On Harmonic Numbers and Nonlinear Euler Sums

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1.1 Harmonic numbers and Euler sums The generalized nth-harmonic numbers of order p are given by ( [23])

n 1 H(p) := , p ∈ N, (1.1) n jp Xj=1 (1) (p) where the Hn ≡ Hn is the classical harmonic number ( [1]) and the empty sum H0 is con- ventionally understood to be zero. In response to a letter from Goldbach in 1742, Euler considered sums of the form (see Berndt [4]) ∞ H(p) S := n , p,q nq n=1 X where p,q are positive integers with q ≥ 2, and w := p+q denotes the weight of sums Sp,q. These sums are called the linear Euler sums today. Euler discovered that in the cases p = 1, p = q and p + q is less than 7 or when p + q is odd and less than 13, the linear sums have evaluations in terms of zeta values, i.e., the values of the Riemann zeta function ( [2])

∞ 1 ζ(s) := , ℜ(s) > 1 ns nX=1 at the positive integer arguments. Moreover, he conjectured that the double linear sums would be reducible to zeta values when p+q is odd, and even gave what he hoped to obtain the general formula. In 1995, Borweins and Girgensohn [12] proved conjecture and formula, and in 1994, Bailey, Borwein and Girgensohn [3] conjectured that the linear sums Sp,q, when p + q > 7 and p + q is even, are not reducible. Hence, the linear sums Sp,q can be evaluated in terms of zeta values in the following cases: p = 1,p = q,p + q odd and p + q = 4, 6 with q ≥ 2. Similarly, the nonlinear Euler sums are the inﬁnite sums whose general term is a product −1 of harmonic numbers of index n and a power of n . Let π := (π1,...,πk) be a partition of integer p and p = π1 + · · · + πk with π1 ≤ π2 ≤···≤ πk. The classical nonlinear Euler sum of index π,q is deﬁned as follows (see [23])

∞ 1 2 H(π )H(π ) · · · H(πk) S := n n n , q ≥ 2 (1.2) π1π2···πk,q nq n=1 X where the quantity w := π1 + · · · + πk + q is called the weight and the quantity k is called the degree. As usual, repeated summands in partitions are indicated by powers, so that for instance

∞ 2 (2) 3 (4) Hn[Hn ] Hn S 2 3 = S = . 1 2 4,q 112224,q nq Xn=1 The nonlinear Euler sums, i.e., Sπ,q with π having two or more parts, are more complicated. Such sums were already considered in [3,23,43,46,47,49,51,52]. In [23], Flajolet and Salvy gave an algorithm for reducing Sπ1π2,q to linear Euler sums when π1 + π2 + q is even and π1,π2,q> 1

2 (see Theorem 4.2 in the reference [23]. In [46, 47], the ﬁrst author jointly with Li, Yan and Shi proved that all quadratic Euler sums of the form

∞ H H(p−1) S = n n (p,q ≥ 2, p + q ≤ 10) 1(p−1),q nq n=1 X can be reduced to polynomials in zeta values and linear sums. Moreover, in the recent papers [43, 49], the ﬁrst author [49] shown that all Euler sums of weight ≤ 8 are reducible to Q-linear combinations of single zeta monomials with the addition of {S2,6} for weight 8. For weight 9, all Euler sums of the form Sπ1···πk,q with q ∈ {4, 5, 6, 7} are expressible polynomially in terms of zeta values. Wang and Lyu [43] shown that all Euler sums of weight eight are reducible to linear sums, and proved that all Euler sums of weight nine are reducible to zeta values. In the most recent papers [44, 51, 52], the ﬁrst author has carried out fruitful cooperation with Wang proved that all Euler sums of weight ten can be expressed as a rational linear combination of 2 2 ζ (10) ,ζ (5) ,ζ (3) ζ (7) ,ζ (2) ζ (3) ζ (5) ,ζ (3) ζ (4) ,ζ (2) S2,6 and S2,8. All Euler sums of weight 11 are reducible to S2,6, S18,2 and zeta values. Similarly, in [23], Flajolet and Salvy introduced the following alternating harmonic numbers

n (−1)j−1 H¯ (p) := and H¯ := H¯ (1), p ∈ N. (1.3) n jp n n Xj=1 Clearly, we have the following relations for harmonic number and alternating harmonic number 1 H¯ (p) = H(p) − H(p) + H(p) , p ≥ 2, n 2[(n+1)/2]−1 2p [n/2] [(n−1)/2] ¯ Hn = Hn − H[n/2], where [x] denotes the greatest integer less than or equal to x. The alternating harmonic number converges to the alternating Riemann zeta value ζ¯(k):

lim H¯ (k) = ζ¯(k), k ∈ N, n→∞ n where the alternating Riemann zeta function is deﬁned by (for more details, see [23])

∞ (−1)n−1 ζ¯(s) := , ℜ(s) ≥ 1. ns n=1 X The generalized alternating Euler sums involving harmonic numbers and alternating harmonic numbers are deﬁned by

∞ (p1) (pl) (r1) (rk) Hn · · · Hn H¯n · · · H¯n S l k := , q> 1 (1.4) nq pi r¯j ,q n=1 iQ=1 jQ=1 X ∞ (p1) (pl) ¯ (r1) ¯ (rk) Hn · · · Hn Hn · · · Hn n−1 S l k := q (−1) , q ≥ 1 (1.5) p r¯ ,q¯ n i j n=1 iQ=1 jQ=1 X l k where the (p1,...,pl) ∈ (N) , (r1, . . . , rk) ∈ (N) with p1 ≤···≤ pl and r1 ≤···≤ rk and the quantity w := p1 + · · · + pl + r1 + · · · + rk + q being called the weight and the quantity l + k being

3 the degree, l ≥ 0 and k ≥ 1. Since repeated summands in partitions are indicated by powers, we denote, for example, the sums

∞ 3 (2) ¯ 2 ¯ (3) ∞ 2 (3) ¯ 3 ¯ (3) HnHn HnHn HnHn HnHn n−1 S 3 ¯2¯ = , S 2 ¯3¯ ¯ = (−1) . 1 21 3,5 n5 1 31 3,4 n4 n=1 n=1 X X It has been discovered in the course of the years that many alternating Euler sums admit expressions involving ﬁnitely the zeta values, ln 2 and polylogarithms. For example, in [23], Flajolet and Salvy gave explicit reductions to zeta values and logarithm for all linear sums

∞ ∞ ∞ ∞ H(p) H¯ (p) H(p) H¯ (p) S = n , S = n , S = n (−1)n−1 and S = n (−1)n−1 p,q nq p,q¯ nq p,q¯ nq p,¯ q¯ nq nX=1 Xn=1 nX=1 nX=1 when p + q is an odd weight. The evaluation of the above four linear sums in terms of values of the Riemann zeta function and polylogarithm function at positive integers is known when (p,q) = (1, 3), (2, 2), or p + q is odd (see [12, 20, 23, 45]). Here are a few explicit evaluations 7 5 S¯ = ζ (3) ln 2 − ζ(4), 1,3 4 16 11 1 2 1 4 7 S ¯ = −2Li (1/2) + ζ(4) + ζ(2)ln 2 − ln 2 − ζ (3) ln 2, 1,3 4 4 2 12 4 3 1 2 1 4 S¯ ¯= ζ (4) + ζ (2) ln 2 − ln 2 − 2Li (1/2) , 1,3 2 2 12 4 51 7 2 1 4 S ¯ = − ζ(4) + 4Li (1/2)+ ζ(3) ln 2 − ζ(2)ln (2) + ln 2. 2,2 16 4 2 6

Here Lip(x) denotes the polylogarithm function, which is deﬁned for |x|≤ 1 by

∞ xn Li (x) := , ℜ(p) > 1. p np n=1 X Furthermore, we consider the nested sum

∞ n yn xk , x,y ∈ [−1, 1) , m,p ∈ N. nm kp n=1 ! X Xk=1 By taking the sum over complementary pairs of summation indices, we obtain a simple reﬂection formula ∞ n ∞ n yn xk xn yk + = Li (x) Li (y)+Li (xy) . nm kp np km p m p+m n=1 ! n=1 ! X Xk=1 X Xk=1 Hence, we readily deduce the following four results of linear sums 13 19 3 S¯ ¯ = ζ (4) , S ¯ = ζ(4) − ζ (3) ln 2, 2,2 16 3,1 16 4 85 2 1 4 7 S¯ = ζ (4) − 4Li (1/2)+ ζ (2) ln 2 − ln 2 − ζ (3) ln 2, 2,2 16 4 6 2 1 3 1 4 1 1 2 S¯ ¯ =2Li + ζ (3) ln 2 + ln 2 − ζ (4) − ζ (2) ln 2. 3,1 4 2 4 12 2 2

4 However, in many other cases we are not able to obtain a formula for the Euler sum constant explicitly in terms of values of the Riemann zeta, logarithm and polylogarithm functions, but we are able to obtain relations involving two of more Euler sum constants of the same weight. In [48], the ﬁrst author jointly with Yang and Zhang proved that the alternating quadratic Euler sums

∞ 2 ∞ ¯ 2 ∞ ¯ Hn n−1 Hn n−1 HnHn n−1 S 2 := (−1) , S 2 = (−1) , S = (−1) , 1 ,2m n2m 1¯ ,2m n2m 11¯,2m n2m n=1 n=1 n=1 X X X ∞ ¯ ∞ 2 ∞ 2 HnHn Hn Hn S ¯ = , S¯2 = and 11,2m n2m 1 ,2m n2m 2m n=1 n=1 n=1 (2n + 1) X X X are reducible to polynomials in zeta values and to linear sums, and are able to give explicit values for certain of these sums in terms of the Riemann zeta function and the polylogarithm function, here m is a positive integer. In [3], Bailey, Borwein and Girgensohn considered sums

∞ ∞ ∞ ¯ q −p ¯ q −p n−1 q −p n−1 Hn(n + 1) , Hn(n + 1) (−1) and Hn(n + 1) (−1) , (p + q ≤ 5) n=1 n=1 n=1 X X X and obtained a number of experimental identities using the experimental method, where p ≥ 2 and q ≥ 1 are positive integers. Euler sums may be studied through a profusion of methods: combinatorial, analytic and algebraic. There are many other researches on Euler sums and Euler type sums. Some related results for Euler sums may be seen in the works of [13, 14, 16, 17, 24, 32, 33, 39–41, 50, 53].

