeta function, the Lerch-type Euler zeta function ℓE,s(x) of the complex exponential argument is deﬁned by ∞ e(2n+1)πix (7.1) ℓ (x)= E,s (2n + 1)s n=0 X for Re(s) > 1 and x R (see [11, p. 1530, (7)]). The Hurwitz-type Euler zeta function and the∈ Lerch-type Euler zeta function are related by the functional equation

Γ(s) − πis πis (7.2) ζ (1 s, x)= e 2 ℓ (x) e 2 ℓ (1 x) , E − πs E,s − E,s − 17 the inverse is

Γ(1 s) πi(1−s) − πi(1−s) (7.3) ℓ (x)= − e 2 ζ (1 s, x) e 2 ζ (1 s, 1 x) . E,s 2π1−s E − − E − − The limiting case x 1− of (7.2) leads to the asymmetric form of the → functional equation for ζE(s) and λ(s) 2Γ(s) πs (7.4) ζ (1 s)= cos λ(s), E − − πs 2 where λ(s) is the the Dirichlet lambda function (2.12). Similarly, in (7.3), 1 with x = 2 , we have 1 iΓ(1 s) π(1 s) 1 (7.5) ℓ = − sin − ζ 1 s, . E,s 2 π1−s 2 E − 2 Theorem 7.1. Let s,s′ R−. Then we have ∈ 0 1 ′ s′−1 ′ πi (1−s′) ′ (7.6) ζ (s , x)ℓ (x)dx = π Γ(1 s )e 2 λ(1 + s s ). E E,s − − Z0 Proof. We have the functional equation for the Hurwitz-type Euler zeta function (see (7.3)):

iΓ(1 s) − πis πis ℓ (x)= − e 2 ζ (1 s, x)+ e 2 ζ (1 s, 1 x) . E,s 2π1−s E − E − − From which, we obtain 1 1 ′ iΓ(1 s) − πis ′ ζ (s , x)ℓ (x)dx = − e 2 ζ (s , x)ζ (1 s, x)dx E E,s 2π1−s E E − 0 0 (7.7) Z 1 Z πis ′ + e 2 ζ (s , x)ζ (1 s, 1 x) . E E − − Z0 Substituting (4.1) and (4.2) in the integral (7.7), we deduce that 1 ′ ′ iΓ(1 s)Γ(s)Γ(1 s ) ′ ζE(s , x)ℓE,s(x)dx = − 2−s′ − λ(1 + s s ) 0 π − Z ′ − πis π(1 s s ) (7.8) e 2 cos − − × 2 ′ πis π(1 s + s ) + e 2 cos − . 2 Now the factor on the right hand side of (7.8) is

′ ′ ′ − πis π(1 s s ) πis π(1 s + s ) − πis e 2 cos − − + e 2 cos − = e 2 sin(πs). 2 2 Therefore, from the reﬂection formula π Γ(s)Γ(1 s)= , − sin(πs) we have

1 ′ ′ ′ iΓ(1 s ) − πis ′ ζ (s , x)ℓ (x)dx = − e 2 λ(1 + s s ). E E,s π1−s′ − Z0 18 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

This completes the proof.

Let φ(x, a, s) be the power Dirichlet series deﬁned for Re(s) > 1, x real, a = negative integer or zero, by the series 6 ∞ e2nπix (7.9) φ(x, a, s)= (n + a)s n=0 X (see [3, p. 161, (1.1)]). It is known that φ(x, a, s) can be extended to whole s-plane by means of the contour integral. The value of φ(x, a, s) when s is 0 or a negative integer can be calculated by applying Cauchy’s residue theorem. For m N , the Apostol’s formula (see [3, p. 164)]) gives ∈ 0 B (a, e2πix) (7.10) φ(x, a, m)= m+1 , − − m +1 where Bm(a, α) are the Apostol-Bernoulli polynomials deﬁned by zeaz ∞ zm (7.11) = B (a, α) . αez 1 m m! m=0 − X We refer to [3, 7, 20, 27, 33] for an account of the properties of Apostol- Bernoulli polynomials Bm(a, α). For a later purpose, we observe that, by (7.1) and (7.9), the following equality is established, 1 (7.12) ℓ (x)=2−seπixφ x, ,s , E,s 2 so that by (7.10) and (7.12),

