Four parametric linear Euler sums

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J. Math. Anal. Appl. ••• (••••) ••••••

Contents lists available at ScienceDirect

Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Four parametric linear Euler sums Horst Alzer a , Junesang Choi b,∗ a b

Morsbacher Str. 10, 51545 Waldbröl, Germany Department of Mathematics, Dongguk University, Gyeongju 38066, Republic of Korea

a r t i c l e

i n f o

Article history: Received 18 June 2019 Available online xxxx Submitted by V. Andriyevskyy Keywords: Euler sums Harmonic numbers Zeta functions Series representations Mathematical constants Analytic continuation

a b s t r a c t In this paper, firstly, we aim to investigate analytic continuations of altogether four types of parametric linear Euler sums, by using the Euler-Maclaurin summation formula and the Euler-Boole summation formula. Secondly, we show how nicely three shuffle relations among the four parametric linear Euler sums can be derived by using an L-summing formula due to Hassani and Rahimpour. Thirdly, using one of the shuffle relations, we present certain series representations of product of Riemann zeta functions. Some special cases and relevant connections of the results presented here with those involving known identities are also demonstrated and indicated. © 2019 Elsevier Inc. All rights reserved.

1. Introduction and preliminaries The harmonic numbers Hn are defined by

Hn :=

n  1 k

(n ∈ N).

(1.1)

k=1

Here and throughout, an empty sum is understood to be nil and so H0 = 0, and let N, Z− , Z, R, R+ , and C be the sets of positive integers, negative integers, integers, real numbers, positive real numbers, and + − + complex numbers, respectively, and let N0 := N ∪ {0}, Z− 0 := Z ∪ {0}, and R0 := R ∪ {0}. The harmonic (s) numbers Hn of order s are defined by

Hn(s) :=

n  1 ks

(n ∈ N, s ∈ C)

k=1

* Corresponding author. E-mail addresses: [email protected] (H. Alzer), [email protected] (J. Choi). https://doi.org/10.1016/j.jmaa.2019.123661 0022-247X/© 2019 Elsevier Inc. All rights reserved.

(1.2)

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(1)

(s)

and Hn = Hn (n ∈ N) are the harmonic numbers. The generalized harmonic numbers Hn (z) are defined by Hn(s) (z) :=

n 

1 (k + z)s

k=1 (s)

(s)



n ∈ N, s ∈ C, z ∈ C \ Z−



(1.3)

(s)

and Hn (0) = Hn . The odd harmonic numbers On of order s are defined by On(s) :=

n  k=1

1 (2k − 1)s

(n ∈ N, s ∈ C)

(1.4)

(n ∈ N, s ∈ C).

(1.5)

(1)

and On := On (n ∈ N). From (1.2) and (1.4), we find On(s) = H2n − 2−s Hn(s) (s)

(s)

The alternating harmonic numbers An of order s are defined by A(s) n :=

n  (−1)k+1 ks

(n ∈ N, s ∈ C)

(1.6)

k=1 (1)

and An := An . The harmonic numbers and the alternating harmonic numbers are connected by the elegant formula (s)

(s) 1−s A(s) H[n/2] , n = Hn − 2

(1.7)

where [x] denotes the greatest integer less than or equal to x. (s) The generalized alternating harmonic numbers An (z) are defined by A(s) n (z) :=

n  (−1)k+1 (k + z)s

(n ∈ N, s ∈ C, z ∈ C \ Z− ).

(1.8)

k=1

The Riemann Zeta function ζ(s) is defined by ζ(s) := lim Hn(s) = n→∞

∞  1 ((s) > 1) ks

(1.9)

k=1

and the closely related Dirichlet eta function η(s) is given by η(s) := lim A(s) n = n→∞

∞  (−1)k+1 k=1

ks

((s) > 0).

(1.10)

The special number ζ(3) = 1.20205 . . . is called Apéry constant. It is named after the French mathematician Roger Apéry, who proved in 1979 that ζ(3) is irrational (see [2]). We also denote O(s) by O(s) := lim On(s) = n→∞

∞  k=1

1 ((s) > 1). (2k − 1)s

(1.11)

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It may be useful to connect the Riemann Zeta function ζ(s) with two other series (see, e.g., [45, p. 164])   O(s) = 1 − 2−s ζ(s)

((s) > 1)

(1.12)

  η(s) = 1 − 21−s ζ(s)

((s) > 0).

(1.13)

and

The Hurwitz (or generalized) zeta function ζ(s, z) is defined by ∞ 

ζ(s, z) := lim Hn(s) (z − 1) = n→∞

1 (k + z)s

k=0



 (s) > 1, z ∈ C \ Z− 0 .

(1.14)

It is easy to find from (1.9) and (1.14) that −1

ζ(s) = ζ(s, 1) = (2s − 1)

  ζ s, 1/2 = 1 + ζ(s, 2).

(1.15)

We define the generalized eta function η(s, z) by η(s, z) := lim A(s) n (z − 1) = n→∞

∞  (−1)k (k + z)s



 (s) > 0, z ∈ C \ Z− 0 .

(1.16)

k=0

From (1.10) and (1.16), we have η(s, 1) = η(s)

and η(s, 2) = 1 − η(s).

(1.17)

The Polylogarithm function Lin (z) is defined by (see, e.g., [45, p. 185]) Lin (z) :=

∞  zk kn

(|z|  1, n ∈ N \ {1})

k=1

z =

(1.18) Lin−1 (t) dt t

(n ∈ N \ {1, 2}).

0

Clearly, we have Lin (1) = ζ(n)

(n ∈ N \ {1}).

(1.19)

The Dilogarithm function Li2 (z) is defined by Li2 (z) :=

∞  zn n2 n=1

z =−

(|z| ≤ 1) (1.20)

log(1 − t) dt. t

0

The psi (or digamma) function ψ(z) is defined by

ψ(z) :=

Γ (z) d {log Γ(z)} = dz Γ(z)

z or

log Γ(z) =

ψ(t) dt, 1

(1.21)

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where Γ is the familiar gamma function (see, e.g., [45, Section 1.1]). The psi function ψ(z) satisfies ∞

ψ(z) = −γ −

1  z + z n=1 n(z + n)

  z ∈ C \ Z− 0 ,

(1.22)

where γ is the Euler-Mascheroni constant (see, e.g., [45, Section 1.2]; see also [31,44]). The polygamma functions ψ (n) (z) are defined by ψ (n) (z) :=



dn+1 dn log Γ(z) = ψ(z) dz n+1 dz n

 n ∈ N, z ∈ C \ Z− 0 .