1.2 Multiple harmonic (star) numbers and multiple zeta (star) values m For m ∈ N, S := (s1,s2, ..., sm) ∈ (N) , and a non-negative integer n, the multiple harmonic number (MHN for short) and the multiple harmonic star number (MHSN for short) are deﬁned by, respectively 1 ζn (s1,s2,...,sm) := s1 sm , (1.6) k · · · km ≤ ··· 1≤ 1 1 km 1, then the limits as n → ∞ exist, and are the well-known multiple zeta and multiple ⋆ star-zeta values ζ (s1,s2,...,sm) and ζ (s1,s2,...,sm) respectively. We put a bar on top of nj sj (j = 1, · · · k) if there is a sign (−1) appearing in the denominator on the right. For example,

(−1)k1+km ζn (¯s1,s2,s3,...,sm−1, s¯m)= s1 sm , (1.8) k1 · · · km 1≤km

5 k2+km ⋆ (−1) ζn (s1, s¯2,s3 ...,sm−1, s¯m)= s1 sm . (1.9) k · · · km ≤ ≤···≤ 1≤ 1 1 km X k n The sums of types (1.8) and (1.9) are called the alternating harmonic numbers and alternating harmonic star numbers, respectively. For alternating MHNs or MHSNs, we have the similar limit cases. For example

(−1)k1+km ζ (¯s1,s2,..., s¯m) := lim ζn (¯s1,s2,..., s¯m)= , (1.10) n→∞ s1 sm k1 · · · km 1≤kmX<···

The sums of types (1.10) and (1.11) (one of more the sj barred) are called the alternating multiple zeta values and alternating multiple zeta star values, respectively. If the ﬁrst index is barred, then the limits as n →∞ exist for all positive-integer indices, e.g.,

ζ (1)¯ = lim ζn (1)¯ = − ln 2. n→∞

Multiple zeta values are certainly interesting and important. The number ζ(6¯, 2)¯ appeared in the quantum field theory literature in 1986 [15], well before the the phrase multiple zeta values had been coined (Multiple zeta values were introduced by the Hoffman [26] and Zagier [55] in the 1990s). The alternating multiple zeta values have been intensely studied since the early 1990s, and extensive tables of them were compiled, first by the group at Lille [6], and later as part of the Multiple Zeta Value Data Mine [7]. In the past several bases have been considered for both the multiple zeta values and alternating multiple zeta values. The vector space of multiple zeta values can be constructed allowing basis elements, which contain besides the zeta values the index of which is a Lyndon word products of this type of zeta values of lower weight. One basis of this kind is (see [7,34,36])

w = 2, ζ(2), w = 3, ζ(3), w = 4, ζ(4), w = 5, ζ(5), ζ(2)ζ(3), w = 6, ζ(6), ζ2(3), w = 7, ζ(7), ζ(2)ζ(5), ζ(3)ζ(4), w = 8, ζ(8), ζ(3)ζ(5), ζ(2)ζ2(3), ζ(5, 3), w = 9, ζ(9), ζ(2)ζ(7), ζ(3)ζ(6), ζ(4)ζ(5), ζ3(3), w = 10, ζ(10), ζ(3)ζ(7), ζ2(5), ζ(2)ζ(3)ζ(5), ζ(4)ζ2(3), ζ(2)ζ(5, 3), ζ(7, 3), etc.

Rational relations among multiple zeta values are known through weight (sum of the indices) 22 and tabulated in the Multiple Zeta Value Data Mine (henceforth MZVDM) [7]. A basis for the Q-vector space spanned by the set of alternating multiple zeta values for 1 ≤ w ≤ 6 should be given

w = 1, ln 2,

6 w = 2, ζ(2), ln2 2, w = 3, ζ(3), ζ(2) ln 2, ln3 2, 2 4 w = 4, Li4(1/2), ζ(4), ζ(3) ln 2, ζ(2) ln 2, ln 2, 2 3 5 w = 5, Li5(1/2), ζ(5), Li4(1/2) ln 2, ζ(2)ζ(3), ζ(4) ln 2, ζ(3) ln 2, ζ(2) ln 2, ln 2, 2 2 w = 6, Li6(1/2), ζ(6), ζ(5¯, 1)¯ , Li5(1/2) ln 2, ζ(5) ln 2, Li4(1/2)ζ(2), Li4(1/2) ln 2, ζ (3), ζ(2)ζ(3) ln 2, ζ(4) ln2 2, ζ(3) ln3 2, ζ(2) ln4 2, ln6 2, etc.

Rational relations among alternating multiple zeta values are also tabulated in the MZVDM (through weight 12 [7]). The relations between multiple zeta (star) values and the values of the Riemann zeta function and polylogarithm function have been studied by many authors. For explicit evaluations of the multiple zeta (star) values or related alternating sums, readers may consult [9–11,22,27,29,32,35, 42,49,55–57]. For example, Zagier [56] proved that the multiple zeta star values ζ⋆ {2}a, 3, {2}b a b and multiple zeta values ζ {2} , 3, {2} , where a, b ∈ N0 := N∪{0} = {0, 1, 2,...}, are reducible to polynomials in zeta values, and gave explicit formulae. Hessami Pilehrood et al. [36, 37] and Li [30] provide three new proofs of Zagier’s formula for ζ⋆ {2}a, 3, {2}b based on a ﬁnite identity for partial sums of the zeta-star series and hypergeometric series computations, respectively.

1.3 Relations between Euler sums and multiple zeta values From the deﬁnitions of harmonic numbers and multiple harmonic numbers, we can ﬁnd that

(p) ¯ (p) Hn = ζn (p) and Hn = −ζn (¯p) , p ∈ N.

According to the rules of the “harmonic algebra” or “stuﬄe product”, it is obvious that the products of any number of harmonic numbers and alternating harmonic number can be expressed in terms of (alternating) multiple harmonic numbers. For example,

¯ 2 2 ¯ ¯ ¯ Hn = ζn (1) = ζ (2) + 2ζn (1, 1) , (3) HnHn = ζn (1) ζn (3) = ζn (1, 3) + ζn (3, 1)+ ζn (4) , HnH¯n = −ζn (1) ζn (1)¯ = −ζn (2)¯ − ζn (1, 1)¯ − ζn (1¯, 1) .

Note that when diﬀerent indices are combined, the combination of a barred and unbarred index is barred, while the combination of two barred indices is unbarred. The series ∞ ζ (i , i , · · · , i ) n 1 2 k , i ∈ N np j nX=1 can be written as the sum of multiple zeta value

ζ (p, i1, i2, · · · , ik)+ ζ (p + i1, i2, · · · , ik) .

Series of the form ∞ ∞ ζ (i , i , · · · , i ) ζ (i , i , · · · , i ) n 1 2 k or n 1 2 k (−1)n np np n=1 n=1 X X 7 with one of more of the ij barred, can be written as a sum of alternating multiple zeta values. It is clear that every (alternating) Euler sum of weight w and degree n is clearly a Q-linear combination of (alternating) multiple zeta values or (alternating) multiple zeta star values of weight w and depth less than or equal to n+1. In other words, (alternating) multiple zeta (star) values are “atomic” quantities into which (alternating) Euler sums decompose. For example,

2 ¯ ¯ ¯ ¯ ¯ ¯ S1¯ ,2¯ = −ζ (2, 2)+2ζ (2, 1, 1) − ζ (4) − 2ζ (3, 1) ,

S13,4 = ζ (4, 1, 3) + ζ (4, 4) + ζ (4, 3, 1)+ ζ (5, 3) + ζ (8) + ζ (7, 1) , ¯ ¯ ¯ ¯ ¯ ¯ S11¯,2 = −ζ (2, 2) − ζ (2, 1, 1) − ζ (2, 1, 1) − ζ (4) − ζ (3, 1) − ζ (3, 1) . Hence, based on the above, we obtain the following two conclusion: 1. All Euler sums of weight ∈ {3, 4, 5, 6, 7, 9} can be expressed as a rational linear combination of products of zeta values. For weight 8, all such sums are the sum of a polynomial in zeta values and a rational multiple of ζ(5, 3). For weight 10, all such sums are reducible to zeta values, ζ(5, 3) and ζ(7, 3). (All explicit formulas can be found in [23, 43, 44, 46, 47, 49]) 2. All alternating Euler sums of weight ≤ 5 can be expressed in terms of zeta values, polylogarithms and ln 2. (Apart of explicit formulas can be found in [3, 12, 33, 45, 48]) In this paper, we will give almost all explicit evaluations of alternating Euler sums with weight less or equal to 5. The evaluation of Euler sums and multiple zeta values has been useful in various areas of theoretical physics, including in support of Feynman diagram calculations and in resolving open questions on Feynman diagram contributions and relations among special functions [18], including the dilogarithm, Clausen function, and generalized hypergeometric function [19]. An array of Euler sums and multiple zeta values is required in calculations of high energy physics. These quantities appear for instance in developing the scattering theory of massless quantum electrodynamics [8]. Harmonic number sums also occur in computer science in the eﬃciency analysis of algorithms [3].

1.4 Purpose and plan The purpose of this paper is to evaluate (alternating) Euler sums and related sums involving (alternating) harmonic numbers. The remainder of this paper is organized as follows. In the second section we evaluate some Euler related sums which involve harmonic numbers (or alternating harmonic numbers), Stirling numbers and Bell numbers. Then we use the evaluations obtained to establish many identities involving two or more Euler sums of the same weight, which can be expressed in terms of Riemann zeta values and polylogarithms. In the third section we use certain integral representations to evaluate several series with harmonic numbers and alternating harmonic numbers. Moreover, we establish many relations involving two or more Euler sums of the same weight. Using these relations obtained we give many closed form of alternating Euler sums with weight ≤ 5. Furthermore, we also consider the following type of Euler-type sums ∞ ζ (m ; x ) · · · ζ (m ; x ) S (r, k; x ,...,x ; x)= n 1 1 n r p xn, p, k, r ∈ N, m1m2...mp 1 p (n + r) (n + k) n=1 X and provide some explicit evaluation in a closed form in terms of zeta values and harmonic numbers.

8 2 Series involving Stirling numbers, harmonic numbers and Bell numbers

In this section, we will establish some explicit relationships that involve Euler sums and zeta values.