1 2πix m πix Bm+1 2 , e (7.13) ℓE,−m(x)= 2 e , − m+1 where m N . ∈ 0 We observe that the function ζE(s, x) in (1.5) is a linear combination of the Hurwitz zeta functions ζ(s, x) in (1.1) when x is a rational number. Theorem 7.2. Let p, q be integers, q N and 1 p q. Then for all s C, we have ∈ ≤ ≤ (7.14)∈ q−1 p 2Γ(s) πs (2r + 1)πp 2r +1 ζ 1 s, = cos ζ s, . E − q (2qπ)s 2 − q 2q r=0 X Remark 7.3. This is an analogue of [5, p. 261, Theorem 12.8], where the Hurwitz zeta function ζ(s, x) instead of ζE(s, x). Proof of Theorem 7.2. Following an idea in [36, p. 337, (8)], we have

∞ q ∞ (7.15) f(n)= f(kq + r) n=1 r=1 X X Xk=0 19 if the involved series is absolutely convergent. Applying (7.15) to the series (7.1), we have (see [11, p. 1530, (8b)] and [10, p. 1487, (3.4)]) q p 1 (2r−1)πi p 2r 1 (7.16) ℓ = e q ζ s, − . E,s q (2q)s 2q r=1 X p Therefore, if we take x = q in the formula (7.2) to (7.16), in considerations of ℓ (1 x)= ℓ (x), we have the following equality E,s − − E,s p Γ(s) − πis p πis p ζ 1 s, = e 2 ℓ + e 2 ℓ E − q πs E,s q E,s −q q 2Γ(s) πs (2r 1)πp 2r 1 = cos − ζ s, − , (2qπ)s 2 − q 2q r=1 X which holds true, by the principle of analytic continuation, for all admissible values of s C. ∈ Corollary 7.4 ([11, p. 1529, Theorem B]). Let p, q be integers, q N and ∈ 1 p q. For m N, the Euler polynomials Em(x) at rational arguments are≤ given≤ by ∈ (7.17) q p 4m! (2r 1)πp mπ 2r 1 E = sin − ζ m +1, − . m q (2qπ)m+1 q − 2 2q r=1 X Proof. By setting s = m +1(m N0) in (7.14) and using (2.7), we obtain the formula in (7.17), which completes∈ the proof. Corollary 7.5. Let p, q be integers, q N and 1 p < q. Then we have ∈ ≤ q 1 2πi p 2(r−1)πi p 2r 1 (7.18) B , e q = qm e q B − , m N . m+1 2 m+1 2q ∈ 0 r=1 X Proof. Letting m N0, then replacing s by m in (7.16), using (1.3) and (7.13), we get our∈ result. − Remark 7.6. This identity is also a special case of the multiplication theorem for Apostol-Bernoulli polynomials (see eg. [28, Eq. (17)]).

Since ℓE,s(x) is a relative of the Hurwitz-type Euler zeta function ζE(s, x) deﬁned (1.5) (see (7.2) and (7.3) above), it is natural to expect a relation involving ζ(s, x): Lemma 7.7 (Generalized Eisenstein formula). Let p, q be integers, q N and 1 p < q. Then we have ∈ ≤ q 2p 1 1 −(2p−1)πi r r (7.19) ζ s, − = (2q)se q ℓ . 2q q E,s q r=1 X Remark 7.8. The equation (7.19) was announced without a proof in [10, p. 1487, (3.3)]. 20 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