(1.23)

We recall a well-known (useful) relationship between the polygamma functions ψ (n)(z) and the Hurwitz zeta function ζ(s, z): ∞ 

1 = (−1)n+1 n! ζ(n + 1, z) (k + z)n+1 k=0   n ∈ N, z ∈ C \ Z− 0 .

ψ (n) (z) = (−1)n+1 n!

(1.24)

It is easy to derive the following expression: ψ (m) (z + n) − ψ (m) (z) = (−1)m m!

n  k=1

1 (z + k − 1)m+1

= (−1)m m! Hn(m+1) (z − 1)

(1.25)

(m, n ∈ N0 ) .

The Bernoulli polynomials Bn (x) are defined by the generating function: ∞  zn z exz = Bn (x) z e − 1 n=0 n!

(|z| < 2π).

(1.26)

The numbers Bn := Bn (0) are called the Bernoulli numbers generated by ∞  z zn = Bn z e − 1 n=0 n!

(|z| < 2π).

(1.27)

It easily follows from (1.26) and (1.27) that n    n Bn (x) = Bk xn−k . k

(1.28)

k=0

It is found that B0 = 1,

1 B1 = − , 2

B2n+1 = 0 (n ∈ N).

(1.29)

The connection between the Bernoulli numbers and the Riemann zeta function is given by Euler’s identity (see, e.g., [45, p. 166]) ζ(2n) = (−1)n+1

(2π)2n B2n 2 (2n)!

(n ∈ N0 ),

(1.30)

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which, in view of (1.27), yields the following special values: π4 π6 π2 , ζ(4) = , ζ(6) = , 6 90 945 π 10 π8 , ζ(10) = , ··· . ζ(8) = 9450 93555

ζ(2) =

¯n (x) (x ∈ R) with period 1 as follows (see, e.g., [3, p. 267]): We recall the nth Bernoulli periodic function B ¯n (x) = Bn (x − [x]) and B ¯n (x) = Bn (x) (0 < x ≤ 1). B

(1.31)

(α)

The generalized Bernoulli polynomials Bn (x) of degree n in x are defined by the generating function: 

z ez − 1

α exz =

∞ 

Bn(α) (x)

n=0

zn n!

(|z| < 2π, 1α := 1)

(1.32)

for arbitrary (real or complex) parameter α. Clearly, we have Bn(α) (x) = (−1)n Bn(α) (α − x),

(1.33)

Bn(α) (α) = (−1)n Bn(α) (0) =: (−1)n Bn(α)

(1.34)

so that

(α)

in terms of the generalized Bernoulli numbers Bn 

z z e −1

α =

∞ 

defined by the generating function:

Bn(α)

n=0

zn n!

(|z| < 2π, 1α := 1).

(1.35)

The Euler polynomials En (x) and the Euler numbers En are defined by the following generating functions: ∞  2exz zn = E (x) n ez + 1 n=0 n!

(|z| < π)

(1.36)

and ∞  2ez zn = sech z = E n e2z + 1 n! n=0



|z| <

π , 2

(1.37)

respectively. It is found that En = 2n En

  1 2

(n ∈ N0 )

and E2n+1 = 0 (n ∈ N0 ).

(1.38)

The first few of the Euler numbers En are given below: E0 = 1, E2 = −1, E4 = 5, E6 = −61, E8 = 1385, E10 = −50521, · · · .

(1.39)

¯n (n ∈ N0 ) with period 2 defined as follows (see, e.g., [13, Eq. (15)]): We recall an antiperiodic function E ¯n (x) = En (x) (0 ≤ x ≤ 1), E ¯n (x) ¯n (x + 2k) = E E

¯n (x + 2k + 1) = −E ¯n (x), E

(k ∈ N0 ) .

(1.40)

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Note that En (0) =

 2  1 − 2n+1 Bn+1 n+1

(n ∈ N0 )

(1.41)

and thus E2n (0) = 0 (n ∈ N).

(1.42)

(α)

(α)

The generalized Euler polynomials En (x) and the generalized Euler numbers En generating functions: 

2 ez + 1

α exp(xz) =

∞ 

En(α) (x)

n=0

zn n!

(|z| < π, 1α := 1)

are defined by the

(1.43)

and 

2 ez e2z + 1

α =

∞ 

En(α)

n=0

zn n!

 π |z| < , 1α := 1 2

(1.44)

for arbitrary (real or complex) parameter α. Clearly, we have En(1) (x) = En (x)

and En(1) = En

(n ∈ N0 ).

(1.45)

The following series involving harmonic numbers ∞  n=1

∞ 1  Hn Hn = = ζ(3) (n + 1)2 2 n=1 n2

(1.46)

was discovered by Euler in 1775 and has a long history (see, e.g., [7, p. 252 et seq.]; see also [15]). By applying Parseval’s identity to a Fourier series and the contour integral to a generating function, D. Borwein and J.M. Borwein [8] established the following interesting identity (see also [6, Eq. (2.16)], [10, p. 280], [20, Eq. (9)]): ∞  n=1

∞ 11  Hn2 11 Hn2 ζ(4). = = 2 2 (n + 1) 17 n=1 n 4

(1.47)

Euler initiated this line of investigation in the course of his correspondence with Goldbach from 1742 and he was the first to consider the linear harmonic sums (see, e.g., [17,23]) Sp,q :=

∞ (p)  Hn . nq n=1

(1.48)

Euler, whose investigations were completed by Nielsen in 1906 (see [35]), showed that the linear harmonic sums in (1.48) can be evaluated in the following cases: p = 1; p = q; p + q odd; p + q even, but with the pair (p, q) being the set {(2, 4), (4.2)}. Of these special cases, in the ones with p = q, if Sp,q is known, then Sq,p can be found by means of the symmetry relation Sp,q + Sq,p = ζ(p) ζ(q) + ζ(p + q)

(1.49)

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and vice versa (see, e.g., [1]). Rather extensive numerical search for linear relations between linear Euler sums and polynomials in zeta values (see [5,23]) strongly suggest that Euler found all the possible evaluations of linear harmonic sums, for example, 2 S1,q = (q + 2) ζ(q + 1) −

q−2 

ζ(q − j) ζ(j + 1) (q ∈ N \ {1}).