2.1 The Bell polynomials, Stirling numbers and Bell numbers

The exponential complete Bell polynomials Yn are deﬁned by (see section 3.3 in the [21] and section 2.8 in the [38]).

tm tk exp x =1+ Y (x ,x , · · · ,x ) . (2.1)  m m! k 1 2 k k! ≥ mX1 Xk≥1   Then Y0 (·)=1 and

k! x1 c1 x2 c2 xk ck Yk (x1,x2, · · · ,xk)= · · · . (2.2) c1!c2! · · · ck! 1! 2! k! 1 2 ··· c +2c +X+kck=k

Additionally, by Eq. (2.44) of [38], the polynomials Yk(·) satisfy the recurrence relation

k−1 k − 1 Y (x ,x , · · · ,x )= x Y (x ,x , · · · ,x ), k ∈ N. (2.3) k 1 2 k j k−j j 1 2 j Xj=0 The (unsigned) Stirling number of the ﬁrst kind s(n, k) is deﬁned by [21]

n x x n!x (1 + x) 1+ · · · 1+ = s(n + 1, k + 1)xk+1, (2.4) 2 n Xk=0 with s(n, k) = 0, if n < k and s(n, 0) = s(0,n) = 0, s(0, 0) = 1, or equivalently, by the generating function: ∞ xn lnk (1 − x) = (−1)kk! s(n, k) , x ∈ [−1, 1) . (2.5) n! nX=k Moreover, the (unsigned) Stirling numbers s(n, k) of the ﬁrst kind satisfy a recurrence relation in the form

s (n, k)= s (n − 1, k − 1)+(n − 1) s (n − 1, k) . (2.6)

By the deﬁnition of s(n, k), we see that we may rewrite (2.4) as

n n x s(n + 1, k + 1)xk = n!exp ln 1+  j  Xk=0 Xj=1  n ∞  xk = n!exp (−1)k−1  kjk  Xj=1 Xk=1 

9   ∞ H(k)xk = n!exp (−1)k−1 n . (2.7) ( k ) Xk=1 k−1 (k) Letting xk = (−1) (k − 1)!Hn in (2.1) and combining (2.7), the following relation holds: n! s (n + 1, k +1) = Y H , −1!H(2), · · · , (−1)k−1 (k − 1)!H(k) . (2.8) k! k n n n Therefore, we know that s(n, k) is a rational linear combination of products of harmonic numbers. Then by (2.3), we have the recurrence

k−1 1 s (n + 1, k +1)= (−1)k−j−1H(k−j)s (n + 1, j + 1), k ∈ N, (2.9) k n Xj=0 which further yields n! s(n + 1, 1) = n!, s(n + 1, 2) = n!H , s(n + 1, 3) = H2 − H(2) , n 2 n n n! h i s(n + 1, 4) = H3 − 3H H(2) + 2H(3) , 6 n n n n n! h i s(n + 1, 5) = H4 − 6H(4) − 6H2H(2) + 3(H(2))2 + 8H H(3) . 24 n n n n n n n h i Throughout this article we deﬁne a Bell number {Yk(n)} by the Bell polynomials

(2) (k) Yk (n)= Yk Hn, 1!Hn , · · · , (k − 1)!Hn , (2.10) Thus, according to the recurrence (2.3), we obtain

k−1 k − 1 Y (n) = 1, Y (n)= (k − j − 1)!H(k−j)Y (n), 0 k j n j Xj=0 from which, we know that Yk (n) is a rational linear combination of products of harmonic numbers, and we have

2 (2) 3 (2) (3) Y1 (n)= Hn,Y2 (n)= Hn + Hn ,Y3 (n)= Hn + 3HnHn + 2Hn ,

4 (3) 2 (2) (2) 2 (4) Y4 (n)= Hn + 8HnHn + 6HnHn + 3(Hn ) + 6Hn ,

2.2 Some lemmas The following lemmas will be useful in the development of the main theorems.

Lemma 2.1 ( [49]) For integers m ≥ 0 and n ≥ 1, then the following integral identity holds:

x m m (−1)l tn−1(ln t)mdt = l! (ln x)m−lxn, x ∈ (0, 1). (2.11) l nl+1 Z0 Xl=0

10 Lemma 2.2 For positive integers m and nonnegative integers k, 1 lnm (1 − x) lnk (x) W (m, k)= dx, (2.12) x Z0 then W (m, k) = (−1)m+km!k!ζ k + 2, {1}m−1 . (2.13) Proof. Expanding out lnm(1 − x) and using the lemma 2.1,

∞ ∞ ∞ 1 xi1+i2+···+im−1lnk (x) W (m, k) = (−1)m · · · dx i1i2 · · · im iX1=1 iX2=1 iXm=1 Z0 ∞ ∞ ∞ k m (−1) k! = (−1) · · · k+1 . i1i2 · · · im(i1 + · · · + im) iX1=1 iX2=1 iXm=1 By Lemma 4.3 in the [26], ∞ ∞ ∞ 1 m! · · · k+1 = k+2 , 1 2 i1i2 · · · im(i1 + · · · + im) 1 2 ··· ≥ n1 n2 · · · nk iX=1 iX=1 iXm=1 n >n >X>nm 1 and the conclusion follows. In [9], Borwein, Bradley and Broadhurst gave the generating function ∞ ∞ xn + yn − (x + y)n ζ m + 1, {1}n−1 xmyn = 1 − exp ζ (n) , (2.14) n n,m=1 n=2 ! X X which implies that for any m, n ∈ N, the multiple zeta value ζ m + 1, {1}n−1 can be repre- sented as a polynomial of zeta values with rational coeﬃcients, and we have the duality formula ζ n + 1, {1}m−1 = ζ m + 1, {1}n−1 .

In particular, one can ﬁnd explicit formulas for small weights. ζ (2, {1}m)= ζ (m + 2) , m m + 2 1 ζ (3, {1}m)= ζ (m + 3) − ζ (k + 1) ζ (m + 2 − k). 2 2 Xk=1 In [46], the ﬁrst author jointly with Yan and Shi gave the following recurrence relation of W (m, k)

m−1 k−1 m − 1 k W (m,k)= − ψ(m+k) (1) / (k + 1) − W (i,j) ψ(m+k−i−j−1) (1) , (2.15) i j Xi=1 Xj=0 where ψ (z) denotes the digamma function (or called Psi function ) which is deﬁned as the logarithmic derivative of the well known gamma function: d Γ′ (z) ψ (z) := (ln Γ (z)) = . dz Γ (z) From (2.14) or (2.15), we see that the values of integrals W (m, k) can be expressed as a rational linear combination of products of zeta values.

11 Lemma 2.3 For positive integers m and nonnegative integers k,

1 lnm (1+ x) lnk (x) V (m, k)= dx, (2.16) x Z0 then

V (m, k) = (−1)m+km!k!ζ k + 2, {1}m−1 . (2.17) Proof. The proof of Lemma 2.3 is similar to the proof of Lemma 2.2.

Lemma 2.4 If m ≥ 1 is a integer and z ∈ [0, 1] , then we have

z lnm (1 + x) 1 1 dx = lnm+1 (1+ z)+ m! ζ (m + 1) − Li x m + 1 m+1 1+ z Z0 m lnm−j+1 (1 + z) 1 − m! Li . (2.18) (m − j + 1)! j 1+ z Xj=1 Proof. We note that the integral in (2.18), which can be rewritten as

1 z 1+z (1+z)− m m m ln (1 + x) t=1+x ln t u=t−1 ln u dx = dt = (−1)m+1 du x t − 1 u − u2 Z0 Z1 Z1 (1+z)−1 (1+z)−1 lnmu lnmu =(−1)m+1 du + du  u 1 − u   Z1 Z1  −1  (1+z)  1  lnmu  = lnm+1 (1 + z) + (−1)m+1 du. (2.19) m + 1 1 − u Z1

Then with help of formula (2.11) we may deduce the desired result.

Lemma 2.5 (see [46]) For integers n ≥ 1 and k ≥ 0, then

1 Y (n) tn−1lnk (1 − t)dt = (−1)k k , (2.20) n Z0

⋆ k where the Bell number Yk (n) is deﬁned by (2.10), and Yk (n)= k!ζn {1} . Lemma 2.6 ([5,25]) For positive integer n,m and any suitable sequences ai ∈ C (i = 1, 2 ...,n),

n ∞ 1 m = Am (n) t , A0 (n) = 1, (1 − ait) Yi=1 mX=0

12 n ∞ m (1 + ait)= A¯m (n) t , A¯0 (n) = 1. m=0 Yi=1 X then

Am (n) := ak1 · · · akm , (2.21) ≤ ≤···≤ 1≤ 1 km X k n ¯ Am (n) := ak1 · · · akm . (2.22) 1≤km

−1 −1 Here |t| < min |a1| , · · · , |an| . n o 2.3 Main results and proofs Theorem 2.7 For positive integer m, then the following symmetric-function identities hold:

m−1 m−1 1 (−1)m−1 h = h p and e = (−1)ie p , (2.23) m m i m−i m m i m−i Xi=0 Xi=0 where hm = hm(x1,x2, · · · ,xn) and em = em(x1,x2, · · · ,xn) are respectively the complete and m m m elementary symmetric functions of degree m, and pm = x1 + x2 + · · · + xn is the mth power sum (for details introductions, see [28], the notation follows [31]).

Proof. From Hoﬀman’s paper [28], we know that

∞ n ∞ n j j 1 E (t)= ejt = (1+ txi) and H (t)= hjt = 1 − txi Xj=0 Yi=1 Xj=0 Yi=1 are the respective generating functions of the elementary and complete symmetric functions. Note that we can rewrite E(t) and H(t) as

∞ k−1 ∞ (−1) k 1 k E (t) = exp pkt and H (t) = exp pkt . k ! k ! Xk=1 Xk=1 −1 −1 k−1 Here |t| < min |x1| , · · · , |xn| . Letting xk = (−1) (k − 1)!pk and (k − 1)!pk in (2.1), then comparingn the coeﬃcients of tok, we can ﬁnd that

k−1 Yk p1, −1!p2, · · · , (−1) (k − 1)!pk Y (p , 1!p , · · · , (k − 1)!p ) e = and h = k 1 2 k . k k! k k! Using the recurrence relation (1.3), by a simple calculation, we obtain the desired results. (Theorem 2.7 can also be found in Entry 12(a) of Chapter 14 in Ramanujan’s second notebook [5]). Hence, the elementary symmetric function ek and the complete symmetric function hk can be expressed in terms of the power sums, i.e., (see [28])

ek = Pk (p1,p2, · · · ,pk) and hk = Q (p1,p2, · · · ,pk) .