Proof of Lemma 7.7. The formula (1.1) could be written in the following form 2p +1 ∞ 1 ζ s, = (2q)s 2q (2(qk + p)+1)s k=0 (7.20) X ∞ s 1 = (2q) s . n=0 (2n + 1) n≡pX(mod q) Note that in fact (see [5, p. 158, Theorem 8.1]) q 1 2πi(n−p) r 1 if n p (mod q) e q = ≡ q 0 otherwise, r=1 ( X and hence q 2πi(n−p) r 2p +1 ∞ 1 e q ζ s, = (2q)s q r=1 2q (2n + 1)s n=0 P (7.21) q X ∞ r (2n+1)πi q 1 e −(2p+1)πi r = (2q)s e q q (2n + 1)s r=1 n=0 X X for 1 p < q, the conclusion is immediate. ≤ Corollary 7.9. Let p, q be integers, q N and 1 p < q. Then we have (7.22) ∈ ≤ q 2p 1 1 −2(p−1)πi r 1 2πi r B − = e q B , e q , m N . m+1 2q qm+1 m+1 2 ∈ 0 r=1 X 1 1 Here, the term Bm+1 2 , 1 is understood as Bm+1 2 (the value of the Bernoulli polynomials B (x) at x = 1 ). m+1 2 Proof. Let m N0, and replace s by m in (7.19). Then, using (1.3) and (7.13), we get∈ our result. − Theorem 7.10. Let p, q be integers, q N and 1 p q. Then we have the identities ∈ ≤ ≤ q 1 −(2p+1)πi r r (2q)se q ℓ q E,s q r=1 (7.23) X Γ(1 s) − πi(1−s) πi(1−s) = − e 2 ℓ (x) e 2 ℓ (1 x) π1−s E,1−s − E,1−s − and n o (7.24) q 1 (2r−1)πi p 2r 1 e q ζ s, − (2q)s 2q r=1 X Γ(1 s) πi(1−s) p − πi(1−s) p = − e 2 ζ 1 s, e 2 ζ 1 s, 1 . 2π1−s E − q − E − − q Proof. From (7.2), (7.3), (7.16) and (7.19), we prove the theorem. 21

We will use [x] to denote the integer part of x R (i.e., [x] is the largest integer x); then, the fractional part of x will be∈ x = x [x]. ≤ { } −

Lemma 7.11. For n N we have ∈

∞ ( 1)n−1e(2k+1)πix + e−(2k+1)πix (7.25) E ( x )=2( i)n−1n! − . n { } − ((2k + 1)π)n+1 Xk=0

Proof. Let x = x [x]. The function E ( x ) is a quasi-periodic function { } − n { } of period 2 (see [20]). The Fourier series of En( x ) are given as follows ([2, p. 805, 23.1.16]): { }

4n! ∞ sin((2k + 1)πx 1 πn) (7.26) E ( x )= − 2 , n { } πn+1 (2k + 1)n+1 Xk=0 where 0 x< 1 if n N and 0

∞ sin((2k + 1)πx) (7.27) E ( x )=4( 1)m(2m)! , 2m { } − ((2k + 1)π)2m+1 Xk=0 and

∞ cos((2k + 1)πx) (7.28) E ( x )=4( 1)m−1(2m + 1)! . 2m+1 { } − ((2k + 1)π)2m+2 Xk=0

Thus (7.25) follows immediately from (7.27) and (7.28).

Theorem 7.12. Let m > 1 be a ﬁxed positive odd integer and let α be an integer such that α 0 (mod m). Then, for n N, we have (7.29) 6≡ ∈ m−1 n−1 r −2πir α r ( 1) n 2α 2α ( 1) e m ℓ = − m E + E (0) − E,1−n m 4 n−1 m − m n−1 r=1 X 1 2α 2α mnB + B (0) . − 2n n m − m n 22 SUHU,DAEYEOULKIM,ANDMIN-SOOKIM

