Multiple zeta values and Euler sums
arXiv:1609.05863v1 [math.NT] 16 Sep 2016
Ce Xu∗
School of Mathematical Sciences, Xiamen University
Xiamen 361005, P.R. China
Abstract In this paper, by using the method of iterated integral representations of series,
we establish some expressions of series involving harmonic numbers and Stirling numbers of
the first kind in terms of multiple zeta values. Furthermore, we can obtain some closed form
representations of sums of products of harmonic numbers through Riemann zeta values and
linear sums, and some explicit relationships between multiple zeta star values and multiple zeta
values are established.
Keywords Multiple zeta value; multiple zeta star value; multiple harmonic number; multiple
star harmonic number; Euler sum.
AMS Subject Classifications (2010): 40B05; 33B15; 11M06; 11M41
1
Introduction
Let R and C denote, respectively the sets of real and complex numbers and let N := {1, 2, 3, . . .}
be the set of natural numbers, and N0 := N ∪ {0} be the set of positive integers and N \ {1} :=
{2, 3, 4, · · · }. For any multi-index S := (s1 , s2 , · · · , sk ) (si ∈ C, k ∈ N, ℜ(s1 ) > 1), the general
multiple zeta value ζ(S) and the multiple zeta star value ζ ⋆ (S) are defined, respectively, by
convergent series ([5,23,36])
ζ (S) = ζ (s1 , s2 , · · · , sk ) :=
X
1
,
· · · nskk
ns1 ns2
n1 >n2 >···>nk ≥1 1 2
ζ ⋆ (S) = ζ ⋆ (s1 , s2 , · · · , sk ) :=
X
1
.
· · · nskk
ns1 ns2
n1 ≥n2 ≥···≥nk ≥1 1 2
(1.1)
(1.2)
where s1 + · · · + sk is called the weight and k is the multiplicity. For convenience, we let {a}k
be the k repetitions of a such that
ζ (5, 3, {1}2 ) = ζ (5, 3, 1, 1) , ζ ⋆ (4, 2, {1}3 ) = ζ ⋆ (4, 2, 1, 1, 1) .
Many papers use the opposite convention, with the ni ’s ordered by n1 < n2 < · · · < nk or
n1 ≤ n2 ≤ · · · ≤ nk , see [10,11,13,14,19]. Multiple zeta values and multiple zeta star values
were introduced and studied by Euler [16] in the old days. The multiple zeta values have
attracted considerable interest in recent years. In the past two decades, many authors have
studied multiple zeta values and multiple zeta star values, and a number of relations among
them have been found [5,7,9-11,13-15,19,20,22-24,27-28,37-39]. There are important properties
∗
Corresponding author. Email: xuce1242063253@163.com (C. XU)
1
for multiple zeta values, so called sum, cyclic sum, and duality formulas. For example, one of
the well known Q-linear relations among multiple zeta values is the sum formula ( see [5, 20]),
which states that
X
ζ (s1 , s2, · · · , sk ) = ζ (n) .
(1.3)
s1 +···sk =n
Each sj ≥1, s1 >1
In [20], M. Igarashi proved a generalization of the sum formula (see Proposition 3 in the reference
[20]). From [5,10,11,13-15], we know that multiple zeta values can be represented by iterated
integrals (or Drinfeld integrals) over a simplex of weight dimension. Thus, we have the alternative
(s1 + s2 + · · · + sk )-dimensional iterated-integral representation
ζ (s1 , s2, · · · , sk ) =
Z1
Ωs1 −1 w1 Ωs2 −1 w2 · · · Ωsk −1 wk , s1 > 1,
(1.4)
0
in which the integrand denotes a string of distinct differential 1-forms of type Ω := dx/x, and
wj is given by
dxj
.
(1.5)
wj :=
1 − xj
By using (1.4), Jonathan M. Borwein, David M. Bradley and David J. Broadhurst [5] proved
the following duality relation
(1.6)
ζ m1 + 2, {1}n1 , . . . , mp + 2, {1}np = ζ np + 2, {1}mp , . . . , n1 + 2, {1}m1 .
A generalization of this duality formula can be found in [10, 11, 14, 15]. On the other hand, the
corresponding property of the duality formula for multiple zeta-star values was not known until
recently. The best result to date are due to Masanobu Kaneko, Yasuo Ohno ([23]) and Chika
Yamazaki ([36]). Kaneko and Ohno proved the following property
(−1)k ζ ⋆ (k + 1, {1}n ) − (−1)n ζ ⋆ (n + 1, {1}k ) ∈ Q [ζ (2) , ζ (3) , . . .]
(1.7)
the right-hand side being the algebra over Q generated by the values of the Riemann zeta
function at positive integer arguments (s > 1). C. Yamazaki ([36]) gave another proof of (1.7).
The subjects of this paper are Multiple zeta values and Euler sums. Next, we give an
introduction to the linear and nonlinear Euler sums. When k = 2 in (1.2), then ζ ⋆ (s1 , s2 ) also
called the classical linear Euler sums, which is defined by
Sp,q :=
∞
X
ζn (p)
n=1
nq
= ζ ⋆ (q, p) , p ∈ N, q ∈ N \ {1},
(1.8)
where ζn (p) stands for the generalized harmonic number defined by
n
X
1
ζn (p) :=
, p, n ∈ N,
jp
(1.9)
j=1
when p = 1, Hn := ζn (1) is classical harmonic number, the empty sum ζ0 (p) is conventionally
understood to be zero. The generalized harmonic number converges to the Riemann zeta function
ζ(s):
lim ζn (p) = ζ (p) , ℜ (p) > 1,
n→∞
2
where the Riemann zeta function is defined respectively by
ζ(p) :=
∞
X
1
, ℜ(p) > 1.
np
(1.10)
n=1
In general, the multiple harmonic number (also called partial sums of multiple zeta values) and
multiple star harmonic number (also called partial sums of multiple zeta star values) are defined
by
X
1
ζn (s1 , s2 , · · · , sk ) :=
,
(1.11)
s1 s2
n1 n2 · · · nskk
n≥n1 >n2 >···>nk ≥1
ζn⋆ (s1 , s2 , · · · , sk ) :=
1
,
· · · nskk
X
ns1 ns2
n≥n1 ≥n2 ≥···≥nk ≥1 1 2
(1.12)
when n < k, then ζn (s1 , s2 , · · · , sk ) = 0, and ζn (∅) = ζn⋆ (∅) = 1. The generalized (nonlinear)
Euler sums are the infinite sums whose general term is a product of harmonic numbers of index
n and a power of n−1 . Namely, for a multi-index S = (s1 , s2 , . . . , sk ) (k, si ∈ N, i = 1, 2, . . . , k)
with s1 ≤ s2 ≤ . . . ≤ sk and q ≥ 2, the nonlinear Euler sums of index S, q is defined by ([17])
SS,q :=
∞
X
ζn (s1 )ζn (s2 ) · · · ζn (sk )
nq
n=1
,
(1.13)
where the quantity s1 + · · · + sk + q is called the weight, the quantity k is called the degree. As
usual, repeated summands in partitions are indicated by powers, so that for instance
S12 23 4,q = S112224,q =
∞
X
H 2 ζ 3 (2) ζn (4)
n n
n=1
nq
.
It has been discovered in the course of the years that many Euler sums admit expressions
involving finitely the “zeta values”, that is to say values of the Riemann zeta function ζ(s)
with the positive integer arguments. Euler started this line of investigation in the course of a
correspondence with Goldbach beginning and he was the first to consider the linear sums Sp,q .
Euler showed this problem in the case p = 1 and gave a general formula for odd weight p + q in
1775. 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 [3], D. Borwein,
J.M. Borwein and R. Girgensohn proved conjecture and formula, and in [1], D.H. Bailey, J.M.
Borwein and R. Girgensohn conjectured that the double linear sums when p+q > 7, p+q is even,
are not reducible. Hence, the evaluation of Sp,q in terms of values of Riemann zeta function at
positive integers is known when p = 1, p = q, (p, q) = (2, 4), (4, 2) or p + q is odd ([1,3,17]). The
relationship between the values of the zeta values and Euler sums has been studied by many
authors, see [1,3,4,8,17,18,26,35] and references therein. For example, in [18], Philippe Flajolet
and Bruno Salvy proved that all Euler sums of the form S1p ,q for weights p + q ∈ {3, 4, 5, 6, 7, 9}
are expressible polynomially in terms of zeta values and gave explicit formula. For weight 8, all
such sums are the sum of a polynomial in zeta values and a rational multiple of S2,6 , but not
the formula. In [35], we showed that all quadratic Euler sums S1p,q can be evaluated in terms of
zeta values and linear sums whenever p + q ≤ 8, p ∈ N, q ∈ N \ {1} and gave explicit formula.
The main purpose of this paper is to establish some relationships between nonlinear Euler
sums and multiple zeta values by using the method of iterated integral representations of series.
We then use these relations to evaluate several series with harmonic numbers. Furthermore, we
can obtain some explicit formulas between multiple zeta star values and multiple zeta values.
3
2
Main results and proofs
The following lemma will be useful in the development of the main theorems.
Lemma 2.1 (see [35]) Let m, k be integers with m ≥ 2, k ≥ 2. we have the recurrence relation
W (m,k) = − ψ
(m+k)
(1) / (k + 1) −
m−1
k−1 XX
i=1 j=0
m−1
i
k
W (i,j) ψ (m+k−i−j−1) (1) ,
j
(2.1)
where the integral W (m,k) is defined by
W (m,k) :=
Z1
lnk (1 − x)(ln x)m
dx,
1−x
0
and ψ (z) stands for digamma function (or called Psi function) defined by
ψ (z) :=
when n ∈ N,
ψ (n) (z)
= (−1)
n+1
n!
∞
X
d
Γ′ (z)
(ln Γ (z)) =
,
dz
Γ (z)
n+1
1/(z + k)
, Γ (z) :=
k=0
function. By a direct calculation, we can deduce that
Z∞
e−t tz−1 dt, ℜ (z) > 0 is gamma
0
W (m, 0) = (−1)m m!ζ (m + 1) , W (1, k) = (−1)k+1 k!ζ (k + 2) ,
From the recurrence relation (3.1), we know that the values of integrals W (m, k) can be expressed as a rational linear combination of products of zeta values. We now give some values of
the integrals W (m, k) for 3 ≤ m + k ≤ 9
1
W (2, 1) = − ζ (4) ,
2
W (3, 1) = 12ζ (5) − 6ζ (2) ζ (3) ,
W (4, 1) = 12ζ 2 (3) − 18ζ (6) ,
W (5, 1) = 360ζ(7) − 120ζ(3)ζ(4) − 120ζ(2)ζ(5),
W (4, 2) = 240ζ (7) − 60ζ (3) ζ (4) − 96ζ (2) ζ (5) ,
W (3, 3) = 180ζ (7) − 45ζ (3) ζ (4) − 72ζ (2) ζ (5) ,
1497
ζ (8) + 576ζ (3) ζ (5) − 144ζ (2) ζ 2 (3) ,
W (4, 3) = −
4
W (3, 4) = −366ζ (8) + 432ζ (3) ζ (5) − 72ζ (2) ζ 2 (3) ,
W (5, 2) = −610ζ (8) + 720ζ (3) ζ (5) − 120ζ (2) ζ 2 (3) ,
W (4, 4) = 8064ζ (9) + 288ζ 3 (3) − 288ζ (2) ζ (7) − 1260ζ (3) ζ (6) − 2016ζ (4) ζ (5) ,
W (3, 5) = 6720ζ (9) + 120ζ 3 (3) − 2160ζ (2) ζ (7) − 1260ζ (3) ζ (6) − 1620ζ (4) ζ (5) ,
84483
ζ (10) − 11520ζ (2) ζ (3) ζ (5) + 28800ζ (3) ζ (7) − 3600ζ (4) ζ 2 (3) + 14400ζ 2 (5) ,
W (5, 4) = −
4
W (4, 5) = −17514ζ (10) − 8640ζ (2) ζ (3) ζ (5) + 23040ζ (3) ζ (7) − 3240ζ (4) ζ 2 (3) + 11520ζ 2 (5) .
4
Theorem 2.2 For n, m ∈ N and x ∈ [−1, 1), the following relation holds:
Zx
tn−1 lnm (1 − t)dt =
(−1)m
1 m
ln (1 − x) (xn − 1) + m!
n
n
0
−
m−1
1 X
(−1)i−1 i!
n
i=1
Proof.
m
i
X
1≤km ≤···≤k1 ≤n
xk m
k1 · · · km
X
lnm−i (1 − x)
1≤ki ≤···≤k1 ≤n
The proof is by induction on m. Define J (n, m; x) :=
Zx
xk i − 1
.
k1 · · · ki
(2.2)
tn−1 lnm (1 − t)dt, for m = 1,
0
by simple calculation, we can arrive at the conclusion that