13 Moreover, from the generating function E(t) and H(t), we have

∞ n ∞ 1 e (x , · · · ,x ) tm = (1 + tx )= e (x , · · · ,x ,x ) tm m 1 n i 1+ tx m 1 n n+1 m=0 i=1 n+1 m=0 X Y∞ m X k k m = (−1) xn+1em−k (x1, · · · ,xn+1) t , m=0 ! X Xk=0 ∞ n+1 ∞ 1 1 h (x , · · · ,x ,x ) tm = = h (x , · · · ,x ,) tm m 1 n n+1 1 − tx 1 − tx m 1 n m=0 i=1 i n+1 m=0 X Y∞ m X k m = xn+1hm−k (x1, · · · ,xn) t . ! mX=0 Xk=0 Then by considering the coeﬃcient of tm, we can get the following relations

m k k em (x1, · · · ,xn)= (−1) xn+1em−k (x1, · · · ,xn+1), k=0 X m k hm (x1, · · · ,xn+1)= xn+1hm−k (x1, · · · ,xn). Xk=0

The polynomials Pn and Qn can be written as determinants (see [29])

y1 1 0 · · · 0 y y 2 · · · 0 2 1 . . . . . n!Pn (y1,...,yn)= ......

y − y − y − · · · n − 1 n 1 n 2 n 3 y y y · · · y n n−1 n−2 1

and y1 −1 0 · · · 0 y y −2 · · · 0 2 1 . . . . . n!Qn (y1,...,yn)= ......

y y y ··· −n + 1 n−1 n−2 n−3 y y y · · · y n n−1 n−2 1

Here are a few explicit evaluations; some of them were given previously by Hoﬀman and Ohno.

y2 − y P (y )= y , P (y ,y )= 1 2 , 1 1 1 2 1 2 2 y3 − 3y y + 2y P (y ,y ,y )= 1 1 2 3 , 3 1 2 3 6 y4 − 6y2y + 8y y + 3y2 − 6y P (y ,y ,y ,y )= 1 1 2 1 3 2 4 , 4 1 2 3 4 24 y5 − 10y3y + 20y2y + 15y y2 − 30y y − 20y y + 24y P (y ,y ,y ,y ,y )= 1 1 2 1 3 1 2 1 4 2 3 5 5 1 2 3 4 5 120

14 and y2 + y Q (y )= y , Q (y ,y )= 1 2 , 1 1 1 2 1 2 2 y3 + 3y y + 2y Q (y ,y ,y )= 1 1 2 3 , 3 1 2 3 6 y4 + 6y2y + 8y y + 3y2 + 6y Q (y ,y ,y ,y )= 1 1 2 1 3 2 4 , 4 1 2 3 4 24 y5 + 10y3y + 20y2y + 15y y2 + 30y y + 20y y + 24y Q (y ,y ,y ,y ,y )= 1 1 2 1 3 1 2 1 4 2 3 5 . 5 1 2 3 4 5 120

By Lemma 2.6, we have the relations

n k1−1 km−1−1

Pm (p1,...,pm)= xk1 xk2 · · · xkm , (2.24) kX1=1 kX2=1 kXm=1 n k1 km−1

Qm (p1,...,pm)= xk1 xk2 · · · xkm . (2.25) kX1=1 kX2=1 kXm=1 Hence, according to the deﬁnitions of Stirling numbers of the ﬁrst kind s(n, k) and Bell numbers Yk(n), we can get the well known identities

(k−1) k−1 s (n, k)= Pk−1 Hn−1,...,Hn−1 = (n − 1)!ζn−1 {1} , (2.26)

(k) ⋆ k Yk (n)= k!Qk Hn,...,Hn = k!ζn {1} , (2.27) which were proved in [49]. Therefore, by (2.26), then ∞ ∞ s (n, k) s (n,p) = = ζ p + 1, {1}k−1 . (2.28) n!np n!nk n=p nX=k X The two series are equal, which is an instance of duality for multiple zeta values. (See also the discussion on pp. 526-7 of Chen [16]). Using Theorem 2.7 and formulas (2.24) and (2.25), we are able to obtain many rela- 1 tions involving multiple zeta-star values and zeta values. For example, taking x = ,p > i ip 1 (i = 1, · · · ,n) in Theorem 2.7, (2.24) and (2.25), then letting n →∞, we deduce the results

m−1 (−1)m−1 ζ ({p}m)= (−1)iζ {p}i ζ (pm − pi), m Xi=0 m−1 1 ζ⋆ ({p}m)= ζ⋆ {p}i ζ (pm − pi). m Xi=0 (−1)i Setting x = and m = 6 in (2.24), then letting n →∞ yields i i 5 1 3 1 1 ζ {1¯}6 = − ζ (6) + ζ2 (3) + ζ (5) ln 2 − ζ (2) ζ (3)ln2+ ζ (4) ln22 128 32 16 8 32 15 1 1 1 + ζ (3) ln32 − ζ (2) ln42+ ln62. 24 48 720 Now we establish some relations between zeta values and series involving Stirling numbers, harmonic numbers and Bell numbers.

Theorem 2.8 ( [49]) For positive integers k and p, we have ∞ s (n, k) Y (n) k + p p = p! ζ (k + p + 1) , (2.29) n!n p n=k X∞ ∞ s (n, k) s (n,p) H(p+1) = H(k+1), (2.30) n!n n n!n n n=k n=p X∞ X ∞ s (n,p) Y (n) s (n, k) Y (n) p! k + k! p = k!p!ζ (k + 1) ζ (p + 1) . (2.31) n!n2 n!n2 n=p X nX=k Proof. Multiplying (2.5) by x−1 lnp(1 − x) and integrating over (0,1), using (2.20), we deduce that 1 ∞ lnp+k (1 − x) s (n, k) Y (n) dx = (−1)k+pk! p x n!n Z0 nX=k By Lemma 2.2, the integral on the left hand side is equal to

1 lnp+k (1 − x) dx = (−1)p+k (p + k)!ζ (p + k + 1) . x Z0 Summing these two contributions yields the result (2.29). To prove (2.30), we need the well known identity

1 xnlnpx dx = (−1)pp! ζ (p + 1) − H(p+1) , p ∈ N. (2.32) 1 − x n Z0 Replacing x by t in (2.5), then dividing (2.5) by t and integrating over the interval (0,x), the result is x ∞ lnk (1 − t) s (n, k) dt = (−1)kk! xn, −1 ≤ x ≤ 1. (2.33) t n!n Z0 nX=k lnpx Multiplying (2.33) by and integrating over the interval (0, 1), by (2.32), we obtain 1 − x

1 x ∞ lnpx lnk (1 − t) s (n, k) dtdx = (−1)k+pk!p! ζ (p + 1) − H(p+1) . (2.34) 1 − x t n!n n Z0 Z0 nX=k Noting that the integral on the left hand side of this equation can be written as

1 x 1 1 1 1 p k p k p k ln x ln (1 − t) x=1−u ln (1 − u) ln y ln (1 − x) ln t dtdx = dydu = dtdx. 1 − x t t=1−y u 1 − y x 1 − t Z0 Z0 Z0 Zu Z0 Zx

16 By integration by parts, we see that

1 1 1 1 x lnp (1 − x) lnkt lnkt lnp (1 − t) dtdx = dt d dt x 1 − t  1 − t   t  Z0 Zx Z0 Zx Z0 1  x 1 1 x lnkt lnp (1 − t) lnkx lnp (1 − t) = dt dt + dtdx  1 − t   t  1 − x t Z Z Z Z x 0 0 0 0  ∞  s (n,p) = (−1)p+kp!k! ζ (k + 1) − H(k+1) . n!n n nX=k Then with the help of formula (2.28) we easily obtain the result. To prove identity (2.31), we consider the following integral

1 x ∞ 1 lnk (1 − x) lnp (1 − t) s (n,p) dtdx = (−1)pp! xn−1lnk (1 − x)dx xt nn! n=p Z0 Z0 X Z0 ∞ s (n,p) Y (n) = (−1)p+kp! k n2n! n=p X x x 1 1 x lnp (1 − t) lnk (1 − t) lnk (1 − t) lnp (1 − x) = dt dt − dtdx  t   t  tx Z Z Z Z 0 0 0 0 0     ∞ s (n, k) Y (n) = (−1)k+pk!p!ζ (k + 1) ζ (p + 1) − (−1)k+ pk! p . n2n! nX=k We thus arrive at (2.31) to complete the proof of Theorem 2.8.

Theorem 2.9 For positive integer m, then ∞ H¯ s (n,m) n = ζ 2, {1} ln 2 − ζ 2¯, {1} ln 2 − (m + 1) ζ (2¯, {1} ) . (2.35) nn! m−1 m−1 m n=m X Proof. Using integration by parts we deduce

n x 1 xj tn−1 ln (1 − t)dt = xn ln (1 − x) − − ln (1 − x) , −1 ≤ x< 1. (2.36) 0 n  j  Z  Xj=1  Taking x = −1 in (2.36) yields  1 (−1)n xn−1 ln (1 + x)dx = (−1)n ln 2 − ln2+ H¯ . n n Z 0 Next, we consider the following integral

1 ∞ 1 lnm+1 (1 + x) (−1)n dx = (−1)mm! s (n,m) xn−1 ln (1 + x)dx x n! n=m Z0 X Z0

17 ∞ ∞ s (n,m) s (n,m) = (−1)mm! ln 2 (−1)n − (−1)mm! ln 2 nn! nn! n=m n=m ∞X X H¯ s (n,m) + (−1)mm! n nn! n=m X By Lemma 2.2 and Lemma 2.3