Proof. It follows, from (7.3), that for odd m> 1, (7.30) m−1 r −2πir α r ( 1) e m ℓ − E,1−n m r=1 X n m−1 (n 1)!i α r r −2πir m = − n ( 1) e ζE n, 2π − m r=1 X m−1 n r −2πir α m r ( 1) ( 1) e m ζ n, − − − − E m r=1 ! X n m−1 (n 1)!i r −2πir α r = − ( 1) e m ζ n, 2πn − E m r=1 X m−1 ′ n r′ −2πir′ α r +( 1) ( 1) e m ζE n, − ′ − m ! rX=1 m−1 ∞ α α n k −2πir m n 2πir m (n 1)!i r ( 1) e +( 1) e = − ( 1) − −n 2πn − k + α r=1 k=0 m X X −2πiα (mk+r) 2πiα (mk+r) m−1 ∞ mk+r m n m (n 1)!inmn ( 1) e +( 1) e = − − − 2πn h (mk + r)n i r=1 X Xk=0 h −2πiα h n 2πiα h (n 1)!inmn ∞ ∞ ( 1) e m +( 1) e m = − − − 2πn h hn i h=1 h=1 h6≡0X (mod m) X h −2πiα h n 2πiα h (n 1)!inmn ∞ ( 1) e m +( 1) e m = − − − 2πn  h hn i Xh=1  ∞ ( 1)mk e−2πikα +( 1)ne2πikα + − − . (km)n k=1 ! X In the following, we will use the identities below: ∞ ( 1)n ∞ 1 ∞ 1 (7.31) − = + ns − (2k 1)s (2k)s n=1 X Xk=1 − Xk=1 and ∞ e2πikx +( 1)ne−2πikx (7.32) B ( x )= ( i)nn! − n { } − − (2πk)n Xk=1 N ′ for n (see [39, p. 13, 3] or [26]). Note that, by (7.25) and En( x ) = nE ∈( x ), we have § { } n−1 { } ∞ ( 1)ne(2k+1)πix + e−(2k+1)πix (7.33) E′ ( x )=2( i)nn! − . n { } − πn(2k + 1)n Xk=0 23

Therefore we have (7.34) m−1 r −2πi r r ( 1) e m ℓ − E,1−n m r=1 X n n ∞ −2πi(2k) α n 2πi(2k) α (n 1)!i m e m +( 1) e m = − n −n 2π (2k) Xk=1 ∞ −2πi(2k−1) α n 2πi(2k−1) α e m +( 1) e m − − (2k 1)n Xk=1 − ∞ e−2πi(2k)α +( 1)ne2πi(2k)α + m−n − (2k)n Xk=1 ∞ −2πi(2k−1)α n 2πi(2k−1)α −n e +( 1) e m − n − (2k 1) ! Xk=1 − ( 1)n−1 2α 2α = − mnE′ + E′ (2α [2α]) 4n n m − m n − 1 2α 2α mnB + B (2α [2α]) − 2n n m − m n − ( 1)n−1 2α 2α = − mnE + E (0) 4 n−1 m − m n−1 1 2α 2α mnB + B (0) . − 2n n m − m n This completes our proof. Remark 7.13. Equation (7.29) is an analogue of Theorem D in [39], which is an extension of the classical Eisenstein formula α α 1 1 m−1 2πγα πγ = sin cot , m − m − 2 −2m m n γ=1 h i X where α is an integer with α 0 (mod m) and m > 2 is a ﬁxed positive integer. 6≡

Acknowledgment We thank the referee for his/her helpful comments and suggestions.

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Department of Mathematics, South China University of Technology, Guangzhou, Guangdong 510640, China E-mail address: [email protected]

Department of Mathematics and Institute of Pure and Applied Mathe- matics, Chonbuk National University, 567 Baekje-daero, deokjin-gu, Jeonju- si, Jeollabuk-do 54896, Korea E-mail address: [email protected]

Division of Mathematics, Science, and Computers, Kyungnam Univer- sity, 7(Woryeong-dong) kyungnamdaehak-ro, Masanhappo-gu, Changwon- si, Gyeongsangnam-do 51767, Korea E-mail address: [email protected]