Z x
n


j
X
1
x
tn−1 ln (1 − t)dt =
J (n, 1; x) =
− ln (1 − x) ,
xn ln (1 − x) −

n
j
0
j=1
and the formula is true. For m > 1 we proceed as follows. By using integration by parts we
have the following recurrence relation
n
J (n, m; x) =
mX
1 m
ln (1 − x) (xn − 1) −
J (k, m − 1; x).
n
n
(2.3)
k=1
Then by the induction hypothesis, we have
n
(−1)m X
1 m
n
J (n, m; x) = ln (1 − x) (x − 1) + m!
n
n
X
j=1 1≤km−1 ≤···≤k1 ≤j
+
−
m
n
m−2
X
(−1)i−1 i!
i=1
m m−1
ln
(1 − x)
n
m−1
i
n
X
xj
j=1
lnm−1−i (1 − x)
X
xkm−1
k1 · · · km−1
1≤ki ≤···≤k1 ≤j
−1
.
j
xk i − 1
k1 · · · ki
(2.4)
By a direct calculation, we can deduce the desired result. This completes the proof of Theorem
2.2.
Letting x approach 1 in (2.2) and using the definition of multiple star harmonic number, we
obtain
Z 1
(−1)m ⋆
ζn ({1}m ) ,
(2.5)
tn−1 lnm (1 − t)dt = m!
n
0
Dividing (2.2) by x and integrating over the interval (0,1), the result is
Z1
x
n−1
(−1)m−1
ln xln (1 − x)dx =m!
n
m
0
+ m!
ζn⋆ ({1}m )
(−1)m−1 ⋆
− ζ (m + 1) + m!
ζn {1}m−1 , 2
n
n
m−1
(−1)m−1 X ⋆
ζn {1}i−1 , 2, {1}m−i − ζ (m − i + 1) ζn⋆ ({1}i ) .
n
i=1
(2.6)
5
On the other hand, in [35], we proved the identity
Z1
tn−1 lnk (1 − t)dt = (−1)k
Yk (n)
, Y0 (n) = 1,
n
(2.7)
0
where Yk (n) := Yk (ζn (1) , 1!ζn (2) , 2!ζn (3) , · · · , (r − 1)!ζn (r) , · · ·) and Yk (x1 , x2 , · · ·) stands for
the complete exponential Bell polynomial defined by (see [21])