1 ∞ lnm (1 + x) s (n,m) dx = (−1)mm! (−1)n = (−1)mm!ζ (2¯, {1} ) , x nn! m−1 n=m Z0 X 1 ∞ lnm (1 − x) s (n,m) dx = (−1)mm! = (−1)mm!ζ (2, {1} ) , x nn! m−1 n=m Z0 X Thus, combining the above identities, we complete the proof of Theorem 2.9. By Lemma 2.3 and Lemma 2.4, we can get the well known alternating multiple zeta values ( [9]) (−1)m 1 ζ 2¯, {1} = lnm+12 + (−1)m ζ (m + 1) − Li m−1 (m + 1)! m+1 2 m (ln 2)m+1−j 1 − (−1)m Li . (m + 1 − j)! j 2 Xj=1 Theorem 2.10 For positive integer m, then ∞ Y (n) 1 m (−1)n−1 = m!Li , (2.37) n m+1 2 n=1 X∞ Y (n)H¯ 1 m−1 n (−1)n−1 = m! ζ (m + 1) − Li , (2.38) n m+1 2 n=1 X∞ H¯ s (n,m) 1 (−1)m n (−1)n−1 = Li − lnm+12. (2.39) n! m+1 2 m! n=m X Proof. To prove the ﬁrst and second identities. Using (2.20), by a direct calculation, it is easy to see that

1 ∞ 1 ∞ lnm (1 − x) Y (n) dx = (−1)n−1 xn−1lnm (1 − x)dx = (−1)m m (−1)n−1. (2.40) 1+ x n n=1 n=1 Z0 X Z0 X Applying the change of variable t → 1 − x to integral above and using (2.11), we obtain

1 lnm (1 − t) 1 dt = (−1)mm!Li . 1+ t m+1 2 Z0 Thus, we deduce (2.37). Using the Cauchy product formula of power series, we can ﬁnd that

∞ ln (1 − x) = (−1)nH¯ xn, −1

1 1 1 lnm (1 − x) lnm (1 − x) lnm (1 − x) dx = dx − dx x (1 + x) x 1+ x Z0 Z0 Z0 ∞ 1 n n−1 m−1 = H¯n(−1) x ln (1 − x)dx n=1 X Z0 ∞ Y (n)H¯ = (−1)m m−1 n (−1)n−1. n n=1 X Then with the help of formulas (2.13) and (2.40) we may deduce the result (2.38). To prove the third identity. By using (2.5), we have

1 ∞ 1 lnm (1 − x) s (n,m) xn dx = (−1)mm! dx 1+ x n! 1+ x n=m Z0 X Z0 ∞ H¯ s (n,m) = (−1)mm! n (−1)n−1 + lnm+12. (2.42) n! n=m X Comparing (2.40) with (2.42) completes the proof of (2.39).

2.4 Some identities on Euler sums

∞ H ζ (2) − ln22 n (−1)n−1 = , (2.43) n 2 n=1 X∞ H2 + H(2) 7 1 1 n n (−1)n−1 = ζ (3) − ζ (2)ln2 + ln32, (2.44) n 8 2 6 n=1 X∞ H H¯ 43 3 1 1 n n = ζ(4) + ζ(2)ln22 − ln42 − 3Li , (2.45) n2 16 4 8 4 2 n=1 X∞ H3 + 3H H(2) + 2H(3) 1 n n n n (−1)n−1 = 6Li , (2.46) n 4 2 n=1 X∞ H¯ 3 + 3H¯ H(2) + 2H¯ (3) 1 3 27 3 n n n n (−1)n−1 = ln42+ ζ (3) ln 2 + ζ (4) + ζ (2) ln22, (2.47) n 4 2 8 2 nX=1 ∞ 2 (2) ¯ Hn + Hn Hn 1 (−1)n−1 = 6 ζ (4) − Li , (2.48) n 4 2 n=1 X ∞ 2 (2) ¯ ∞ ∞ Hn − Hn Hn H H¯ H¯ (−1)n−1 − 2 n n (−1)n−1 + 2 n (−1)n−1 n n2 n3 n=1 n=1 n=1 X X X 1 1 = 2Li + ln42, (2.49) 4 2 3

19 ∞ 3 (2) (3) ¯ Hn + 3HnHn + 2Hn Hn 1 (−1)n−1 = 24 ζ (5) − Li , (2.50) n 5 2 n=1 X 2 ∞ 4 (3) 2 (2) (2) (4) Hn + 8HnHn + 6HnHn + 3 Hn + 6Hn 1 (−1)n−1 = 24Li , (2.51) n 5 2 nX=1 ∞ 2 (2) ¯ ∞ ∞ Hn − Hn Hn H H¯ H¯ 3 − 6 n n + 6 n n2 n3 n4 n=1 n=1 n=1 X X X 5 11 21 1 = 12ζ (4) ln 2 − ζ (2) ln32+ ln52+ ζ (3) ln22 + 18Li ln 2 2 20 4 4 2 1 + 24Li − 24ζ (5) , (2.52) 5 2 2 ∞ ¯ 4 ¯ ¯ (3) ¯ 2 (2) (2) (4) Hn + 8HnHn + 6HnHn + 3 Hn + 6Hn (−1)n−1 n nX=1 9 27 1 = ζ (5) + 3ζ (2) ζ (3) + ζ (4) ln 2 + 3ζ (3) ln22 + 2ζ (2) ln32+ ln52, (2.53) 2 2 5 ∞ 3 (2) (3) ¯ ∞ 2 (2) ¯ Hn − 3HnHn + Hn Hn Hn − Hn Hn (−1)n−1 − 3 (−1)n−1 n n2 n=1 n=1 X ∞ ∞ X H H¯ H¯ 1 1 + 6 n n (−1)n−1 − 6 n (−1)n−1 = 6Li − ln52. (2.54) n3 n4 5 2 4 n=1 n=1 X X These identities can be obtained from the main theorems which are presented in the section 2.3. For example, setting m = 1, 2, 3, 4 in (2.37) respectively yields (2.43), (2.44), (2.46) and (2.51). (−1)k−1 Taking x = and m = 4, 5 in (2.25) and Theorem 2.7, then letting n →∞, we obtain k k the results (2.47) and (2.53). etc

3 Some explicit evaluation of Euler sums

In the section, by constructing the integrals of polylogarithm functions, we establish some relations involving two or more Euler sums of the same weight. Then we use the relations obtained to evaluate alternating Euler sums, and give explicit formulas. First, we need the following lemma.

Lemma 3.1 ( [48]) For positive integers n,m and real x ∈ [−1, 1), we have

x m−1 n m n k n−1 i−1 x (−1) x t Lim (t) dt = (−1) i Lim+1−i (x) − m n n k ! Z0 Xi=1 Xk=1 (−1)m + ln (1 − x) (xn − 1) . (3.1) nm

20 3.1 Main theorems and corollaries Theorem 3.2 Let m,p be integers with m,p > 0, |x| , |y| , |z|∈ (0, 1), we have

∞ Li (x) ζ (p; y) − Li (y) ζ (m; x) m n p n zn n n=1 X ∞ ∞ ζ (1; z) ζ (1; z) = Li (y) n xn − Li (x) n yn p nm m np n=1 n=1 X X + Lim (x) Lip+1 (zy) − Lip (y) Lim+1 (zx) , (3.2) where the partial sum ζn (m; a) is deﬁned by n ak ζ (m; a)= , (a ∈ C). n km Xk=1 By convenience, we let ζ0 (m; a) = 0. It is obvious that (m) ¯ (m) ζn (m;1) = Hn and ζn (m; −1) = −Hn .

Proof. Let us consider the N-th partial sum

N Li (x) ζ (p; y) − Li (y) ζ (m; x) S (x,y,z)= m n p n zn N n n=1 X N N ζ (p; y) ζ (m; x) = Li (x) n zn − Li (y) n zn m n p n n=1 n=1 X X N n N n ykzn xkzn = Li (x) − Li (y) m nkp p nkm Xn=1 Xk=1 Xn=1 Xk=1 N N N N ykzn xkzn = Li (x) − Li (y) m nkp p nkm n=1 Xk=1 nX=k X nX=k N N ζ (1; z) − ζ (1; z) ζ (1; z) − ζ (1; z) = Li (x) N k−1 yk − Li (y) N k−1 xk m kp p km Xk=1 Xk=1 = ζN (1; z)(Lim (x) ζN (p; y) − Lip (y) ζN (m; x)) N N ζ (1; z) ζ (1; z) + Li (y) k−1 xk − Li (x) k−1 yk. p km m kp Xk=1 Xk=1 Then letting N →∞ in above equation, we get the required identity (3.1). The proof of Theorem 3.2 is thus completed. Taking (x,y,z) → (1, 1, 1), (1, −1, 1) and (−1, −1, 1) in Theorem 3.2, we can get the following corollary. Corollary 3.3 For positive integers m and p, then

∞ ζ (m) H(p) − ζ (p) H(m) n n = ζ (p) S − ζ (m) S n 1,m 1,p n=1 X 21 + ζ (m) ζ (p + 1) − ζ (p) ζ (m + 1) , (m,p ≥ 2) (3.3) ∞ ζ (m) H¯ (p) − ζ¯(p) H(m) n n = ζ¯(p) S − ζ (m) S n 1,m 1,p¯ n=1 X + ζ (m) ζ¯(p + 1) − ζ¯(p) ζ (m + 1) , (m ≥ 2) (3.4) ∞ ζ¯(m) H¯ (p) − ζ¯(p) H¯ (m) n n = ζ¯(p) S − ζ¯(m) S n 1,m¯ 1,p¯ n=1 X + ζ¯(m) ζ¯(p + 1) − ζ¯(p) ζ¯(m + 1) . (3.5)

The main Theorems of this section can be stated as follows.

Theorem 3.4 For positive integers m,p and real a, b, x with the conditions |a| = |b| > 1 and |x| < |a|−1 or |b|−1, we have

∞ ∞ ζ (1; ax) ζ (m; b) ζ (1; bx) ζ (p; a) (−1)p n n − (−1)m n n annp bnnm n=1 n=1 X X p−1 ∞ m−1 ∞ ζ (m; b) ζ (p; a) = (−1)i−1Li (ax) n xn − (−1)i−1Li (bx) n xn p+1−i ni m+1−i ni i=1 n=1 i=1 n=1 X ∞ X X ∞X ζ (m; b) ζ (p; a) + (−1)p ln (1 − ax) n xn − a−n − (−1)m ln (1 − bx) n xn − b−n . np nm n=1 n=1 X X (3.6)

Proof. Applying the deﬁnition of polylogarithm function Lip(x) and using the Cauchy product of power series, we can ﬁnd that

∞ Li (at) p = ζ (p; a)tn, t ∈ (−1, 1) . 1 − t n n=1 X Then using above identity it is easily see that

x ∞ x ∞ x Li (at) Li (bt) p m dt = ζ (p; a) tn−1Li (bt)dt = ζ (m; b) tn−1Li (at)dt. t (1 − t) n m n p Z0 nX=1 Z0 nX=1 Z0 Then with the help of formula (3.1) and the elementary transform

x bx 1 tn−1Li (bt) dt = tn−1Li (t) dt m bn m Z0 Z0 by a direct calculation, we may deduce the result.