(1.50)

j=1

The nonlinear harmonic sums involve products of at least two (generalized) harmonic numbers. Let P = (p1 , . . . , pk ) be a partition of an integer p into k summands, so that p = p1 +· · ·+pk and p1 ≤ p2 ≤ · · · ≤ pk . The Euler sum of index P, q is defined by SP ;q :=

∞ (p ) (p ) (p )  H n 1 H n 2 · · · Hn k , nq n=1

(1.51)

where the quantity q + p1 + · · · + pk is called the weight, the quantity k is the degree. For simplicity, repeated summands in partitions are denoted by powers, for example, S13 ,22 ,6;q = S1,1,1,2,2,6;q

(2) 2 (6) ∞  Hn3 Hn Hn = . q n n=1

For a broad survey of works on multiple zeta values and Euler sums of arbitrary degree, the interested reader can refer to a survey-cum-expository paper [11]. Since then and up to now, the research subject on multiple zeta values and Euler sums has been actively investigated (see, e.g., [21,22,24,25,32,34,36–38,40, 42,43,46–55]). Flajolet and Salvy [23] introduced the following notations for altogether four types of linear Euler sums: ++ Sp,q =

∞ (p)  Hk , kq

+− Sp,q =

k=1 −+ Sp,q =

∞  k=1

(p) Ak , kq

∞ 

(p)

(−1)k+1

Hk , kq

(−1)k+1

Ak . kq

k=1 −− Sp,q =

∞ 

(p)

k=1

++ Clearly Sp,q = Sp,q . Analogously, we introduce altogether four types of parametric linear Euler sums:

++ Sp,q (a, b) :=

∞ (p)  Hk (a) , (k + b)q

+− Sp,q (a, b) :=

k=1 −+ (a, b) Sp,q

:=

∞  k=1

(p) Ak (a) , (k + b)q

∞ 

(p)

(−1)k+1

Hk (a) , (k + b)q

k+1

Ak (a) , (k + b)q

k=1 −− Sp,q (a, b)

:=

∞  k=1

(p)

(−1)

(1.52)

where a, b ∈ C \ Z− and p, q ∈ C are adjusted so that the involved defining series can converge. Obviously ++ ++ +− +− −+ −+ −− −− Sp,q (0, 0) = Sp,q , Sp,q (0, 0) = Sp,q , Sp,q (0, 0) = Sp,q , Sp,q (0, 0) = Sp,q .

Certain parametric Euler sums have been investigated (see, e.g., [9,32,37,38,48]). In this paper, firstly, we aim to investigate analytic continuations of the four types of parametric linear Euler sums in (1.52), by using the Euler-Maclaurin summation formula (2.14) and the Euler-Boole summation formula (2.15). Secondly, we establish three shuffle relations among the four parametric linear Euler sums, by using the following L-summing formula due to Hassani and Rahimpour [30]:

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ajk =

n  k 

(ajk + akj ) −

k=1 j=1

j,k=1

n 

akk

(n ∈ N)

(1.53)

k=1

(see also [27–29]). Thirdly, using one of the shuffle relations, we establish certain series representations of product of Riemann zeta functions. Some special cases and relevant connections of the results presented here with those involving known identities are also demonstrated and indicated. 2. Analytic continuations of the four parametric linear Euler sums The Riemann zeta function ζ(s) in (1.9) is an analytic function of s in the half-plane (s) > 1. The ζ(s) can be continued meromorphically to the whole complex s-plane with a simple pole only at s = 1 with residue 1. It is known that ⎧ 1 ⎪ (n = 0) ⎨ −2 (2.1) ζ(−n) = ⎪ ⎩ − Bn+1 (n ∈ N) n+1 and ζ(−2n) = 0 (n ∈ N).

(2.2)

The Hurwitz (or generalized) zeta function ζ(s, z) in (1.14) is an analytic function of s in the half-plane (s) > 1. The ζ(s, z) can be continued meromorphically to the whole complex s-plane with a simple pole only at s = 1 with residue 1. It is known that ζ(−, x) = −

B+1 (x) +1

( ∈ N0 ).

(2.3)

The eta function η(s) in (1.10) is an analytic function of s in the half-plane (s) > 0. The function η(s)   can be continued entirely to the whole complex s-plane. In view of the relation (1.13), the factor 1 − 21−s is an entire function of s and ζ(s) is an analytic function in the whole complex s-plane except for s = 1. The Laurent series of ζ(s) in a neighborhood of its pole s = 1 has the form (see, e.g., [45, p. 165]): ζ(s) =

∞  1 +γ+ an (s − 1)n , s−1 n=1

(2.4)

with  an = lim

m→∞

m  (log m)n+1 (log k)n − k n+1

 (n ∈ N).

(2.5)

k=1

Using the expansion (2.4), we compute   1 − 21−s = log 2 and so lim 1 − 21−s ζ(s) = lim s→1 s→1 s − 1

η(1) = log 2.

(2.6)

From (1.13), (2.1) and (2.2), we find η(0) =

1 , 2

η(−2n) = 0,

η(1 − 2n) = −

B2n 2n

(n ∈ N).