m
X
X
t
tk
(2.8)
xm  = 1 +
exp 
Yk (x1 , x2 , · · ·) .
m!
k!
m≥1
k≥1
Comparing (2.5) with (2.7), we have the following relation
ζn⋆ ({1}m ) =
1
Ym (n) , n, m ∈ N0 .
m!
(2.9)
By using the definition of the complete exponential Bell polynomial, we can deduce that
Y1 (n) = Hn , Y2 (n) = Hn2 + ζn (2) , Y3 (n) = Hn3 + 3Hn ζn (2) + 2ζn (3) ,
Y4 (n) = Hn4 + 8Hn ζn (3) + 6Hn2 ζn (2) + 3ζn2 (2) + 6ζn (4) .
Y5 (n) = Hn5 + 10Hn3 ζn (2) + 20Hn2 ζn (3) + 15Hn ζn2 (2) + 30Hn ζn (4) + 20ζn (2) ζn (3) + 24ζn (5) .
Moreover, Yk (n) is a rational linear combination of products of harmonic numbers. Further, we
know that the multiple star harmonic number ζn⋆ ({1}m ) can be expressed as a rational linear
combination of harmonic numbers. In fact, using Eulerian beta integral, we can obtain the more
general recurrence relation:
Theorem 2.3 For integers m, k, n ∈ N, then we have the following recurrence relation
I (n, m, k) =
m−1
X
i=0
+
k−1 X
j=0
−
k−1 X
j=0
+
(m − i − 1)!
(−1)m−i
I (n, i, k)
nm−i
k
j
(−1)m+k−j (m + k − j − 1)!ζn (m + k − j) I (n, 0, j)
k
j
(−1)m+k−j (m + k − j − 1)!ζ (m + k − j) I (n, 0, j)
k−1 m−1
XX
i=1 j=0
−
m−1
i
k−1 m−1
XX
i=1 j=0
m−1
i
k
(−1)m+k−i−j (m + k − i − j − 1)!ζn (m + k − i − j) I (n, i, j)
j
m−1
i
k
(−1)m+k−i−j (m + k − i − j − 1)!ζ (m + k − i − j) I (n, i, j).
j
(2.10)
6
where I (n, m, k) is defined by the integral
I (n, m, k) :=
Z1
xn−1 lnm xlnk (1 − x)dx.
(2.11)
0
1
1
, I (n, i, 0) = (−1)i i! i+1 .
n
n
When m = 0, then I (n, 0, k) = J (n, k, 1).
I (n, 0, 0) =
Proof.
Applying the definition of Beta function B (α, β), we can find that
I (n, m, k) :=
Z1
xn−1 lnm xlnk (1 − x)dx =
∂ m+k B (α, β)
∂αm ∂β k
0
,
(2.12)
α=n,β=1
where the Eulerian Beta function is defined by
B (α, β) :=
Z1
xα−1 (1 − x)β−1 dx =
Γ (α) Γ (β)
, ℜ (α) > 0, ℜ (β) > 0.
Γ (α + β)
(2.13)
0
By using (2.13) and the definition of ψ(x), it is obvious that
∂B (α, β)
= B (α, β) [ψ (α) − ψ (α + β)] .
∂α
Therefore, differentiating m − 1 times this equality, we can deduce that
m−1
i
X m − 1 ∂ i B (α, β) h
∂ m B (α, β)
(m−i−1)
(m−i−1)
=
·
ψ
(α)
−
ψ
(α
+
β)
.
i
∂αm
∂αi
(2.14)
i=0
Since B(α, β) = B(β, α), the change of variable α 7→ β, β 7→ α, then we also have
m−1
i
X m − 1 ∂ i B (α, β) h
∂ m B (α, β)
(m−i−1)
(m−i−1)
=
·
ψ
(β)
−
ψ
(β
+
α)
.
i
∂β m
∂β i
(2.15)
i=0
Putting α = n, β = 1 in (2.15), we arrive at the conclusion that
I (n, 0, m) =
m−1
X
(−1)
m−i
(m − i − 1)!
i=0
m−1
i
I (n, 0, i)ζn (m − i).
Furthermore, by using (2.14), the following identity is easily derived
∂ k ∂ m B (α, β)
∂ m+k B (α, β)
=
∂αm ∂β k
∂β k
∂αm
m−1
i
X m − 1 ∂ i+k B (α, β) h
(m−i−1)
(m−i−1)
·
ψ
(α)
−
ψ
(α
+
β)
=
i
∂αi ∂β k
i=0
7
(2.16)
−
k−1 j
X
k ∂ B (α, β)
j=0
−
∂β j
j
k−1 m−1
XX
i=1 j=0
m−1
i
ψ (m+k−j−1) (α + β)
i+j
k ∂ B (α, β) (m+k−i−j−1)
ψ
(α + β) .
j
∂αi ∂β j
(2.17)
From the definition of digamma function, we know that if α = n, β = 1, then we have
ψ (m−i−1) (n) − ψ (m−i−1) (n + 1) = (−1)m−i (m − i − 1)!
1
,
nm−i
ψ (m+k−j−1) (n + 1) = (−1)m+k−j (m + k − j − 1)! (ζ (m + k − j) − ζn (m + k − j)) .
(2.18)
(2.19)
Hence, taking α = n, β = 1 in (2.17), then substituting (2.18) (2.19) into (2.17) respectively, we
can obtain (2.1). The proof of Theorem 2.3 is finished.
From (2.10) and (2.16), we can gives the following identities: for n ∈ N,
Z1
xn−1 ln (1 − x) dx = −
Z1
xn−1 ln2 (1 − x) dx =
I (n, 1, 1) =
Z1
xn−1 ln x ln (1 − x) dx =
I (n, 0, 3) =
Z1
xn−1 ln3 (1 − x) dx = −
I (n, 1, 2) =
Z1
xn−1 ln xln2 (1 − x) dx = −
I (n, 0, 1) =
Hn
,
n
0
I (n, 0, 2) =
Hn2 + ζn (2)
,
n
0
Hn ζ (2) − ζn (2)
,
−
n2
n
0
Hn3 + 3Hn ζn (2) + 2ζn (3)
,
n
0
Hn2 + ζn (2)
ζ (3) − ζn (3)
ζ (2) − ζn (2)
+2
+2
Hn .
2
n
n
n
0
By using integration by parts, we can find the relation
I (n, m, 1) = −
ζ (m + 1) − ζn (m + 1)
m
I (n, m − 1, 1) + (−1)m m!
.
n
n
(2.20)
Furthermore, by simple calculation, we obtain the result
I (n, m, 1) =
Z1
m
x
n−1
m
m+1
ln x ln (1 − x) dx = (−1)
X ζ (j + 1) − ζn (j + 1)
Hn
m! m+1 + (−1)m m!
.
n
nm−j+1
j=1
0
(2.21)
Therefore, from Theorem 2.3, we know that the integral I (n, m, k) is a rational linear combination of products of harmonic numbers and zeta values. Taking m = 1 in (2.10), we can give the
the following corollary.
8
Corollary 2.4 For integer n, k ∈ N, then we have
k−1 X
1
I (n, 1, k) = − I (n, 0, k)−
n
j=0
k
j
(−1)k+1−j (k − j)! (ζ (k + 1 − j) − ζn (k + 1 − j))I (n, 0, j) .
(2.22)
Comparing (2.6) with (2.22), we can obtain the conclusion: For integers m, n ∈ N, then the sums
m
X
ζ ⋆ {1}i−1 , 2, {1}m−i can be expressed by the harmonic
of multiple star harmonic number
numbers, and we have the formula
m
X
ζn⋆
i=1
{1}i−1 , 2, {1}m−i =
i=1
m
X
i=1
ζn (m + 2 − i) ζn⋆ {1}i−1 .
(2.23)
Theorem 2.5 For integer k > 0 and x ∈ [−1, 1), then have
lnk (1 − x) = (−1)k k!
∞
X
xn
n=1
n
ζn−1 {1}k−1 ,
s (n, k) = (n − 1)!ζn−1 {1}k−1 .
(2.24)
(2.25)
where s (n, k) is called (unsigned) Stirling number of the first kind (see [21]).
(n − 1)! 2
Hn−1 − ζn−1 (2) ,
2
(n − 1)! 3
s (n, 4) =
Hn−1 − 3Hn−1 ζn−1 (2) + 2ζn−1 (3) ,
6
(n − 1)! 4
2
2
s (n, 5) =
Hn−1 − 6ζn−1 (4) − 6Hn−1
ζn−1 (2) + 3ζn−1
(2) + 8Hn−1 ζn−1 (3) .
24
s (n, 1) = (n − 1)!, s (n, 2) = (n − 1)!Hn−1 , s (n, 3) =
The 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) , n, k ∈ N,
with s (n, k) = 0, n < k, s (n, 0) = s (0, k) = 0, s (0, 0) = 1.
Proof. To prove the first identity we proceed by induction on k. Obviously, it is valid for k = 1.
For k > 1 we use the equality
k+1
ln
(1 − x) = − (k + 1)
Zx
lnk (1 − t)
dt
1−t
0
and apply the induction hypothesis, by using Cauchy product formula, we arrive at
k+1
ln
(1 − x) = − (k + 1)
Zx
lnk (1 − t)
dt
1−t
0
= (−1)k+1 (k + 1)!
∞
X
n=1
9
n
1 X ζi−1 {1}k−1 n+1
x
n+1
i
i=1
= (−1)k+1 (k + 1)!
∞
X
ζn ({1}k ) n+1
x
.
n+1
n=1
Nothing that ζn ({1}k ) = 0 when n < k. We can deduce (2.24). To prove the second identity of
our theorem, we use the following equation ([21])
lnk (1 − x) = (−1)k k!
∞
X
s (n, k) n
x , −1 ≤ x < 1.
n!
(2.26)
n=k
Comparing (2.24) and (2.26), we can derive (2.25). The proof of Theorem 2.5 is thus completed.
From (2.1) and (2.26), we have the result
W (k, m) =
Z1
lnk (1 − x) lnm x
dx
x
0
Z1
∞
X
ζn−1 {1}k−1
= (−1) k!
xn lnm xdx
n
n=1
0
∞
X ζn−1 {1}k−1
m+k
= (−1)
m!k!
nm+2
n=1
= (−1)m+k m!k!ζ m + 2, {1}k−1 ,
k
(2.27)
which implies that for any m, k ∈ N, the multiple zeta value ζ(m + 1, {1}k−1 ) can be represented
as a polynomial of zeta values with rational coefficients. For example:
ζ (2, {1}m ) = ζ (m + 2) ,
m
1X
m+2
ζ (m + 3) −
ζ (3, {1}m ) =
ζ (k + 1) ζ (m + 2 − k).
2
2
k=1
By integration by parts, we can deduce that
W (m, k − 1) =
m
W (k, m − 1) .
k
(2.28)
Applying (2.27) into (2.28), we can obtain the well known duality formula
ζ k + 1, {1}m−1 = ζ m + 1, {1}k−1 .
Obviously, the equation above is special cases of equation (1.6). From (2.25)-(2.28), we get the
result
∞
X
s (n, k)
= ζ k + 1, {1}m−1 = ζ m + 1, {1}k−1 .
(2.29)
m
n!n
n=p
Now we state our main results. The main results of this paper are the following theorems.
Theorem 2.6 For integers p ∈ N and m ∈ N0 . Then
∞
X
Hn s (n, p)
n=1
n!nm+1
= (p + 1) ζ (p + 2, {1}m ) +
m
X
i=1
10
ζ p + 1, {1}i−1 , 2, {1}m−i .
(2.30)
Proof.
Applying (2.26), we can show that
p+1
(−1)
p!
∞
X
Hn s (n, p)
n!nm+1
n=1
=
Z1
0
=
ln (1 − t1 )
dt1
t1
Z
Zt1
0
1
dt2 · · ·
t2
tZ
m−1
0
1
dtm
tm
Ztm
0
lnp (1 − tm+1 )
dtm+1
tm+1
ln (1 − t1 ) lnp (1 − tm+1 )
dt1 dt2 · · · dtm+1 . (2.31)
t1 t2 · · · tm+1
0<tm+1 <tm <···<t1 <1
Applying the change of variables
ti 7→ 1 − tm+2−i , i = 1, 2, · · · , m + 1.
to the above multiple integral, we get the identity
Z
ln (1 − t1 ) lnp (1 − tm+1 )
dt1 dt2 · · · dtm+1
t1 t2 · · · tm+1
0<tm+1 <tm <···<t1 <1
Z
=
0<tm+1 <tm <···<t1 <1
=
Z1
0
=
Zt1
lnp (t1 )
dt1
1 − t1
∞
X
∞
X
1
dt2 · · ·
1 − t2
0
···
nm+1 =1 nm =1
ln (tm+1 ) lnp (t1 )
dt1 dt2 · · · dtm+1
(1 − tm+1 ) (1 − tm ) · · · (1 − t1 )
∞
X
Z1
tZ
m−1
0
1
dtm
1 − tm
n
−1
t1 m+1 lnp (t1 ) dt1
n1 =1 0
Zt1
Ztm
0
ln (tm+1 )
dtm+1 ,
1 − tm+1
tn2 m −1 dt2 · · ·
0
tZ
m−1
tnm2 −1 dtm
0
Ztm
1 −1
tnm+1
ln (tm+1 )dtm+1 .
0
(2.32)
By using integration by parts, we can deduce that, for n, m ∈ N,
Zx
t
n−1
0
m
l
X
m (−1)
(ln x)m−l xn , x ∈ (0, 1),
l!
(ln t) dt =
l
nl+1
m
(2.33)
l=0
Taking m = 2 in (2.33), we get
Zx
tn−1 ln tdt =
xn
1 n
x ln x − 2 .
n
n
0
Substituting (2.34) into (2.32) and combining (2.31), we arrive at the conclusion that
(−1)
p+1
p!
∞
X
Hn s (n, p)
n=1
=
∞
X
n!nm+1
(p + 1)!(−1)p+1
p+2
n1 ,n2 ,···nm+1 =1 n1 (n1 + n2 ) · · · (n1 + · · · + nm ) (n1 + · · · + nm + nm+1 )
−
∞
X
p!(−1)p
2
p+1
n1 ,n2 ,···nm+1 =1 n1 (n1 + n2 ) · · · (n1 + · · · + nm ) (n1 + · · · + nm + nm+1 )
11
(2.34)
− ···
−
∞
X
(−1)p p!
p+1
2
n1 ,n2 ,···nm+1 =1 n1 (n1 + n2 ) · · · (n1 + · · · + nm ) (n1 + · · · + nm + nm+1 )
=(−1)p+1 (p + 1)!ζ (p + 2, {1}m ) + (−1)p+1 p!
m
X
i=1
ζ p + 1, {1}i−1 , 2, {1}m−i .
This completes the proof of Theorem 2.6.
(2.35)
Theorem 2.7 For integers m, p ∈ N and r ∈ N \ {1}. Then
∞
X
s (n, m) ζn (r)
n!np
n=1
Proof.
Z1
=
tZr−1
0
1
dtr
tr
0<tp+1 <···<t1 <1
Ztr
0
ln (1 − tr+1 )
dtr+1
tr+1
tZr+1
0
1
tr+2
dtr+2 · · ·
tZp−1
0
1
dtp
tp
Ztp
0
lnm (1 − tp+1 )
dtp+1
tp+1
ln (1 − tr+1 ) lnm (1 − tp+1 )
dt1 dt2 · · · dtp+1 .
t1 t2 · · · tp+1
Applying the change of variables
ti 7→ 1 − tp+2−i , i = 1, 2, · · · , p + 1.
to the above multiple integral, we deduce the identity
Z
ln (1 − tr+1 ) lnm (1 − tp+1 )
dt1 dt2 · · · dtp+1
t1 t2 · · · tp+1
0<tp+1 <···<t1 <1
Z
=
0<tp+1 <···<t1 <1
=
Z1
0
lnm (t1 )
dt1
1 − t1
tp−r+1
Z
×
0
ln (tp−r+1 ) lnm (t1 )
dt1 dt2 · · · dtp+1
(1 − tp+1 ) (1 − tp ) · · · (1 − t1 )
Zt1
0
1
dt2 · · ·
1 − t2
tp−r−1
Z
0
1
dtp−r+2 · · ·
1 − tp−r+2
Ztp
0
1
dtp−r
1 − tp−r
tZp−r
0
ln (tp−r+1 )
dtp−r+1
1 − tp−r+1
1
dtp+1 ,
1 − tp+1
∞
X
(m + 1)!(−1)m+1
=
n · · · (n1 + · · · + np ) (n1 + · · · + np + np+1 )m+2
n1 ,n2 ,···np+1 =1 1
−
(2.36)
Similarly to the proof of Theorem 2.6, we consider the multiple integral
1
dt1 · · ·
t1
Z
0
= ζ (r) ζ m + 1, {1}p−1 − ζ m + 1, {1}p−1 , 2, {1}r−2 .
∞
X
m!(−1)m
2
m+1
n1 ,n2 ,···np+1 =1 n1 · · · (n1 + · · · + np−1 ) (n1 + · · · + np ) (n1 + · · · + np + np+1 )
− ···
12
(2.37)
∞
X
−
(−1)m m!
n1 ,n2 ,···np+1 =1
n1 · · · (n1 + · · · + nr ) (n1 + · · · + nr+1 )2 · · · (n1 + · · · + np + np+1 )m+1
p−r−1
X ζ m + 1, {1}i , 2, {1}p−1−i .
=(−1)m+1 (m + 1)!ζ m + 2, {1}p + (−1)m+1 m!
(2.38)
i=0
On the other hand, by using (2.21), we have
Z1
0
1
dt1 · · ·
t1
tZr−1
0
1
dtr
tr
Ztr
0
(−1)r
ln (1 − tr+1 )
dt
=
r+1
r!
t1−n
r+1
Z1
t1n−1 lnr (t1 ) ln (1 − t1 ) dt1
0
r
=−
X ζn (j + 1) − ζ (j + 1)
Hn
−
.
nr+1
nr−j+1
(2.39)
j=1
Substituting (2.39) into (2.37) and combining (2.26), we conclude that
Z
ln (1 − tr+1 ) lnm (1 − tp+1 )
dt1 dt2 · · · dtp+1
t1 t2 · · · tp+1
0<tp+1 <···<t1 <1