Theorem 3.5 For positive integers m,p and real a, b, x with the conditions |a| = |b| > 1 and |x| < |a|−1 or |b|−1, then the following identity holds

∞ ∞ ζ (2; ax) ζ (m; b) ζ (2; bx) ζ (p; a) (−1)p n n − (−1)m n n annp bnnm n=1 n=1 X X 22 p−1 p−i m−1 m−i i+j i+j = (−1) Lip+2−i−j (ax) Lii+j,m (x, b) − (−1) Lim+2−i−j (bx) Lii+j,p (x, a) Xi=1 Xj=1 Xi=1 Xj=1 p −1 + (−1) p ln (1 − ax) Lip+1,m (x, b) − Lip+1,m a , b m −1 − (−1) m ln (1 − bx) Lim+1,p (x, a) − Lim+1,p b , a ∞ ∞ p ζn (1; ax)ζn (m; b) m ζn (1; bx) ζn (p; a) − (−1) p + (−1) m annp+1 bnnm+1 n=1 n=1 p X −1 m X −1 + (−1) Li2 (ax) Lip,m a , b − (−1) Li2 (bx) Lim,p b , a , (3.7) where Lis,t (x,y) is deﬁned by the double sum

∞ n xn yk Li (x,y) := , |x|≤ 1, |y| > 0, |xy| < 1,s,t > 0 s,t ns kt nX=1 Xk=1 Of course, if s> 1, then we can allow |xy| = 1.

Proof. Replacing x by t in (3.6), then dividing it by t and integrating over the interval (0,x) with the help of formulae (3.1) and

x ln (1 − at) dt = −Li (ax) , t 2 Z0 by a simple calculation, we have the result. Letting (a, b, x) → (−1, 1, 1) , (−1, −1, 1) and (p, a, b, x) → (1, −1, 1, −1) , (1, −1, −1, −1) in The- orem 3.4, then with the help of Corollary 3.3 we obtain the following corollaries.

Corollary 3.6 For positive integers p ≥ 2 and m ≥ 2, then

m−1 p−1 m p i−1 i−1 ¯ (−1) S1¯p,m + (−1) Sm1¯,p¯ = (−1) ζ (m + 1 − i) Sp,i¯ − (−1) ζ (p + 1 − i) Sm,i Xi=2 Xi=2 + ζ¯(p) S1,m − ζ (m) S1,p¯ + ζ (m) ζ¯(p + 1) − ζ¯(p) ζ (m + 1) , p + (−1) (Sm,p + Sm,p¯) ln 2, (3.8) m−1 p−1 p m i−1 ¯ i−1 ¯ (−1) S1¯m, ¯ p¯ − (−1) S1¯¯p,m¯ = (−1) ζ (m + 1 − i) Sp,i¯ − (−1) ζ (p + 1 − i) Sm,i¯ Xi=2 Xi=2 + ζ¯(p) S1,m¯ − ζ¯(m) S1,p¯ + ζ¯(m) ζ¯(p + 1) − ζ¯(p) ζ¯(m + 1) p m + (−1) (Sm,p¯ + Sm,¯ p¯) ln 2 − (−1) (Sp,m¯ + Sp,¯ m¯ ) ln 2. (3.9)

Corollary 3.7 For integer m> 1, then

m−1 m−1 i−1 ¯ m−1 (−1) S1¯2,m + S1m,1¯ = (−1) ζ (m + 1 − i) S1¯,¯i + (−1) S1¯,m + S1¯,m¯ ln 2, (3.10) Xi=1 m−1 m i−1 (−1) S11¯,m¯ + S1m, ¯ 1¯ = (−1) ζ (m + 1 − i) S1¯,¯i. (3.11) Xi=1

23 Similarly, letting (a, b, x) → (1, 1, 1) and (−1, −1, −1) in Theorem 3.5, we can get the following two corollaries.

Corollary 3.8 For positive integers m> 1 and p> 1, we have

p−1 p−i p m i+j (−1) S2m,p − (−1) S2p,m = (−1) ζ (p + 2 − i − j)Sm,i+j Xi=1 Xj=1 m−1 m−i i+j − (−1) ζ (m + 2 − i − j)Sp,i+j Xi=1 Xj=1 p m − (−1) pS1m,p+1 + (−1) mS1p,m+1 p m + (−1) ζ (2) Sm,p − (−1) ζ (2) Sp,m. (3.12)

Corollary 3.9 For positive integers m and p, then

m−1 m−i m p i+j (−1) S2¯p,m¯ − (−1) S2m, ¯ p¯ = (−1) ζ (m + 2 − i − j) Sp,¯ i+j Xi=1 Xj=1 p−1 p−i i+j − (−1) ζ (p + 2 − i − j) Sm,¯ i+j Xi=1 Xj=1 m−1 p−1 + (−1) mS1¯p,m+1 − (−1) pS1m, ¯ p+1 m−1 p−1 − (−1) ζ (2) Sp,¯ m¯ + (−1) ζ (2) Sm,¯ p¯. (3.13)

Theorem 3.10 For positive integer k and 1 ≤ i1 ≤ i2 ≤··· ≤ ik with xj ∈ [−1, 1], then the following relation holds:

k k−1 n m m k−j+1 ζm (il1 ; xl1 ) · · · ζm ilj ; xlj x1 · · · xk ζn (ij; xj)= (−1) m m , (3.14) i1+···+ik− il1 +···+il x · · · x j=1 j=0 1 m=1 j il1 il Y X 1≤l

Proof. Consider the generating function

∞ k F (x; X ) := ζ (i ; x ) − ζ (i + · · · + i ; x · · · x ) xn, (x ∈ (−1, 1)) , k k  n j j n 1 k 1 k  n=1 X jY=1  where Xk = (x1, · · · ,xk).  Using the deﬁnition of the numbers ζn (m; a), we have

∞ k k (1 − x) F (x; X )= ζ (i ; x ) − ζ i + · · · + i ; x xn k k  n j j n  1 k j nX=1 jY=1 jY=1  ∞ k  k k    n − ζ − (i ; x ) − ζ − i ; x x  n 1 j j n 1  j j n=1 X jY=1 Xj=1 jY=1      24 k n ∞ k k n xj xj j=1 n =  ζn (ij; xj) − ζn (ij; xj) − −  x  nij niQ1+···+ik  n=1   X jY=1 jY=1 

 k j  k  xn  n ∞ j k−1 ζn (ilr ; xlr ) xa j=1 = −  + (−1)k−j r=1 a=1  xn i1+···+i Q k 1 Q··· − ··· Qj n i + +ik il1 + +ilj  n=1  j=0 1≤l1<···

3.2 Identities for alternating Euler sums It is clear that from Theorem 3.4, 3.5, 3.10 and Corollary 3.6-3.9, we can establish many identities involving two or more Euler sums of the same weight. Some examples on alternating Euler sums with weight ≤ 5 follows: ∞ ∞ H H¯ H H¯ (2) 5 1 n n (−1)n−1 + n n (−1)n−1 = ζ (4) + ζ (2) ln22, (3.16) n2 n 4 2 n=1 n=1 X∞ X∞ H H(2) H¯ 2 5 7 11 n n (−1)n−1 − n = ζ (4) + ζ (3) ln 2 − ζ (2) ln22, (3.17) n n2 8 8 4 n=1 n=1 X 2 X ∞ (2) ∞ Hn H(2)H¯ 43 15 35 (−1)n−1 + 2 n n = − ζ (5) + ζ (2) ζ (3) + ζ (4) ln 2, (3.18) n n2 16 8 8 n=1 n=1 X 2 X ∞ ¯ (2) ∞ Hn H¯ H¯ (2) 19 1 5 (−1)n−1 + 2 n n (−1)n−1 = ζ (5) − ζ (2) ζ (3) + ζ (4) ln 2, (3.19) n n2 16 4 2 n=1 n=1 X∞ ∞X H H¯ (3) H H¯ 1 9 15 n n (−1)n−1 − n n (−1)n−1 = ζ (3) ln22+ ζ (2) ζ (3) − ζ (4) ln 2, (3.20) n n3 2 8 4 n=1 n=1 X∞ ∞ X H H¯ (2) H(2)H¯ 1 1 7 n n + n n (−1)n−1 = 4Li ln2+ ln52+ ζ (3) ln22 − ζ (2) ln32 n2 n2 4 2 6 2 n=1 n=1 X X 5 23 + ζ (2) ζ (3) − ζ (4) ln 2, (3.21) 8 16 ∞ ∞ H¯ H¯ (2) H H(2) 1 1 7 n n + n n (−1)n−1 = −4Li ln 2 − ln52 − ζ (3) ln22+ ζ (2) ln32 n2 n2 4 2 6 2 n=1 n=1 X X 3 49 + ζ (2) ζ (3) + ζ (4) ln 2, (3.22) 16 8 ∞ ∞ H H(3) H¯ 2 1 1 17 1 n n (−1)n−1 + n = −2Li ln 2 − ln52+ ζ (3) ln22+ ζ (2) ln32 n n3 4 2 12 8 2 nX=1 nX=1 11 11 + ζ (2) ζ (3) − ζ (4) ln 2, (3.23) 16 16

25 ∞ ∞ ∞ H(2)H¯ H(2)H¯ (2) H H¯ (2) n n (−1)n−1 + n n (−1)n−1 + n n (−1)n−1 n2 n n2 nX=1 nX=1 nX=1 1 1 37 7 1 5 = −4Li + ln52+ ζ (5) + ζ (3) ln22 − ζ (2) ln32+ ζ (2) ζ (3) − 3ζ (4) ln 2. 5 2 30 8 4 3 8 (3.24)

These identities can be obtained from the Theorem 3.4, 3.5, 3.10 and Corollary 3.6-3.9 which are presented in the section 3.1. For example, taking k = 3, (i1, i2, i3) = (2, 2, 1) and (x1,x2,x3)= (1, 1, −1) or (−1, −1, −1) in Theorem 3.10, then letting n → ∞ yields the formulas (3.18) and (3.19). In (2.5), setting k = 3, 4 and x = −1, using (2.18) we can get the following two identities