(2.7)

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The n-ple (or, simply, multiple) Hurwitz zeta function ζn (s, a) :=

∞ 

−s

((s) > n, a ∈ C \ Z− 0)

(a + k1 + · · · + kn )

(2.8)

k1 , ··· , kn =0

is an analytic function of s in the half-plane (s) > n. The function ζn (s, a) in (2.8) can be continued meromorphically to the whole complex s-plane with simple poles at s = k (k = 1, . . . , n) (see, e.g., [45, Section 2.1]). It is found (see [45, p. 151]) that ζn (−, x) = (−1)n

! (n) B (x) ( ∈ N0 ), (n + )! n+

(2.9)

whose special case n = 1 yields (2.3). The multiple Hurwitz-Euler eta function ηn (s, a) defined by ηn (s, a) :=

∞ 

  (s) > 0, a > 0, n ∈ N

(−1)k1 +···+kn s (a + k1 + · · · + kn )

k1 , ··· , kn =0

(2.10)

is an analytic function of s in the half-plane (s) > 0. The function ηn (s, a) in (2.8) can be continued entirely to the whole complex s-plane (see [18]). It is known (see [18]) that ηn (−, a) =

(−1) (n) E (n − a) 2n

( ∈ N0 , n ∈ N) .

(2.11)

In particular, for the generalized eta function η1 (s, a) = η(s, a), η(−, a) =

(−1) E (1 − a) 2

( ∈ N0 ) .

(2.12)

++ Apostol and Vu [4] and Matsuoka [33] showed that S1,s can be continued meromorphically to the whole s-plane with a second-order pole at s = 1 and simple poles at s = 0, −1, −3, · · · . +− −− −+ , S1,s and S1,s can be analytically extended to the whole sBoyadzhiev et al. [14] showed that S1,s plane with poles at negative integers whose residues are given in terms of Bernoulli and Euler numbers: +− The function S1,s admits an extension to the closed left half plane (s) ≤ 0 as an entire function. The −− function S1,s admits an analytic extension to the closed left half plane (s) ≤ 0 with simple poles at −+ s = 0, −1, −3, · · · . The function S1,s extends as an analytic function to the whole complex s-plane with a simple pole at s = 1 with residue η(1) = log 2. Here, modifying the proofs given in Apostol and Vu [4] and Boyadzhiev et al. [14], we give analytic continuations of the following four parametric linear Euler sums: ++ Sp,s (a, b),

+− Sp,s (a, b),

−+ Sp,s (a, b),

−− Sp,s (a, b)

  p ∈ N, a, b ∈ R+ 0

(2.13)

when a = b. The following summation formulas are required. Euler-Maclaurin summation formula (see, e.g., [45, p. 220]): Let f be a function defined for t ≥ x with continuous derivatives of order 2q + 1 and such that f (k) (t) → 0 as t → ∞ (k = 0, 1, . . . , 2r + 1). Then ∞  k=0

∞ f (x + k) =

f (t) dt + x

+

1 (2r + 1)!

r  1 B2m (2m−1) f (x) − f (x) 2 (2m)! m=1

(2.14)

∞ ¯2r+1 (t + x) f (2r+1) (x + t) dt. B 0

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Euler-Boole summation formula (see, e.g., [13, Eq. (13)]; see also [12, Eq. (3.30)]): Let f be a function defined for t ≥ x with continuous derivatives of order 2r + 1 and such that f (k) (t) → 0 as t → ∞ (k = 0, 1, . . . , 2r + 1). Then ∞  k=0

r−1 1  E2m+1 (0) (2m+1) 1 f (−1)k f (x + k) = f (x) + (x) 2 2 m=0 (2m + 1)!

1 + 2 (2r)!

(2.15)

∞ ¯2r (−t) f E

(2r+1)

(x + t) dt.

0

Theorem 2.1. Let p ∈ R+ , b ∈ R+ 0 . Also let (s) > −2r and (s) > −2r − p + 1 for any r ∈ N. Then ++ Sp,s (b, b) =

ζ(s + p − 1, b + 1) 1 + ζ(s + p, b + 1) s−1 2 r  B2m ζ(s + p + 2m − 1, b + 1) (s)2m−1 + (2m)! m=1 −

(2.16)

(s)2r+1 α(s; b, p, r), (2r + 1)!

where ∞ ∞ 

α(s; b, p, r) :=

n=1 0

¯2r+1 (t + n) B dt. (n + b)p (n + t + b)s+2r+1

(2.17)

Here (λ)ν denotes the Pochhammer symbol which is defined (for λ, ν ∈ C), in terms of the gamma function Γ, by Γ(λ + ν) = (λ)ν := Γ(λ)

 1

(ν = 0, λ ∈ C \ {0})

λ(λ + 1) · · · (λ + n − 1)

(ν = n ∈ N, λ ∈ C),

(2.18)

it being understood conventionally that (0)0 := 1.   Proof. Taking f (x) = (x + b)−s (s) > 0, b ∈ R+ 0 , we find f (m) (x) = (−1)m s(s + 1) · · · (s + m − 1) m

= (−1)

1 (x + b)s+m

(s)m . (x + b)s+m

(2.19)

Changing the order of summations, we get ++ Sp,s (b, b) =

∞  n=1

=

∞  n=1

n ∞ ∞    1 1 1 1 = p (n + b)s (k + b)p (n + b) (k + b)s n=1 k=1

1 (n + b)p

∞  k=0

k=n

1 . (k + n + b)s

(2.20)

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H. Alzer, J. Choi / J. Math. Anal. Appl. ••• (••••) ••••••

11

Applying (2.19) with x = n to (2.14), we obtain ∞  k=0

r  1 (s)2m−1 (n + b)1−s 1 B2m + = + (k + n + b)s s−1 2(n + b)s m=1 (2m)! (n + b)s+2m−1

(s)2r+1 − (2r + 1)!

∞

¯2r+1 (t + n) B dt. (n + t + b)s+2r+1

(2.21)

0

Multiplying both sides of (2.21) by find (2.16) with (2.17). 

1 (n+b)p

and summing both sides of the resulting from n = 1 to ∞, we

¯2r+1 (t + n) is bounded on R, choose Remark 2.1. α(s; b, p, r) converges under given conditions. Since B + ¯ M ∈ R such that |B2r+1 (t + n)| ≤ M (t ∈ R). Then ∞    ∞ ¯   1 B2r+1 (t + n)   dt  (n + t + b)s+2r+1 dt ≤ M (s)+2r+1 (n + t + b)   0

0

=

M M ≤ ((s) + 2r)(n + t + b)(s)+2r ((s) + 2r)(n + b)(s)+2r

for any t ∈ R+ 0 . Then |α(s; b, p, r)| ≤

∞  M 1 < ∞. (s) + 2r n=1 (n + b)(s)+2r+p

++ When p ∈ R+ \ {1}, Sp,s (b, b) has simple poles at

s = 1, −p + 2, −p + 1, −p − 2m + 2 (m = 1, . . . , r). ++ S1,s (b, b) has a pole of order 2 at s = 1 and simple poles at s = 0, 1 − 2m (m = 1, . . . , r). ++ Since we can take r ∈ N as r → ∞, Sp,s (b, b) can be continued meromorphically to the whole complex s-plane. ++ (b, b) ( ∈ N) can be expressed in terms of Using (2.3) or the case n = 1 of (2.9), when p ∈ N, Sp,− Bernoulli numbers and Bernoulli polynomials.