r
∞

X
X
ζn (j + 1) − ζ (j + 1) 
s (n, m)
Hn
+
= (−1)m+1 m!
.

n!np−r  nr+1
nr−j+1
n=1
(2.40)
j=1
Combining (2.38) and (2.40), we get


∞
r
X
X
s (n, m)  Hn
ζn (j + 1) − ζ (j + 1) 
+

n!np−r  nr+1
nr−j+1
n=1
j=1
p−r−1
X ζ m + 1, {1}i , 2, {1}p−1−i .
= (m + 1) ζ m + 2, {1}p +
(2.41)
i=0
Applying (2.29) and (2.30) into (2.41) yields
r X
∞
X
s (n, m) ζn (j + 1)
j=1 n=1
n!np−j+1
=
r n
o
X
ζ (j + 1) ζ m + 1, {1}p−j − ζ m + 1, {1}p−j , 2, {1}j−1 .
j=1
(2.42)
Replacing r by r − 1 in (2.42), then substituting it into (2.42), we obtain
∞
X
s (n, m) ζn (r + 1)
n=1
n!np−r+1
= ζ (r + 1) ζ m + 1, {1}p−r − ζ m + 1, {1}p−r , 2, {1}r−1 .
The proof of Theorem 2.7 is finished.
Remark 2.8 By considering the following multiple integral
Z1
0
lnr (t1 )
dt1
1 − t1
Zt1
0
1
dt2 · · ·
t2
tZp−1
0
1
dtp
tp
13
Ztp
0
lnm (1 − tp+1 )
dtp+1 ,
tp+1
and using the elementary integral identity
Z1
xn lnm x
dx = (−1)m m! (ζ (m + 1) − ζn (m + 1)) , m ∈ N.
1−x
0
then combining (2.25), we also obtain (2.36). When p = 1, then the integral above can be written
as
Z1
lnm x
1−x
0
Zx
lnr (1 − t)
dtdx
t
0
= (−1)r+m r!m!
∞
X
s (n, r)
n=r
=
Z1
lnp x
1−x
Zx
(ζ (m + 1) − ζn (m + 1)),
n!n
lnm (1 − t)
dtdx
t
0
0
= (−1)m+r m!r!
∞
X
s (n, m)
n=r
n!n
(ζ (r + 1) − ζn (r + 1)).
Applying (2.29), we deduce the following duality relation
∞
X
s (n, m)
n!n
n=1
∞
X
s (n, r)
ζn (r + 1) =
n=1
n!n
ζn (m + 1),
with s (n, k) = 0, n < k, s (n, 0) = s (0, k) = 0, s (0, 0) = 1.
In the same manner, we obtain the more general identity, see the following Theorem 2.8.
Theorem 2.9 For integers k, m ∈ N0 and p ∈ N. Then
 


m
X
∞
m
X
X
X
ij 
Yk (n) s (n, p)
p + k −
ij , im + 1, · · · , i1 + 1,
= k!
 ζ p + k + 1 −

m+1
j=1
n!n
n=1
j=1
0≤i1 +···im ≤k
p
ij ≥0,j=1,2··· ,m
(2.43)
where Yk (n) is complete exponential Bell number defined by (2.8). When m = 0, then
∞
X
Yk (n) s (n, p)
p+k
= k!
ζ (p + k + 1) .
k
n!n
n=1
Proof.
that
Similarly to the proof of Theorem 2.6 and 2.7, using (2.7) and (2.26), we can deduce
p+k
(−1)
p!
∞
X
Yk (n) s (n, p)
n=1
n!nm+1
=
Z1
lnk (1 − t1 )
dt1
t1
=
Z
0
Zt1
0
1
dt2 · · ·
t2
k
tZ
m−1
0
1
dtm
tm
p
Ztm
0
lnp (1 − tm+1 )
dtm+1
tm+1
ln (1 − t1 ) ln (1 − tm+1 )
dt1 dt2 · · · dtm+1 .
t1 t2 · · · tm+1
0<tm+1 <tm <···<t1 <1
(2.44)
14
Using (2.33), by a similar argument as in the proof of Theorem 2.6 and 2.7, we have
p!
∞
X
Yk (n) s (n, p)
n!nm+1
n=1
X
=
0≤i1 +···im ≤k
ij ≥0,j=1,2··· ,m