∞ H3 − 3H H(2) + 2H(3) 3 1 21 1 n n n n (−1)n−1 = 6ζ (4) + ζ (2) ln22 − ln42 − ζ (3) ln 2 − 6Li , n 2 2 4 4 2 n=1 X (3.25) 2 ∞ 4 (3) 2 (2) (2) (4) Hn + 8HnHn − 6HnHn + 3 Hn − 6Hn (−1)n−1 n n=1 X 1 1 21 = 24Li + 24Li ln 2 + ln52+ ζ (3) ln22 − 24ζ (5) − 4ζ (2) ln32. (3.26) 5 2 4 2 2 3.3 Closed form of alternating Euler sums In this subsection, we use our equation system to obtain explicit evaluations for alternating Euler sums with weight ≤ 5. Note that the linear sums Sp,q¯ , Sp,q¯ and Sp,¯ q¯ are reducible to zeta values and polylogarithms when p+q is odd (see [12,20,23,45]). The quadratic sums S12,2m, S1¯2,2m, S11¯,2m,S11¯,2m, S1¯2,2m can be computed by the paper [48]. The quadratic and cubic sums S12,3¯, S1¯2,3, S1¯2,3¯, S13,2¯, S1¯3,2 and S1¯3,2¯ can be evaluated by Bailey, Borwein and Girgensohn [3], we get

∞ H3 1 1 1 21 n (−1)n−1 = 6Li + 6Li ln2+ ln52+ ζ (3) ln22 n2 5 2 4 2 5 8 n=1 X 9 27 − ζ (5) − ζ (2) ln32 − ζ (2) ζ (3) , (3.27) 4 16 ∞ H2 1 1 2 7 n (−1)n−1 = 4Li + 4Li ln2+ ln52+ ζ (3) ln22 n3 5 2 4 2 15 4 nX=1 19 2 11 − ζ (5) − ζ (2) ln32 − ζ (2) ζ (3) , (3.28) 32 3 8 ∞ H¯ 2 1 1 7 167 1 n = 4Li − ln52+ ζ (3) ln22 − ζ (5) + ζ (2) ln32 n3 5 2 30 4 32 3 n=1 X 3 19 + ζ (2) ζ (3) + ζ (4) ln 2, (3.29) 4 8 ∞ H¯ 2 1 1 19 n (−1)n−1 = −4Li ln 2 − ln52 − ζ (5) + ζ (2) ln32 n3 4 2 6 32 n=1 X 26 3 19 + ζ (2) ζ (3) + ζ (4) ln 2. (3.30) 8 8 Hence, from identities (2.46)-(2.49), (3.16), (3.17) and (3.25), we deduce the following six closed form

∞ H(2)H¯ 43 1 1 1 n n (−1)n−1 = ζ (4) − ζ (2) ln22 − ln42 − 3Li , (3.31) n 16 4 8 4 2 n=1 X∞ H2H¯ 53 1 1 1 n n (−1)n−1 = ζ (4) + ln42+ ζ (2) ln22 − 3Li , (3.32) n 16 8 4 4 2 n=1 X∞ H¯ 3 1 59 13 11 n (−1)n−1 = 5Li − ζ (4) + ζ (2) ln22+ ln42, (3.33) n 4 2 16 4 24 n=1 X∞ H H¯ (2) 1 1 1 9 n n (−1)n−1 = 3Li + ln42 − ζ (2) ln22 − ζ (4) , (3.34) n 4 2 8 4 16 n=1 X∞ H H(2) 1 1 7 1 n n (−1)n−1 = 2Li + ln42+ ζ (3) ln 2 − ζ (2) ln22 − ζ (4) , (3.35) n 4 2 12 8 4 n=1 X∞ H3 5 3 1 9 n (−1)n−1 = ζ (4) + ζ (2) ln22 − ln42 − ζ (3) ln 2. (3.36) n 8 4 4 8 n=1 X Note that formula (3.36) was also proved in [33] by another method. In (2.5) and (3.8) of [48], letting s = 3 and (p, l) = (1, 3), respectively, and combining (3.28)- (3.30) of this paper, we give

∞ H H¯ 1 1 37 7 n n (−1)n−1 =2Li − ln52 − ζ (5) − ζ (3) ln22 n3 5 2 60 16 8 Xn=1 1 1 + ζ (2) ln32 + 4ζ (4) ln 2 − ζ (2) ζ (3) , (3.37) 6 8 ∞ H H¯ 1 1 193 7 n n =2Li − ln52 − ζ (5) − ζ (3) ln22 n3 5 2 60 64 8 n=1 X 1 3 + ζ (2) ln32 + 4ζ (4) ln 2 + ζ (2) ζ (3) . (3.38) 6 8 Combining (3.20) with (3.37), and (3.23) with (3.29), we have the two results

∞ H H¯ (3) 1 1 3 37 1 n n (−1)n−1 =2Li − ln52 − ζ (3) ln22 − ζ (5) + ζ (2) ln32 n 5 2 60 8 16 6 n=1 X 1 + ζ (2) ζ (3) + ζ (4) ln 2, (3.39) 4 ∞ H H(3) 1 1 1 3 167 n n (−1)n−1 = − 4Li − 2Li ln 2 − ln52+ ζ (3) ln22+ ζ (5) n 5 2 4 2 20 8 32 n=1 X 1 1 49 + ζ (2) ln32 − ζ (2) ζ (3) − ζ (4) ln 2. (3.40) 6 16 16

27 By (2.18), letting m = 4, z = 1, we arrive at the conclusion that

1 ∞ ln4 (1 + x) H3 − 3H H(2) + 2H(3) dx = 4 n n n n (−1)n+1 x 2 n=1 (n + 1) Z0 X 1 1 4 21 = −24Li − 24Li ln 2 − ln52 − ζ (3) ln22 + 24ζ (5) + 4ζ (2) ln32. 5 2 4 2 5 2 (3.41)

Substituting (3.27) into (3.41) yields

∞ H H(2) 1 1 2 7 n n (−1)n−1 = −4Li − 4Li ln 2 − ln52 − ζ (3) ln22 n2 5 2 4 2 15 4 n=1 X 23 2 15 + ζ (5) + ζ (2) ln32+ ζ (2) ζ (3) . (3.42) 8 3 16 Combining (3.22) with (3.42) gives

∞ H¯ H¯ (2) 1 1 7 23 1 n n = 4Li − ln52 − ζ (3) ln22 − ζ (5) + ζ (2) ln32 n2 5 2 30 4 8 3 n=1 X 3 49 − ζ (2) ζ (3) + ζ (4) ln 2. (3.43) 4 8 Theorem 3.11 For positive integer m, then the integrals

1 1 ln2mxln2 (1 − x) ln2mxln2 (1+ x) dx and dx 1+ x 1 − x Z0 Z0 are expressible in terms of ln 2, zeta values and linear sums.

Proof. Using (2.32) and the well known identity

1 xnlnmx dx = (−1)m+nm! ζ¯(m + 1) − H¯ (m+1) , m ∈ N, 1+ x n Z0 we can ﬁnd that

1 m 2 ∞ ¯ (m+1) ∞ ¯ (m+1) ln xln (1 − x) m HnHn n−1 Hn n−1 dx = 2(−1) m! (−1) − 2 (−1) 1+ x ( n n ) Z0 nX=1 nX=1 + (−1)mm!ζ¯(m + 1) ln22, (3.44)

1 ∞ ∞ lnmxln2 (1 + x) H H(m+1) H(m+1) dx = 2(−1)mm! n n (−1)n−1 − n (−1)n−1 1 − x n n2 (n=1 n=1 ) Z0 X X + (−1)mm!ζ (m + 1) ln22. (3.45)

28 From Corollary 3.7, we see that the quadratic sums S1(2m),1¯ and S12m,1¯ are reducible to linear sums and zeta values. Thus, Theorem holds. Setting m = 2 in (3.44) and (3.45), and combining (3.39) and (3.40), we obtain

1 ln2xln2 (1 − x) 1 1 15 2 13 dx = 8Li − ln52+ ζ (5) + ζ (2) ln32 − ζ (2) ζ (3) + ζ (4) ln 2, 1+ x 5 2 15 2 3 2 Z0 (3.46) 1 ln2xln2 (1 + x) 1 1 1 7 47 dx = −16Li − 8 ln 2Li − ln52+ ζ (3) ln22+ ζ (5) 1 − x 5 2 4 2 5 2 2 Z0 2 13 49 + ζ (2) ln32 − ζ (2) ζ (3) − ζ (4) ln 2. (3.47) 3 4 4 Now, we consider the generating function ∞ ¯ 2 n y = Hnx , (−1

x ∞ ln (1 − t) xn y = xy + 2 dt + . (3.48) 1+ t n2 Z0 nX=1 By integrating, we may rewrite (3.48) as ∞ H(2) − H¯ 2 x ln (1 − t) x ln2 (1 − t) n n xn+1 = 2 ln (1 − x) dt − 2 dt. (3.49) n + 1 1+ t 1+ t n=1 0 0 X Z Z ln x Multiplying (3.49) by and integrating over the interval (0,1). The result is x ∞ ∞ H(2)H¯ 5 15 9 1 H¯ 2 n n (−1)n−1 = ζ (5) + ζ (4) ln 2 − ζ (2) ζ (3) + n n2 4 4 8 2 3 n=1 n=1 (n + 1) X X ∞ 1 H H¯ 1 ln2xln2 (1 − x) − n n (−1)n−1 − dx. (3.50) n3 2 1+ x n=1 X Z0 Substituting (3.29), (3.37) and (3.46) into (3.50), we have ∞ H(2)H¯ 1 1 7 29 n n (−1)n−1 = −4Li + ln52+ ζ (3) ln22 − ζ (5) n2 5 2 30 4 64 n=1 X 1 15 23 − ζ (2) ln32+ ζ (2) ζ (3) − ζ (4) ln 2. (3.51) 3 8 16 ln x Applying the same argument as above, replacing x by −x in (3.49), then dividing this by x and integrating over the interval (0, 1), we arrive at the conclusion that ∞ H¯ H¯ (2) 1 1 1 35 n n (−1)n−1 = 8Li + 4Li ln2+ ln52 − ζ (5) n2 5 2 4 2 10 4 n=1 X 29 1 1 49 − ζ (2) ln32+ ζ (2) ζ (3) + ζ (4) ln 2. (3.52) 3 16 8 Letting (a, x, b, p, m) = (1, 1, −1, 2, 1), (−1, −1, 1, 1, 2), (−1, 1, −1, 1, 2) and (−1, −1, −1, 2, 1) in Theorem 3.5, and combining identities (2.51), (3.18), (3.19), (3.21), (3.26), (3.37), (3.38), (3.40), (3.42), (3.43), (3.51) and (3.52), we obtain the following eight evaluations