Using (2.15), the proofs of the following three theorems would run parallel to the proof of Theorem 2.1. We omit the details. Theorem 2.2. Let p ∈ R+ , b ∈ R+ 0 . Also let (s) > −2r and (s) > −2r − p + 1 for any r ∈ N. Then r−1 1  E2m+1 (0) 1 +− (s)2m+1 η(s + p + 2m + 1, b + 1) Sp,s (b, b) = η(s + p, b + 1) − 2 2 m=0 (2m + 1)!

(2.22)

(s)2r+1 + β(s; b, p, r), 2 (2r)! where β(s; b, p, r) :=

∞ ∞  n=1 0

¯2r (−t) (−1)n E dt. p (n + b) (t + n + b)s+2r+1

(2.23)

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H. Alzer, J. Choi / J. Math. Anal. Appl. ••• (••••) ••••••

Remark 2.2. In (2.10), η1 (s, a) = η(s, a) can be extended to an entire function to the whole complex s-plane. +− As r → ∞, Sp,s (b, b) extends to an entire function to the whole complex s-plane. +− In view of (2.12), Sp,− (b, b) can be expressed in terms of Euler polynomials when p,  ∈ N with − + p, − − + p + 2m + 1 ∈ Z in (2.22). Theorem 2.3. Let p ∈ R+ , b ∈ R+ 0 . Also let (s) > −2r and (s) > −2r − p + 1 for any r ∈ N. Then r−1 1  E2m+1 (0) 1 −− (s)2m+1 ζ(s + p + 2m + 1, b + 1) Sp,s (b, b) = ζ(s + p, b + 1) − 2 2 m=0 (2m + 1)!

(2.24)

(s)2r+1 − γ(s; b, p, r), 2 (2r)! where

γ(s; b, p, r) :=

∞ ∞  n=1 0

¯2r (−t) E dt. (n + b)p (n + t + b)s+2r+1

(2.25)

+− Remark 2.3. As r → ∞, Sp,s (b, b) extends to an analytic function to the whole complex s-plane except for simple poles at s = 1 − p and s = −p − 2m (m = 0, . . . , r − 1). −− In view of (2.3) or the case n = 1 of (2.9), Sp,− (b, b) can be expressed in terms of Euler polynomials and Bernoulli polynomials when p,  ∈ N with − + p, − + p + 2m + 1 ∈ Z− in (2.24).

Theorem 2.4. Let p ∈ R+ , b ∈ R+ 0 . Also let (s) > −2r and (s) > −2r − p + 1 for any r ∈ N. Then 1 −+ Sp,s (b, b) = η(s + p, b + 1) 2 r−1 1  E2m+1 (0) − (s)2m+1 η(s + p + 2m + 1, b + 1) 2 m=0 (2m + 1)! +

(2.26)

(s)2r+1 δ(s; b, p, r), 2 (2r)!

where

δ(s; b, p, r) :=

∞ ∞  n=1 0

¯2r (−t) (−1)n E dt. p (n + b) (t + n + b)s+2r+1

(2.27)

Remark 2.4. As in Remark 2.2, η1 (s, a) = η(s, a) can be extended to an entire function to the whole complex −+ s-plane. As r → ∞, Sp,s (b, b) extends to an entire function to the whole complex s-plane. −+ In view of (2.12), Sp,− (b, b) can be expressed in terms of Euler polynomials when p,  ∈ N with − + p, − − + p + 2m + 1 ∈ Z in (2.26). 3. Shuffle (or reciprocity) relations We present three shuffle relations among the four parametric linear Euler sums in the following three theorems.

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H. Alzer, J. Choi / J. Math. Anal. Appl. ••• (••••) ••••••

13

Theorem 3.1. The following formulas hold. ++ ++ Sp,q (a, b) + Sq,p (b, a) = ζ(p, a + 1) ζ(q, b + 1) +



∞ 

(k +  −

k=1

min{(p), (q)} > 1, a, b ∈ C \ Z

1 (k + b)q

a)p

(3.1)

and ++ ++ Sp,q (a, a) + Sq,p (a, a) = ζ(p, a + 1) ζ(q, a + 1) + ζ(p + q, a + 1)   min{(p), (q)} > 1, a ∈ C \ Z− .

(3.2)

Here ∞  k=1

  1 q p + q − 2 ψ (b + 1) − ψ (a + 1) =(−1) (k + a)p (k + b)q (a − b)p+q−1 p−1   q−1  1 p+q−j−2 ψ (j) (b + 1) + (−1) j! (a − b)p+q−1−j p−1 j=1 q

+ (−1)p 

(3.3)

  p−1  ψ (j) (a + 1) 1 p+q−j−2 j! q−1 (b − a)p+q−1−j j=1

 a, b ∈ C \ Z− , a = b, p, q ∈ N .

Proof. In (1.53), setting ajk =

1 1 (j + a)p (k + b)q



 a, b ∈ C \ Z− , p, q ∈ C ,

we obtain n n n (p) (q)   Hk (a)  Hk (b) 1 (p) (q) + = H (a) H (b) + . n n q p p (k + b) (k + a) (k + a) (k + b)q

k=1

k=1

(3.4)

k=1

Taking the limit as n → ∞ on both sides of (3.4) and using (1.14), we get (3.1). Choosing a = b in (3.1), we have (3.2). For Equation (3.3), we use (1.22) to derive the following known identity (see, e.g., [26, Entry (6.1.18)]): ∞  k=0

1 1 = {ψ (b) − ψ (a)} (k + a) (k + b) b−a



 a, b ∈ C \ Z− 0 , a = b .