(k)i1 (k − i1 )i2 · · · k −
× ζ p + k + 1 −
m
X
j=1
m−1
X
j=1

ij 

im
p + k −
m
X
j=1


ij !
ij , im + 1, · · · , i1 + 1 ,
(2.45)
where (m)l := m(m − 1) · · · (m − l + 1). By a direct calculation, we find that


 


m
X
m−1
m
X
X
ij 
p + k −
(k)i1 (k − i1 )i2 · · · k −
ij  p + k −
ij ! = k!p! 
.
j=1 
j=1
j=1
p
im
Combining (2.45) and (2.46), we can obtain the formula (2.43).
Moreover, by simple calculation, we obtain the following result.
(2.46)
Theorem 2.10 For integers k, m ∈ N0 and p ∈ N. Then
p!
∞
X
Yk (n) s (n, p)
n=1
= p!k!
n!nm+1
m
X
i=1
Proof.
+ (−1)m−1 k!
∞
X
Yp (n) s (n, k)
n!nm+1
n=1
(−1)i−1 ζ i + 1, {1}k−1 ζ m + 2 − i, {1}p−1 .
(2.47)
First, by using multiply integral, we can find that
Z1
0
1
dt1 · · ·
t1
tZ
m−1
0
1
dtm
tm
Ztm
0
∞
X s (n, p)
lnp (1 − tm+1 )
dtm+1 = (−1)p p!
tm+1
n!nm+1
n=1
= (−1)p p!ζ m + 2, {1}p−1 .
Then applying formula (2.48) to the right hand side of (2.44), we can show that
(−1)p+k p!
∞
X
Yk (n) s (n, p)
n!nm+1
=(−1)p+k p!k!ζ 2, {1}k−1 ζ m + 1, {1}p−1
n=1
−
Z1
0
1
dt1
t1
Zt1
0
lnk (1 − t2 )
dt2
t2
Zt1
0
1
dt2 · · ·
t2
tZ
m−2
0
1
tm−1
dtm−1
tZ
m−1
0
lnp (1 − tm )
dtm
tm
=···
=(−1)
p+k
p!k!
m
X
i=1
(−1)i−1 ζ i + 1, {1}k−1 ζ m + 2 − i, {1}p−1
15
(2.48)
+ (−1)
m
Z1
0
=(−1)p+k p!k!
lnp (1 − t1 )
dt1
t1
m
X
i=1
+ (−1)p+k+mk!
Zt1
0
tZ
m−1
1
dt2 · · ·
t2
0
1
dtm
tm
Ztm
0
lnk (1 − tm+1 )
dtm+1
tm+1
(−1)i−1 ζ i + 1, {1}k−1 ζ m + 2 − i, {1}p−1
∞
X
Yp (n) s (n, k)
n=1
n!nm+1
.
Thus, formula (2.47) holds.
Setting m = 1 in the above equation we obtain
p!
∞
X
s (n, p) Yk (n)
n!n2
n=1
3
+ k!
∞
X
s (n, k) Yp (n)
n!n2
n=1
= k!p!ζ (k + 1) ζ (p + 1) .
(2.48)
Relations between multiple zeta values and multiple zeta star
values
In this section, we will establish some explicit relationships between multiple zeta star values
and multiple zeta values. Furthermore, we give closed formed for several classes of nonlinear
Euler sums in terms of zeta values and linear sums.
Lemma 3.1 For integers n, p ∈ N, then
Z1
xn−1 Lip (x)dx =
Z1
xn−1 ln xLip (x)dx =
p−1
X
(−1)i−1
ni
i=1
0
p−1 X
p−i
X
ζ (p + 1 − i) +
(−1)i+j−1
i=1 j=1
0
+ (−1)p p
(−1)p−1
Hn ,
np
(3.1)
ζ (p + 2 − i − j)
ni+j
ζn (2) − ζ (2)
Hn
+ (−1)p
,
np+1
np
(3.2)
where Lip (x) is polylogarithm function defined by
Lip (x) =
∞
X
xn
n=1
np
, ℜ(p) > 1, |x| ≤ 1,
with Li1 (x) = − ln(1 − x), x ∈ [−1, 1).
Proof.
Zx
0
Using integration by parts, we deduce that, for x ∈ [−1, 1),
tn−1 Lip (t) dt =
p−1
X
i=1
(−1)p
(−1)p
xn
n
ln
(1
−
x)
(x
−
1)
−
(−1)i−1 i Lip+1−i (x) +
n
np
np
16
n
X
xk
k=1
k
!
.
(3.3)
Letting x approach 1, the result is (3.1). Multiplying (3.3) by
interval (0, 1), we have the following recurrence relation
Z1
x
n−1
m
ln xLip (x)dx =m
Z1
p−1
X
(−1)i
i=1
0
+ m!
ni
lnm−1 x
and integrating over the
x
ζn (m + 1)
np