∞ H H¯ (2) 125 13 n n (−1)n−1 = ζ (5) − ζ (2) ζ (3) , (3.53) n2 32 8 n=1 X∞ H(2)H¯ (2) 75 3 25 n n (−1)n−1 = ζ (5) + ζ (2) ζ (3) − ζ (4) ln 2, (3.54) n 64 8 16 n=1 X∞ H H¯ (2) 1 1 2 29 7 n n = 4Li + 4Li ln2+ ln52+ ζ (5) + ζ (3) ln22 n2 5 2 4 2 15 64 4 n=1 X 2 5 − ζ (2) ln32 − ζ (2) ζ (3) , (3.55) 3 4 ∞ H2H(2) 1 1 7 1 n n (−1)n−1 = −2Li ln 2 − ln52 − ζ (3) ln22+ ζ (2) ln32 n 4 2 12 8 3 n=1 X 3 7 + ζ (2) ζ (3) + ζ (4) ln 2, (3.56) 8 8 ∞ H(2)H¯ 1 1 1 27 1 n n = −8Li − 4Li ln 2 − ln52+ ζ (5) + ζ (2) ln32 n2 5 2 4 2 10 4 3 n=1 X 1 23 + ζ (2) ζ (3) − ζ (4) ln 2, (3.57) 2 16 2 ∞ (2) Hn 1 1 1 259 2 (−1)n−1 = 16Li + 8Li ln2+ ln52 − ζ (5) − ζ (2) ln32 n 5 2 4 2 5 16 3 n=1 X 5 29 + ζ (2) ζ (3) + ζ (4) ln 2, (3.58) 8 4 2 ∞ ¯ (2) Hn 1 1 1 299 2 (−1)n−1 = −16Li − 8Li ln 2 − ln52+ ζ (5) + ζ (2) ln32 n 5 2 4 2 5 16 3 nX=1 1 39 − ζ (2) ζ (3) − ζ (4) ln 2, (3.59) 4 4 ∞ H4 1 1 3 9 83 n (−1)n−1 = 8Li + 4Li ln2+ ln52+ ζ (3) ln22 − ζ (5) n 5 2 4 2 10 4 16 nX=1 4 11 11 − ζ (2) ln32 − ζ (2) ζ (3) + ζ (4) ln 2. (3.60) 3 8 4 In many other cases we are also able to obtain a formula for other Euler sum of weight ﬁve explicitly in terms of values of zeta, logarithm and polylogarithm functions. For example, we also use the method of the present paper to obtain explicit evaluations for the following sums

∞ H(3)H¯ 1 1 3 193 1 n n (−1)n−1 = −2Li + ln52 − ζ (3) ln22 − ζ (5) − ζ (2) ln32 n 5 2 60 8 64 6 nX=1

30 11 5 + ζ (4) ln 2 − ζ (2) ζ (3) , (3.61) 16 8 ∞ H¯ H¯ (3) 1 1 3 19 1 n n (−1)n−1 = 2Li ln2+ ln52+ ζ (3) ln22 − ζ (5) − ζ (2) ln32 n 4 2 12 8 32 2 n=1 X 11 1 + ζ (4) ln 2 + ζ (2) ζ (3) , (3.62) 16 4 ∞ H2H¯ 1 1 1 165 1 n n = 4Li + 2Li ln2+ ln52 − ζ (5) − ζ (2) ln32 n2 5 2 4 2 20 32 6 nX=1 1 109 + ζ (2) ζ (3) + ζ (4) ln 2, (3.63) 2 16 ∞ H2H¯ 1 1 7 197 1 n n (−1)n−1 = 2Li − ln52 − ζ (3) ln22 − ζ (5) + ζ (2) ln32 n2 5 2 60 8 64 6 n=1 X 3 109 − ζ (2) ζ (3) + ζ (4) ln 2, (3.64) 4 16 ∞ H¯ 2H¯ (2) 1 7 7 899 2 n n (−1)n−1 = 14Li − ln52 − ζ (3) ln22 − ζ (5) − ζ (2) ln32 n 5 2 60 8 64 3 n=1 X 1 193 + ζ (2) ζ (3) + ζ (4) ln 2, (3.65) 4 16 ∞ H(2)H¯ 2 1 1 1 1 n n (−1)n−1 = −6Li ln 2 − ln52 − ζ (2) ln32+ ζ (2) ζ (3) n 4 2 4 6 8 n=1 X 9 + ζ (4) ln 2, (3.66) 2 ∞ H3H¯ 1 11 9 55 7 n n (−1)n−1 = −2Li − ln52 − ζ (3) ln22+ ζ (5) + ζ (2) ln32 n 5 2 60 16 16 12 n=1 X 1 37 − ζ (2) ζ (3) + ζ (4) ln 2, (3.67) 16 8 ∞ H H(2)H¯ 1 1 7 155 1 n n n (−1)n−1 = −6Li + ln52+ ζ (3) ln22+ ζ (5) − ζ (2) ln32 n 5 2 20 16 32 12 nX=1 7 + ζ (2) ζ (3) − 2ζ (4) ln 2. (3.68) 16 Some of the above identities can be found in [45]. Moreover, we use Mathematica tool to check numerically each of the speciﬁc identities listed. The numeric results prove that the obtained formulas are correct.

3.4 Some evaluation of Euler-type sums In this subsection, we will establish some explicit relationships which involve Euler type sums of the form ∞ ζ (m ; x ) · · · ζ (m ; x ) S (r, k; x ,...,x ; x)= n 1 1 n r p xn, p, k, r ∈ N m1m2...mp 1 p (n + r) (n + k) n=1 X and (alternating) harmonic numbers, where x, xj = ±1 and r ≤ k. We develops an approach to evaluation of Euler-type sums Sm1m2...mp (r, k; x1,...,xp; x). The approach is based on integral

31 computations.

Theorem 3.12 For integers 0 ≤ r < k and i1 ≤ i2 ≤···≤ ip, then the following relation holds:

(k − r) Si1···ip (r, k; x1, · · · ,xp; 1) − S(i1+···+ip) (r, k; x1 · · · xp; 1) k−r ∞ xn · · · xn = − 1 p ni1+···+ip (n + r + i − 1) n=1 Xi=1 X j p n k−r p−1 ζn (ilr ; xlr ) xa −  (−1)p−j r=1 a=1 . (3.69) 1 ··· − Q ··· Qj  i + +ip il1 + +ilj  i=1 j=0 1≤l1<···

∞ p ζ (i ; x ) − ζ (i + · · · + i ; x · · · x ) xn  n j j n 1 p 1 p  n=1 X jY=1  p j p  xn  n ∞ j p−1 ζn (ilr ; xlr ) xa n  j=1 p−j r=1 a=1  x = − i1+···+i + (−1) . (3.70)  Q p 1 Q··· − ··· Qj  n i + +ip il1 + +ilj  1 − x n=1  j=0 1≤l1<···

Corollary 3.13 For positive integers m1, m2, k and r ≥ 0, we have

∞ 1 2 1 2 H(m )H(m ) − H(m +m ) (k − r) n n n (n + r) (n + k) nX=1 k−r ∞ 1 2 H(m ) H(m ) 2 = n + n − , (3.71) m2 m1 m1+m2 (n (n + r + j − 1) n (n + r + j − 1) n (n + r + j − 1)) Xj=1 nX=1 ∞ 1 2 1 2 H¯ (m )H¯ (m ) − H(m +m ) (k − r) n n n (n + r) (n + k) n=1 X k−r ∞ 1 2 H¯ (m )(−1)n−1 H¯ (m )(−1)n−1 2 = n + n − , (3.72) nm2 (n + r + j − 1) nm2 (n + r + j − 1) nm1+m2 (n + r + j − 1) n=1 ( ) Xj=1 X ∞ 1 2 1 2 H(m )H¯ (m ) − H¯ (m +m ) (k − r) n n n (n + r) (n + k) nX=1 k−r ∞ 1 2 H(m )(−1)n−1 H¯ (m ) 2(−1)n−1 = n + n − , (3.73) nm2 (n + r + j − 1) nm1 (n + r + j − 1) nm1+m2 (n + r + j − 1) n=1 ( ) Xj=1 X

32 ∞ ∞ H(m) H¯ (m) where the Euler-type sums n and n (p,m,j ∈ N) can be evaluated by np (n + j) np (n + j) n=1 n=1 the following formula X X

∞ p−1 ∞ ∞ f (n) (−1)i−1 f (n) (−1)p−1 f (n) = + . np (n + j) ji np+1−i jp−1 n (n + j) n=1 n=1 n=1 X Xi=1 X X Note that the closed forms of the following Euler-type sums can be found in [39, 53, 54].

∞ ∞ H(m) H¯ (m) n , n , r 6= k. (n + r) (n + k) (n + r) (n + k) nX=1 nX=1

Corollary 3.14 For integers m1 > 0,m2 > 0,m3 > 0 and k > r ≥ 0, we have

∞ 1 2 3 1 2 3 H(m )H(m )H(m ) − H(m +m +m ) (k − r) n n n n (n + r) (n + k) n=1 X k−r ∞ 1 2 1 3 2 3 H(m )H(m ) H(m )H(m ) H(m )H(m ) = n n + n n + n n nm3 (n + r + j − 1) nm2 (n + r + j − 1) nm1 (n + r + j − 1) n=1 ( ) Xj=1 X k−r ∞ 1 2 3 H(m ) H(m ) H(m ) − n + n + n . nm2+m2 (n + r + j − 1) nm1+m3 (n + r + j − 1) nm1+m2 (n + r + j − 1) n=1 ( ) Xj=1 X It is possible that some of other Euler type sums can be obtained using techniques of the present paper. Acknowledgments. The authors would like to thank the anonymous referee for his/her helpful comments, which improve the presentation of the paper.

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