(3.5)

By using Leibniz’s rule for higher-order differentiation of product of two functions, we differentiate p − 1 times both sides of (3.5) with respect to a and differentiate q − 1 times both sides of the resulting identity with respect to b. Then we divide both sides by (−1)p+q (p − 1)!(q − 1)! and simplify the expression on the right-hand side. Finally, we replace a by a + 1 and b by b + 1. This leads to (3.3).  Remark 3.1. In view of relation (1.24), the Equation (3.3) can be expressed in terms of the Hurwitz zeta function.

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H. Alzer, J. Choi / J. Math. Anal. Appl. ••• (••••) ••••••

14

Theorem 3.2. The following formulas hold. ∞ 

(−1)k+1 (k + a)p (k + b)q k=1   (p) > 1, (q) > 0, a, b ∈ C \ Z−

+− −+ Sp,q (a, b) + Sq,p (b, a) = ζ(p, a + 1) η(q, b + 1) +

(3.6)

and +− −+ Sp,q (a, a) + Sq,p (a, a) = ζ(p, a + 1) η(q, a + 1) + η(p + q, a + 1)   (p) > 1, (q) > 0, a ∈ C \ Z− .

(3.7)

Here ∞  k=1

 p−1   q+j−1 (−1)k+1 −q (a − b)−j η(p − j, a + 1) =(b − a) q − 1 (k + a)p (k + b)q j=0  q−1   p+j−1 −p + (a − b) (b − a)−j η(q − j, b + 1) p − 1 j=0   p, q ∈ N, a, b ∈ C \ Z− , a = b .

(3.8)

Proof. In (1.53), setting ajk =

1 (−1)k+1 (j + a)p (k + b)q



 a, b ∈ C \ Z− , p, q ∈ C ,

we get n n n (q)    Ak (b) (−1)k+1 (p) (−1)k+1 (p) (q) H (a) + = H (a) A (b) + . n n (k + b)q k (k + a)p (k + a)p (k + b)q

k=1

k=1

(3.9)

k=1

Taking the limit as n → ∞ on both sides of (3.9), we obtain (3.6). Choosing a = b in (3.6), we find (3.7). Equation (3.8) is a modified version of a known identity (see, e.g., [26, Entry (6.9.11)]).  Theorem 3.3. The following formulas hold. ∞ 

1 (k + (k + b)q k=1   min{(p), (q)} > 0, (p + q) > 1, a, b ∈ C \ Z−

−− −− Sp,q (a, b) + Sq,p (b, a) = η(p, a + 1) η(q, b + 1) +

a)p

(3.10)

and −− −− Sp,q (a, a) + Sq,p (a, a) = η(p, a + 1) η(q, a + 1) + ζ(p + q, a + 1)   min{(p), (q)} > 0, (p + q) > 1, a ∈ C \ Z− .

Here, when p, q ∈ N, a, b ∈ C \ Z− with a = b, the summation in (3.10) is given in (3.3).

(3.11)

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H. Alzer, J. Choi / J. Math. Anal. Appl. ••• (••••) ••••••

15

Proof. In (1.53), setting ajk =

(−1)j+1 (−1)k+1 (j + a)p (k + b)q



 a, b ∈ C \ Z− , p, q ∈ C ,

we obtain n n n    (−1)k+1 (p) (−1)k+1 (q) 1 (p) (q) A (a) + A (b) = A (a) A (b) + . n n k k q p p (k + b) (k + a) (k + a) (k + b)q

k=1

k=1

k=1

Now, similarly as in the proof of Theorems 3.1 and 3.2, we can get (3.10) and (3.11).  Setting a = 0 in (3.2), (3.7) and (3.11) and combining the resulting identities, we obtain four symmetric series identities involving the zeta and eta functions. It turns out that the coefficients of the series can be (s) (s) expressed in terms of Hk and Ak . Corollary 3.1. Let min{(p), (q)} > 1. Then ∞  (p) (q)   Pk Pk = ζ(p) ζ(q) + η(p) η(q), + kq kp

(3.12)

k=1

where (z)

Pk

(z)

(z)

:= Hk−1 + (−1)k+1 Ak =

k  1 1 + (−1)k+j − z z j k j=1

(k ∈ N, z ∈ C) ;

∞  (p) (q)   Qk Qk = ζ(p) ζ(q) − η(p) η(q), + kq kp

(3.13)

(3.14)

k=1

where (z)

(z)

(z)

Qk := Hk + (−1)k Ak =

k  1 − (−1)k+j jz j=1

(k ∈ N, z ∈ C) ;

∞  (p) (q)   Uk Uk = ζ(p) η(q) + ζ(q) η(p), + p kq k

(3.15)

(3.16)

k=1

where (z)

Uk

(z)

(z)

:= (−1)k+1 Hk−1 + Ak = −

k  (−1)j + (−1)k (−1)k + z j kz j=1

(k ∈ N, z ∈ C) ;

∞  (p) (q)   Vk Vk = ζ(p) η(q) − ζ(q) η(p), − kq kp

(3.17)

(3.18)

k=1

where (z)

Vk

(z)

(z)

:= (−1)k+1 Hk − Ak =

k  (−1)j − (−1)k jz j=1

(k ∈ N, z ∈ C) .

(3.19)

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H. Alzer, J. Choi / J. Math. Anal. Appl. ••• (••••) ••••••

16

4. Series representation of product of the Riemann zeta functions The identity ζ 2 (t) =

∞  d(k) k=1

(t > 1),

kt

(4.1)

where d(k) denotes the number of divisors of k, is a known result in analytic number theory (see, e.g., [3, p. 229]). Here, we use the identity (3.18) to establish new series representations for ζ(s)ζ(t) (s = t) and ζ 2 (t). Theorem 4.1. Each of the following representations holds:  (s)  ∞ (t)  Vk Vk 1 ζ(s)ζ(t) = 1−s − s 2 − 21−t kt k

(s, t > 1 with s = t) ,

(4.2)

k=1

(z)

where Vk

is given in (3.19). ∞

ζ 2 (t) =

2t−1  Ck log 2 kt

(t)

(t > 1) ,

(4.3)

 log(k/j) . jt

(4.4)

k=2

where (t) Ck

:=

k−1 



(−1)k − (−1)j

j=1

In particular, ζ(3) = (−1)n

∞  (3) (2n)   Vk Vk 8 (2n)! − (23−2n − 1)(2π)2n B2n k2n k3

(n ∈ N)

(4.5)

k=1

and π 4n =

∞ (2n)  Ck {(2n)!}2  2 2n 22n−1 B2n log 2 k=2 k

(n ∈ N) .