m−1
H

X
ζ
(j
+
1)
−
ζ
(j
+
1)
n
n
−
−
ζ
(m
+
1)
.
 nm

nm−j
xn−1 lnm−1 xLip+1−i (x)dx + m!(−1)m+p−1
0
(−1)m+p−1
np
j=1
(3.4)
Taking m = 1 in (3.4), we obtain (3.2). This completes the proof of Lemma 3.1.
Theorem 3.2 For integers p, m ∈ N, then we have the relation
⋆
ζ (p + 1, {1}m ) =
p−1
X
(−1)i−1 ζ (p + 1 − i) ζ m + 1, {1}i−1
i=1
+ (−1)p+1
p−1 X
ζ m + 1, {1}i−1 , 2, {1}p−1−i
i=1
Proof.
+ (−1)p+1 (m + 1) ζ m + 2, {1}p−1 .
(3.5)
First, we consider the integral
Z1
0
∞
X 1
Lip (x) lnm (1 − x)
dx =
x
np
n=1
Z1
xn−1 lnm (1 − x)dx
0
∞
X
ζ ⋆ ({1}m )
= m!
np+1
n=1
= m!ζ ⋆ (p + 1, {1}m ) .
(3.6)
On the other hand, by using (3.1), we conclude that
Z1
0
1
Z
∞
X
Lip (x) lnm (1 − x)
s (n, m)
m
dx = (−1) m!
xn−1 Lip (x)dx
x
n!
n=m
0
m
= (−1) m!
p−1
X
(−1)
i−1
i=1
m+p−1
+ (−1)
m!
∞
X
s (n, m)
ζ (p + 1 − i)
n!ni
n=m
∞
X
Hn s (n, m)
n=1
np n!
.
(3.7)
Combining (3.6) and (3.7) and using (2.29), we arrive at
⋆
ζ (p + 1, {1}m ) =
p−2
X
(−1)i ζ (p − i) ζ (m + 1, {1}i ) + (−1)p−1
∞
X
Hn s (n, m)
n=1
i=0
np n!
.
(3.8)
The formula (3.8) also immediately follows from (2.47). Substituting (2.30) into (3.8) yields the
desired result.
17
Theorem 3.3 For integers p, m ∈ N, then we have the relation
m
X
m−1
X
ζ (m − i + 1) ζ ⋆ (p + 1, {1}i )
ζ ⋆ p + 1, {1}i−1 , 2, {1}m−i + ζ ⋆ (p + 2, {1}m ) −
i=1
p−i
p−1 X
X
=
i=1 j=1
i=0
(−1)i+j ζ (p + 2 − i − j) ζ m + 1, {1}i+j−1 + (−1)p+1 p (m + 1) ζ m + 2, {1}p
p+1
+ (−1)
p−1 X
ζ m + 1, {1}i , 2, {1}p−1−i − (−1)p+1 ζ m + 1, {1}p−1 , 2 .
p
(3.9)
i=0
Proof. Similarly to the proof of Theorem 3.2, first, by using (2.6), we obtain the following
identity
Z1
0
∞
X 1
Lip (x) ln xlnm (1 − x)
dx =
x
np
n=1
=m!(−1)
Z1
xn−1 ln xlnm (1 − x)dx
0
m−1
+ m!(−1)
m
X
ζ ⋆ p + 1, {1}i−1 , 2, {1}m−i
i=1
m−1 ⋆
ζ (p + 2, {1}m )
− m!(−1)m−1
m−1
X
ζ (m − i + 1) ζ ⋆ (p + 1, {1}i ).
(3.10)
i=0
Then by (2.26) and (3.2), we can deduce the result
Z1
Lip (x) ln xlnm (1 − x)
dx
x
0
1
m
=(−1) m!
Z
∞
X
s (n, m)
n!
n=1
=(−1)m m!
xn−1 lnxLip (x)dx
0
p−i
p−1 X
X
i=1 j=1
+ (−1)m+p pm!
(−1)i+j−1 ζ (p + 2 − i − j) ζ m + 1, {1}i+j−1
∞
X
Hn s (n, m)
n=1
n!np+1
+ (−1)m+p m!
∞
X
s (n, m) (ζn (2) − ζ (2))
n=1
n!np+1
Combining (2.30), (2.36), (3.10) and (3.11), we can obtain (3.9).
Putting p = 1 in (3.9), we can give the following Corollary.
.
(3.11)
Corollary 3.4 For positive m, we have
m
X
ζ ⋆ 2, {1}i−1 , 2, {1}m−i + ζ ⋆ (3, {1}m )
i=1
m−1
X
=
ζ (m − i + 1) ζ ⋆ (2, {1}i ) + (m + 1) ζ (m + 2, 1) .
i=0
18
(3.12)
Setting p = 1, 2 in (3.5) we obtain
ζ ⋆ (2, {1}m ) = (m + 1) ζ (m + 2) ,
(3.13)
ζ ⋆ (3, {1}m ) = ζ (2) ζ (m + 1) − (m + 1) ζ (m + 2, 1) − ζ (m + 1, 2) .
(3.14)
At the end of this section we give some closed form for several classes of nonlinear Euler sums.
In [9], J.M. Borwein, R. Girgensohn proved that all ζ (q, p, r) with r + p + q is even or less than
or equal to 10 or r + p + q = 12 were reducible to zeta values and linear sums. From [9], we have
1
3
ζ (5, 1) = ζ (6) − ζ 2 (3) ,
4
2
61
1
ζ (6, 1, 1) = ζ (8) − 3ζ (3) ζ (5) + ζ (2) ζ 2 (3) ,
24
2
499
ζ (5, 1, 1, 1) =
ζ (8) − 4ζ (3) ζ (5) + ζ (2) ζ 2 (3) ,
192
73
9
3
ζ (5, 1, 2) = − ζ (8) + ζ (3) ζ (5) − ζ (2) ζ 2 (3) − S2,6 ,
72
2
2
541
7
7
ζ (5, 2, 1) = −
ζ (8) + ζ (3) ζ (5) − ζ (2) ζ 2 (3) + S2,6 .
144
2
2
Putting p = 3, m = 3 in (3.5), we can deduce that
ζ ⋆ (5, 1, 1, 1) = −
385
3
ζ (8) + 5ζ (3) ζ (5) − ζ (2) ζ 2 (3) − S2,6 .
192
4
Taking p = 3, m = 4 in (3.5), we get
ζ ⋆ (4, {1}4 ) =ζ (3) ζ (5) − ζ (2) ζ (5, 1) + 5ζ (6, 1, 1) + ζ (5, 2, 1) + ζ (5, 1, 2)
107
1
3
=
ζ (8) − 6ζ (3) ζ (5) + ζ (2) ζ 2 (3) + S2,6 ,
16
2
4
(3.15)
where S2,6 = ζ ⋆ (6, 2). On the other hand, we note that
ζ ⋆ (4, {1}4 ) =
∞
∞
X
ζn⋆ ({1}4 )
1 X Y4 (n)
=
n4
24
n4
n=1
n=1
∞
1 X Hn4 + 8Hn ζn (3) + 6Hn2 ζn (2) + 3ζn2 (2) + 6ζn (4)
.
=
24
n4
(3.16)
n=1
In [35], we proved the results
∞
X
Hn ζn (3)
n=1
∞
X
n=1
n4
=−
511
25
ζ (8) + 7ζ (3) ζ (5) + ζ (2) ζ 2 (3) − S2,6 ,
144
4
ζn2 (2)
457
= 11S2,6 +
ζ (8) + 6ζ (2) ζ 2 (3) − 40ζ (3) ζ (5) ,
4
n
18
(3.17)
(3.18)
and
∞
X
6H 2 ζn (2) − H 4
n
n=1
n
n4
= −17S2,6 − 10ζ (2) ζ 2 (3) + 104ζ (3) ζ (5) −
19
5911
ζ (8) ,
72
(3.19)
Substituting (3.17) and (3.18) into (3.17) respectively, we have
∞
X
H 4 + 6H 2 ζn (2)
n
n
n4
n=1
=
956
ζ (8) − 80ζ (3) ζ (5) − 14ζ (2) ζ 2 (3) + 35S2,6 .
9
(3.20)
Hence, combining (3.19) and (3.10), we obtain the identities
∞
X
H4
n=1
∞
X
n=1
n
n4
=
13559
ζ (8) − 92ζ (3) ζ (5) − 2ζ (2) ζ 2 (3) + 26S2,6 ,
144
Hn2 ζn (2)
193
3
=
ζ (8) + 2ζ (3) ζ (5) − 2ζ (2) ζ 2 (3) + S2,6 .
4
n
96
2
(3.21)
(3.22)
In fact, using the method of this paper, it is possible to evaluate other Euler sums involving
harmonic numbers. For example, we have used our method to obtain the following explicit
evaluations,
∞
X
H 2 ζn (2)
n
n=1
∞
X
n2
=
41
ζ (6) + 2ζ 2 (3) ,
12
19
Hn2 ζn (2)
= −7ζ (7) + ζ (3) ζ (4) − 2ζ (2) ζ (5) ,
3
n
2
n=1
∞
X
H 3 ζn (2)
n
n=1
∞
X
n=1
∞
X
n2
=
83
27
5
ζ(7) + ζ(3)ζ(4) − ζ(2)ζ(5),
16
2
2
19
Hn2 ζn (2)
= −7ζ(7) + ζ(3)ζ(4) − 2ζ(2)ζ(5),
n3
2
Hn2 ζn (3)
9
329
ζ(7) − 6ζ(3)ζ(4) − ζ(2)ζ(5),
=
2
n
16
2
n=1
∞
X
Hn ζ 2 (2)
n
n=1
∞
X
n2
=−
217
13
ζ(7) + 5ζ(3)ζ(4) + ζ(2)ζ(5),
16
2
Hn4 ζn (2)
1289
ζ (8) − 11ζ (3) ζ (5) + 5S2,6 ,
=
2
n
96
n=1
∞
X
H 2 ζn (3)
n
n3
n=1
∞
X
=−
443
9
3
23
ζ (8) + ζ (3) ζ (5) + ζ (2) ζ 2 (3) − S2,6 ,
288
2
2
4
Hn5
393
15
235
60499
ζ (8) −
ζ (3) ζ (5) − ζ (2) ζ 2 (3) +
S2,6 ,
=
3
n
288
2
2
4
n=1
∞
X
H 3 ζn (2)
n
n=1
∞
X
n=1
∞
X
n3
=−
2159
93
3
53
ζ (8) + ζ (3) ζ (5) + ζ (2) ζ 2 (3) − S2,6 ,
48
2
2
4
6313
43
1
17
Hn ζn2 (2)
=−
ζ (8) + ζ (3) ζ (5) + ζ (2) ζ 2 (3) − S2,6 ,
3
n
288
2
2
4
Hn2 ζn (2)
295
1481
ζ (9) − 3ζ 3 (3) − 5ζ (2) ζ (7) +
ζ (3) ζ (6) + 18ζ (4) ζ (5) .
=−
5
n
72
24
n=1
20
In [23], Masanobu Kaneko and Yasuo Ohno proved that
(−1)k ζ ⋆ (k + 1, {1}n ) − (−1)n ζ ⋆ (n + 1, {1}k )
= kζ k + 2, {1}n−1 − nζ n + 2, {1}k−1
k−2
X
k
(−1)j ζ (k − j) ζ n + 1, {1}j
+ (−1)
j=0
− (−1)n
n−2
X
j=0
(−1)j ζ (n − j) ζ k + 1, {1}j .
(3.23)
Combining (2.30), (3.5), (3.8) and (3.23), we obtain the following results
∞
X
Hn s (n, p)
n=1
p−1
X
i=1
n!nm
−
∞
X
Hn s (n, m)
n=1
n!np
= pζ p + 2, {1}m−1 − mζ m + 2, {1}p−1 ,
m−1
X
ζ p + 1, {1}i−1 , 2, {1}m−1−i
ζ m + 1, {1}i−1 , 2, {1}p−1−i −
(3.24)
i=1
= ζ p + 2, {1}m−1 − ζ m + 2, {1}p−1 .
(3.25)
Evaluation of the alternating multiple zeta star values ζ ⋆ (1̄, {1}m, 1̄)
and ζ ⋆ (2, {1}m, 1̄)
4
In this section, we will show that the alternating multiple zeta star values ζ ⋆ (1̄, {1}m , 1̄) and
ζ ⋆ (2, {1}m , 1̄) satisfy certain recurrence relations that allow us to write them in terms of zeta
values, polylogarithms and ln 2. In (1.1), (1.2), (1.11), (1.12), we put a bar on top of sj (j =
1, 2, · · · , m) if there is a sign (−1)kj appearing in the denominator on the right. For example
X
⋆
ζ (s̄1 , s2 , · · · , sk−1 , s̄k ) :=
n1 ≥n2 ≥···≥nk ≥1
ζn⋆ (s̄1 , s2 , · · · sk−1 , s̄k ) :=
X
(−1)n1 +nk
ns11 ns22 · · · nskk
(−1)n1 +nk
· · · nskk
ns1 ns2
n≥n1 ≥n2 ≥···≥nk ≥1 1 2
Now, we prove two lemmas which will be useful in proving Theorem 4.3 and 4.4.
Lemma 4.1 For integer m > 0, the following identity holds,
Z1
0
lnm (1 + x) ln (1 − x)
1
dx =
lnm+2 2 − ζ (2) lnm 2
1+x
m+1
 k 
X