(4.6)

Proof. Setting p = s, q = t in (3.18) and using (1.13) in the resulting identity, we obtain (4.2). Next, we prove (4.3). Let

Ft (s) :=

∞  k=1



(s)

(t)

Vk V − ks kt k

 (s, t > 1) .

(4.7)

Here we consider s as a variable and t as a fixed number. Note that k k     1 1  (s)  = 2 Hk ≤ 2 Vk  ≤ 2 s j j j=1 j=1

(s > 1, k ∈ N).

(4.8)

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H. Alzer, J. Choi / J. Math. Anal. Appl. ••• (••••) ••••••

17

Using (4.8),  |Ft (s)| ≤ 2

∞ ∞  Hk  Hk + kt ks

k=1

 <∞

(s, t > 1) .

(4.9)

k=1

The series Ft (s) converges absolutely for s, t > 1. Let (s)

fk (s) :=

(t)

Vk V − ks t k k

(s, t > 1, k ∈ N) .

(4.10)

We find that each fk (s) (k ∈ N) is differentiable for s > 1: fk (s) =

k (t) Vk 1  (−1)k − (−1)j log j + log k kt j=1 js ks

(s, t > 1, k ∈ N) .

(4.11)

Let a, b be real numbers with 1 < a < b. Then, for k ≥ 1 and a ≤ s ≤ b, |fk (s)| ≤

2 2 4 Hk log k + a Hk log k ≤ c Hk log k t k k k

(4.12)

with c := min{t, a}. We have Hk ≤ 3 log k

(k ≥ 2).

(4.13)

Let c∗ := (c − 1)/2. Then, c − c∗ > 1. There exists an integer k0 ≥ 2 such that log k ≤ kc

∗

/2

(k ≥ k0 ).

(4.14)

It follows that ∞ ∞ ∞    Hk log k (log k)2 1 ≤ 3 ≤ 3 < ∞. c c c−c k k k ∗

k=k0

k=k0

(4.15)

k=k0

∞ From (4.11) and (4.14), we conclude that k=1 fk (s) converges uniformly on every compact interval [a, b] ⊂ (1, ∞). Thus, the series Ft (s) can be differentiated term-by-term at each s > 1, Ft (s) =

∞ 

fk (s).

(4.16)

k=1

From (4.2) and (4.7), since Ft (t) = 0, by using l’Hôpital’s rule, we get ζ 2 (t) = lim

F  (t) Ft (s) . = − 1−tt 1−t −2 2 log 2

s→t 21−s

(4.17)

From (4.11), we obtain fk (t) = −

k−1 1  (−1)k − (−1)j log(k/j) kt j=1 jt

(t > 1, k ∈ N).

(4.18)

Then, using (4.16), (4.17) and (4.18), we obtain the identity (4.3). Setting s = 3, t = 2n (n ∈ N) in (4.2) and t = 2n (n ∈ N) in (4.3), respectively, and using (1.30), we find the representations (4.5) and (4.6). 

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18

5. Remarks and special cases We give certain special cases of the main identities along with some suitable remarks. +− Example 1. Sitaramachandra Rao [41, Theorem 3.7] evaluated S1,2q (q ∈ N) as a linear combination of +− 5 products of Riemann zeta functions. For example, S1,2 = 8 ζ(3). By using contour integral representations and residue computations, Flajolet and Salvy [23, Theorem 7.2] evaluated, when p + q is odd, −+ Sp,q ,

+− Sp,q ,

−− Sp,q

in terms of linear combinations of products of Riemann zeta functions. Example 2. Setting a = 0 in (3.2), we obtain the reflection formula (1.49), which has been proved in various ways (see, e.g., [1,16,19,23]). Rao and Sarma [39] showed the reciprocity relation (1.49) for p, q ∈ N \ {1}. Apostol and Vu [4] argued that the reciprocity relation (1.49) holds for min{(p), (q)} > 1 (see (3.2) with a = 0) together with many involved identities and some useful and informative comments. Example 3. Setting a = 0 in (3.7) and (3.11), with the aid of (1.10), (1.13), (1.15) and (1.16), respectively, we obtain     +− −+ Sp,q + Sq,p = 1 − 21−q ζ(p) ζ(q) + 1 − 21−p−q ζ(p + q)

(5.1)

((p) > 1, (q) > 0) and    −− −− Sp,q + Sq,p = 1 − 21−p 1 − 21−q ζ(p) ζ(q) + ζ(p + q)

(5.2)

(min{(p), (q)} > 0, (p) + (q) > 1) . The relations (5.1) and (5.2) analogous to (1.49) are recorded in [23, p. 33]. In view of (2.6), the case q = 1 of (5.1) and the case p = q = 1 of (5.2), respectively, give +− −+ Sp,1 + S1,p = ζ(p) log 2 + η(p + 1)

((p) > 1)

(5.3)

and −− S1,1 =

1 1 log2 2 + ζ(2). 2 2

(5.4)

Recall the following known formula (see, e.g., [23, Theorem 7.1]; see also [41]) 1 −+ S1,p =ζ(p) log 2 − p ζ(p + 1) + η(p + 1) 2 p  1 η(k) η(p − k + 1) (p ∈ N \ {1}). + 2

(5.5)

k=1

Using (5.3) and (5.5), we get p 1  1 +− Sp,1 = p ζ(p + 1) − η(k) η(p − k + 1) (p ∈ N \ {1}). 2 2 k=1

(5.6)

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H. Alzer, J. Choi / J. Math. Anal. Appl. ••• (••••) ••••••

19

Example 4. Setting p = q in (3.11), we have −− Sp,p (a, a) =

1 2 1 η (p, a + 1) + ζ(2p, a + 1) 2 2

 (p) >

 1 , a ∈ C \ Z− . 2

(5.7)

Further setting a = 0 in (5.7), we obtain −− Sp,p

2 1  1 1 − 21−p ζ 2 (p) + ζ(2p) = 2 2



1 (p) > 2

 .