1 
k
m


X m
l!
(ln 2)m−l Lil+2
k+1
l
.
(−1)
−
2
k



 l=1
m−k
k=1
−k!(ln 2)
ζ (k + 2)
21
(4.1)
Proof. To prove (3.1), noting that the integral in (4.1), by change the variable x 7→ 2t − 1,
which can be rewritten as
Z1
lnm (1 + x) ln (1 − x) 1+x=2t
dx =
1+x
0
=
Z1
lnm (2t) ln (2 − 2t)
dt
t
1/2
m X
m
k
k=0
= (ln 2)m+2
m−k+1
(ln 2)
m X
m
k
m k
X
m (−1)
k
k=0
Zx
0
m
X
(ln t)k
dt +
t
k=0
1/2
k=0
Noting that
Z1
k+1
t
n−1
k
(−1)
−
k+1
=
m X
k=0
m
k
m
k
Z1
m−k
(ln 2)
(ln t)k ln (1 − t)
dt
t
1/2


(ln 2)m−k 
∞
X
n=1
1
n
Z1
1/2


tn−1 (ln t)k dt.
(4.2)
1
and using the following elementary integral identity
m+1
m
l
X
m (−1)
(ln x)m−l xn , x ∈ (0, 1),
l!
(ln t) dt =
l
nl+1
m
l=0
we can deduce the following results
Z1
tn−1 (ln t)m dt = m!
(−1)m
,
nm+1
(4.3)
0
Z1/2
m
X
1
m
m
m
n−1
t
(ln t) dt = (−1)
l!
(ln 2)m−l n l+1 ,
l
2 n
0
Z1
1/2
(4.4)
l=0
m
t
n−1
X
(−1)m
l!
(ln t) dt = m! m+1 + (−1)m−1
n
m
l=0
m
l
(ln 2)m−l
1
2n nl+1
.
(4.5)
Substituting (4.5) into (4.2) respectively, we can obtain (2.9). This completes the proof of
Lemma 4.1.
Lemma 4.2 If m ≥ 1 is a integer and z ∈ [0, 1] , then have
Zz
0
lnm (1 + x)
1
1
m+1
dx =
ln
(1 + z) + m! ζ (m + 1) − Lim+1
x
m+1
1+z
− m!
m
X
lnm−j+1 (1 + z)
j=1
Proof.
(m − j + 1)!
Lij
1
1+z
.
(4.6)
We note that the integral in (4.6), which can be rewritten as
Zz
0
lnm (1 + x) t=1+x
dx =
x
1+z
Z
lnm t u=t−1
dt = (−1)m+1
t−1
−1
(1+z)
Z
1
1
22
lnm u
du
u − u2
=(−1)m+1
=

(1+z)−1

 Z
lnm u


u
du +
1
−1
(1+z)
Z
1
1
lnm+1 (1 + z) + (−1)m+1
m+1



m
ln u
du
1−u 

−1
(1+z)
Z
lnm u
du.
1−u
(4.7)
1
Applying (2.33), we arrive at the conclusion that
−1
(1+z)
Z
1
∞
X
lnm u
du =
1−u
n=1
−1
(1+z)
Z
un−1 lnm udu =(−1)m+1 m! ζ (m + 1) − Lim+1 (1 + z)−1
1
+ m!(−1)
m
m
X
lnm−j+1 (1 + z)
j=1
(m − j + 1)!
Lij (1 + z)−1 .
(4.8)
Substituting (4.8) into (4.7) yields the desired result.
Theorem 4.3 For integer m > 1, we have the recurrence relation
(−1)m−1
ζ ⋆ 1̄, {1}m−1 , 1̄ =
ζ (2) lnm−1 2
(m − 1)!
m−1
(−1)m−1 X
m
i+1
(−1) i!
(ln 2)m−i ζ ⋆ 1̄, {1}i−1 , 1̄ − ζ ⋆ (1̄, {1}i )
−
i
m!
i=1
 k 
X

1 
k
m−1


(−1)m−1 X m − 1
l!
(ln 2)m−1−l Lil+2
k+1
l
+
.
(−1)
2
k


(m − 1)!

 l=1
m−1−k
k=1
−k!(ln 2)
ζ (k + 2)
(4.9)
Proof.
Multiplying (2.2) by (−1)n−1 and summing with respect to n, the result is
Zx
0
∞
X
lnm (1 − t)
(−1)n−1
dt =
1+t
n=1
= lnm (1 − x)
∞
X
xn
n=1
−
m−1
X
i=1
(−1)i−1 i!
Zx
tn−1 lnm (1 − t)dt
0
∞
X
−1
(−1)n−1
(−1)n−1 + m!(−1)m
n
n
n=1
m
i
lnm−i (1 − x)
∞
X
n=1
n−1
(−1)
n
X
1≤km ≤···≤k1 ≤n
X
1≤ki ≤···≤k1 ≤n
xk i − 1
.
k1 · · · ki
On the other hand, by integration by parts, we obtain the formula
 x

Z lnm (1 − t)

dt − lnm (1 − x) ln (1 + x)
lim
x→−1 

1+t
0
23
xk m
k1 · · · km
(4.10)