(5.8)

Example 5. We find ++ ++ Sp,q (0, 1) = Sp,q − ζ(p + q),

+− +− Sp,q (0, 1) = −Sp,q + η(p + q),

−+ −+ (0, 1) = Sp,q − η(p + q), Sp,q

−− −− Sp,q (0, 1) = −Sp,q + ζ(p + q).

Example 6. Coffey and Lubbers [19, p. 696, Eq. (57)] presented the following Euler sum (see also [23,41]): +− S2,2 (0, 1)

65 1 ζ(4) + ζ(2) log2 2 − log4 2 − 4 Li4 = 16 6

  1 7 − ζ(3) log 2. 2 2

(5.9)

From Example 5, we have +− +− +− S2,2 (0, 1) = −S2,2 + η(4) = −S2,2 +

7 ζ(4). 8

(5.10)

Applying (5.9) and (5.10), we get +− S2,2

51 1 = − ζ(4) − ζ(2) log2 2 + log4 2 + 4 Li4 16 6

  7 1 + ζ(3) log 2. 2 2

(5.11)

From (5.1), we have +− −+ S2,2 + S2,2 =

17 ζ(4). 8

(5.12)

Using (5.11) in (5.12), we obtain −+ S2,2 =

85 1 ζ(4) + ζ(2) log2 2 − log4 2 − 4 Li4 16 6

  7 1 − ζ(3) log 2. 2 2

(5.13)

Example 7. Setting a = 0 and b = 1 in Theorems 3.1, 3.2 and 3.3 and choosing to use some suitable identities given up to here, we obtain certain shuffle relations in the following corollary. Corollary 5.1. The following relations hold. ++ ++ Sp,q + Sq,p (1, 0) =ζ(p + q) + ζ(p) ζ(q) − ζ(p) +

∞ 

1 , kp (k + 1)q

(5.14)

∞  (−1)k+1 , kp (k + 1)q

(5.15)

k=1 −+ +− (1, 0) − Sp,q = ζ(p) − ζ(p) η(q) − η(p + q) + Sq,p

k=1 −− −− (1, 0) − Sp,q = η(p) − η(p) η(q) − ζ(p + q) + Sq,p

∞  k=1

kp (k

1 , + 1)q

(5.16)

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where ∞  k=1

 q−1   p+q−j−2 1 p−1 =(−1) p−1 kp (k + 1)q j=0  q−1   p+q−j−2 + (−1) ζ(j + 1) p−1 j=1 p

p−1 

+ (−1)p

(5.17)

 (−1)j+1

j=1

 p+q−j−2 ζ(j + 1) q−1

(p, q ∈ N) and ∞  k=1

   q−1   (−1)k+1 p+j−1 p p+1 p + q − 2 =(−1) + 2(−1) log 2 kp (k + 1)q p−1 p−1 j=0    q+j−1  1 − 21−p+j ζ(p − j) (−1) + q − 1 j=0 p−2 

j

+ (−1)p+1

(5.18)

 q−2    p+j−1  1 − 21−q+j ζ(q − j) p−1 j=0 (p, q ∈ N) .

It is interesting to compare the formulas in (5.17) and (5.18) with those in [26, Entries (6.2.12) and (6.2.14)], respectively. Example 8. We set p = q = 3 in (3.12) and (3.16) and make use of (1.13). Then ∞

ζ 2 (3) =

∞

32  Pk 4  Uk = . 3 25 k 3 k3 (3)

k=1

(3)

(5.19)

k=1

Applying (3.14) with p = q = 3 leads to the representations 32  Qk 64  X2k Y2k−1 ζ (3) = = + 7 k3 7 (2k)3 (2k − 1)3 ∞

∞

(3)

2

k=1

(5.20)

k=1

with k  1 Xk = 3 j j=1

k  1 and Yk = . 3 j j=1

(5.21)

j even

j odd

Example 9. From (3.14) and (3.16) with p = q = 2, we obtain the following counterparts of (1.47): π 4 = 90ζ(4) = 96

∞ (2)  Q k=1

k k2

= 72

∞ (2)  U k

k=1

k2

.

(5.22)

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Example 10. Setting p = 2, q = 4, a = 0 in (3.2) and (3.11), we obtain π6 =

∞ ∞ (2) (4) (4)  (2) Hk 60480  Ak 3780  Hk k+1 Ak + (−1) + = , 11 k4 k2 113 k4 k2 k=1

(5.23)

k=1

whereas from (3.12), (3.14), (3.16) and (3.18) with p = 2, q = 4, we get π6 =

∞ ∞  (2) (2) (4) (4)  Pk Qk Qk 8640  Pk + + = 960 23 k4 k2 k4 k2 k=1

4320 = 11

k=1

∞   k=1

(2) Uk k4

+

(4) Uk k2

= 1440

∞  (2)  V k=1

k k4

(4) V − k2 . k

(5.24)

Example 11. Note that (s)

(s)

V2k = −2 Ok ,

(s)

(s)

V2k−1 = 21−s Hk−1

(k ∈ N).

(5.25)

Rearranging the series in (4.2), even and odd indices, to make two series, and using (5.25) for both series, we obtain  ∞ ∞ (t) (s)   Ok Ok 1 1−s 1−t ζ(s)ζ(t) = 1−s 2 − 2 2 − 21−t ks kt k=1 k=1 (5.26)  (s) (t) ∞ ∞   Hk−1 Hk−1 1−s 1−t +2 −2 (2k − 1)t (2k − 1)s k=1

k=1

(s, t > 1 with s = t) . Example 12. From (1.53) with ajk = (j + k)−s , we get Hn(s) = 2s+1

n  k=1

(s)

Hk (k) − 2s

n 

Hn(s) (k)

(n ∈ N, s ∈ C).

(5.27)

k=1

References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]

[12] [13]

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