= lim
x→−1 
=m
Z−1
m
Zx
ln
m−1
0

(1 − t) ln (1 + t) 
dt

1−t
lnm−1 (1 − t) ln (1 + t)
dt
1−t
0
= −m
Z1
lnm−1 (1 + t) ln (1 − t)
dt,
1+t
0
Hence, letting x approach −1 in (4.10) and combining (4.1), we can deduce (4.9).
Noting that, taking x → 1 in (4.7), we have the result
Z1
lnm (1 − t)
dt = m!(−1)m−1 ζ ⋆ (1̄, {1}m ) = (−1)m m!Lim+1
1+t
0
1
.
2
Therefore, we conclude that
1
.
ζ (1̄, {1}m ) = −Lim+1
2
⋆
(4.11)
Theorem 4.4 For integer m > 1, we have the recurrence relation
m+1
1
m+1 m+2
m+1
⋆
ζ 2, {1}m−1 , 1̄ =
(−1)
ln
2 + (m + 1) (−1)
ζ (m + 2) − Lim+2
(m + 2)!
2
m+1
m+1
m+2−j
X (ln 2)
1
3 (−1)
Lij
(ln 2)m ζ (2)
− (m + 1) (−1)m+1
−
(m + 2 − j)!
2
2
m!
j=1
−
Proof.
m+1 m−1
X
(−1)
m!
i=1
(−1)i−1 i!
m
i
(ln 2)m−i ζ ⋆ 2, {1}i−1 , 1̄ − ζ ⋆ (2, {1}i ) .
(4.12)
Similarly to the proof of Theorem 4.3, we consider the integral
Z−1
0
−1
Z
∞
X
lnm+1 (1 − t)
1
dt = −
tn−1 lnm (1 − t) dt.
t
n
n=1
(4.13)
0
Putting x → −1 in (2.2), we deduce that
Z −1
(−1)m ⋆
1
ζn ({1}m , 1̄)
tn−1 lnm (1 − t)dt = lnm 2 ((−1)n − 1) + m!
n
n
0
m−1
1 X
m
i−1
−
(−1) i!
lnm−i 2 ζn⋆ {1}i−1 , 1̄ − ζn⋆ ({1}i ) . (4.14)
i
n
i=1
Setting x → 1 in (4.6) and combining (4.13) with (4.14), we can obtain (4.12).
By considering the case m = 2 in (4.9) and (4.12) we get
1
1
1
ζ ⋆ (1̄, 1, 1̄) = ζ (3) + ζ (2) ln 2 − ln3 2,
8
2
6
24
1
ζ ⋆ (2, 1, 1̄) = ln4 2 + 3Li4
8
7
1
3
− 3ζ (4) − ζ (2) ln2 2 + ζ (3) ln 2.
2
2
8
Therefore, from Theorem 3.2 and Theorem 3.3, we know that the alternating multiple zeta star
values ζ ⋆ (1̄, {1}m , 1̄) and ζ ⋆ (2, {1}m , 1̄) can be expressed as a rational linear combination of zeta
values, polylogarithms and ln 2.
5
Some identities for H (a, b; m, p) and H ⋆ (a, b; m, p)
In this section, we will prove the sums H (a, b; m, p) and H ⋆ (a, b; m, p) can be expressed in
terms of the Riemann zeta values, where a, b ∈ N0 , the sums H (a, b; m, p) and H ⋆ (a, b; m, p)
are defined by
X
H (a, b; m, p) :=
ζ ({p}a , p + 1, {p}b ), m, p ∈ N,
a+b=m−1
X
H ⋆ (a, b; m, p) :=
ζ ⋆ ({p}a , p + 1, {p}b ), m, p ∈ N.
a+b=m−1
We define the parametric Hurwitz zeta function and parametric Hurwitz zeta star function by
ζ (s1 , s2 , · · · , sm ; a + 1) :=
X
1
,
(k1 + a) (k1 + a)s2 · · · (km + a)sm
X
1
,
(k1 + a)s1 (k1 + a)s2 · · · (km + a)sm
s1
1≤k1 <···<km
ζ ⋆ (s1 , s2 , · · · , sm ; a + 1) :=
1≤k1 ≤···≤km
where ℜ(s1 ) > 1, si ≥ 1, m ∈ N, a 6= −1, −2, · · · . When m = 1, then the Hurwitz zeta function
( or parametric Hurwitz zeta star function ) reduces to the classical Hurwitz zeta function, which
is defined by
∞
X
1
ζ (s, a + 1) =
s.
(n
+
a)
n=1
Theorem 5.1 Define two sequence Am (n) and Bm (n) by
m−1
X
Am (n) = (m − 1)!
i=0
Bm (n) =
n
X
xk 1
k1 =1
k1
X
n
Ai (n) X m−i
xk , A0 (n) = 1, (xk ∈ C, k = 1, 2, · · · , n) ,
i!
k=1
km−1
xk 2 · · ·
k2 =1
X
xkm , B0 (n) = 1, (xk ∈ C, k = 1, 2, · · · , n) .
km =1
Then
Am (n) = m!Bm (n).
(5.1)
Proof.
By induction, we can tell that Am (n) and Bm (n) are polynomials of degree m with n
n
X
variables x1 , x2 , · · · xn , moreover, Bm (n) has coefficient 1 at xs11 xs22 · · · xsnn , where
sk = m,
k=1
and sk ≥ 0. Suppose cs1 ,··· ,sn is the coefficient of Am (n) at xs11 xs22 · · · xsnn , then in order to get
the conclusion, all we need to do is prove that cs1 ,··· ,sn = m! by induction. For convenience, we
25
can suppose ct1 ,··· ,tn = 0 if there is a ti < 0, otherwise ct1 ,··· ,tn is the coefficient of At1 +···+tn (n)
at xt11 xt22 · · · xtnn . For m = 0, the conclusion holds. If the conclusion holds for any i ≤ m − 1. For
n
m−1
XX
cs1 ,··· ,sk −(m−i),··· ,sn
, since for
m, by the recursion formula of Am (n), cs1 ,··· ,sn = (m − 1)!
i!
i=0 k=1
n
Ai (n) X m−i
s −(m−i)
0 ≤ i ≤ m−1,
xk at xs11 xs22 · · · xsnn = xs11 · · · xkk
, where 1 ≤ k ≤ n,
· · · xsnn ·xm−i
k
i!
k=1
cs ,··· ,sk −(m−i),··· ,sn
. So
provides coefficient 1
i!
n m−1
X
X cs1 ,··· ,s −(m−i),··· ,sn
k
cs1 ,··· ,sn = (m − 1)!
i!
k=1 i=0
m−1
n
X cs1 ,··· ,s −(m−i ),··· ,sn
X
k
k
= (m − 1)!
ik !
k=1 ik =m−sk
n
m−1
X
X
= (m − 1)!
1
k=1 ik =m−sk
n
X
((m − 1) − (m − sk ) + 1)
= (m − 1)!
k=1
n
X
= (m − 1)! sk
k=1
= m!
since cs1 ,··· ,sk −(m−ik ),··· ,sn = ik ! by the inductive assumption. So the conclusion holds by induction. So far we have completed the proof of Theorem 5.1.
n
X
xm
Let Xn (m) =
i in Theorem 5.1, we can deduce that
i=1
Xn2 (1) + Xn (2)
,
2
X 3 (1) + 3Xn (1) Xn (2) + 2Xn (3)
,
B3 (n) = n
3!
X 4 (1) + 8Xn (1) Xn (3) + 3Xn2 (2) + 6Xn2 (1) Xn (2) + 6Xn (4)
,
B4 (n) = n
4!
5
1 Xn (1) + 10Xn3 (1) Xn (2) + 20Xn2 (1) Xn (3) + 15Xn (1) Xn2 (2)
B5 (n) =
,
5! +30Xn (1) Xn (4) + 20Xn (2) Xn (3) + 24Xn (5)
B0 (n) = 1, B1 (n) = Xn (1) , B2 (n) =
m−1
1 X
Bi (n) Xn (m − i).
Bm (n) =
m
(5.2)
i=0
Furthermore, we are able to obtain relations involving multiple zeta-star values and zeta values.
1
For example, taking xi = p , p > 1 (i = 1, · · · , n) and substituting it into Theorem 2.3, letting
i
n → ∞, we have
m−1
1 X ⋆
⋆
ζ ({p}m ) =
ζ ({p}i ) ζ (pm − pi).
(5.3)
m
i=0
26
where {p}m denotes the m-tuple {p, ..., p}, ζ ⋆ ({p}0 ) = 1. Hence ,we obtain
31
127
7
ζ (6) , ζ ⋆ ({2}4 ) =
ζ (8) ,
ζ ⋆ ({2}2 ) = ζ (4) , ζ ⋆ ({2}3 ) =
4
16
64
1 2
1 3
ζ ⋆ ({3}2 ) =
ζ (3) + ζ (6) , ζ ⋆ ({3}3 ) =
ζ (3) + 3ζ (3) ζ (6) + 2ζ (9) ,
2
3!
1 4
⋆
2
ζ ({3}4 ) =
ζ (3) + 8ζ (3) ζ (9) + 3ζ (6) + 6ζ 2 (3) ζ (6) + 6ζ (12) .
4!
By a similar argument as in the proof of Theorem 5.1, we can gives the following Theorem.
Theorem 5.2 Define two sequence Ām (n) and B̄m (n) by
m−1
Ām (n) = (m − 1)!(−1)
m−1
X
(−1)
i=0
B̄m (n) =
n
X
xk 1
k1 =1
i Āi (n)
i!
n
X
xm−i
, Ā0 (n) = 1,
k
k=1
km−1 −1
kX
1 −1
xk 2 · · ·
k2 =1
X
xkm , B̄0 (n) = 1.
km =1
Then
Ām (n) = m!B̄m (n).
(5.4)
Proof.
By induction, we can tell that Ām (n) and B̄m (n) are polynomials of degree m with n
n
X
s1 s2
s
n
variables x1 , x2 , · · · xn , moreover, B̄m (n) has coefficient 1 at x1 x2 · · · xn , when
sk = m,
k=1
and 0 ≤ sk ≤ 1, otherwise, the coefficient is 0. Suppose cs1 ,··· ,sn is the coefficient of Ām (n) at
xs11 xs22 · · · xsnn , then in order to get the conclusion, all we need to do is prove that cs1 ,··· ,sn = m!
if 0 ≤ sk ≤ 1, and cs1 ,··· ,sn = 0 if there is a k such that sk ≥ 2 by induction. For convenience, we
can suppose ct1 ,··· ,tn = 0 if there is a ti < 0, otherwise ct1 ,··· ,tn is the coefficient of Āt1 +···+tn (n)
at xt11 xt22 · · · xtnn .
For m = 0, the conclusion holds. If the conclusion holds for any i ≤ m − 1. For m, by the
n
m−1
XX
cs ,··· ,sk −(m−i),··· ,sn
, since
recursion formula of Ām (n), cs1 ,··· ,sn = (m − 1)!(−1)m−1
(−1)i 1
i!
i=1 k=1
n
Āi (n) X m−i
s −(m−i)
for 1 ≤ i ≤ m − 1,
xk
at xs11 xs22 · · · xsnn = xs11 · · · xkk
, where
· · · xsnn · xm−i
k
i!
k=1
cs ,··· ,sk −(m−i),··· ,sn
. So, If there are k1 , k2 such that sk1 , sk2 ≥ 2,
1 ≤ k ≤ n, provides coefficient 1
i!
then cs1 ,··· ,sk −(m−i),··· ,sn = 0 for any 0 ≤ i ≤ m − 1, 0 ≤ k ≤ n, so cs1 ,··· ,sn = 0. If there is only
one k such that sk ≥ 2, say j, then cs1 ,··· ,sk −(m−i),··· ,sn = 0 for any 0 ≤ i ≤ m − 1, 0 ≤ k ≤ n and
27
k 6= j, so
m−1
cs1 ,··· ,sn = (m − 1)!(−1)
= (m − 1)!(−1)m−1
m−1
n
XX
(−1)i
i=1 k=1
n
m−1
X
X
cs1 ,··· ,sk −(m−i),··· ,sn
i!
(−1)ik
k=1ik =m−sk
= (m − 1)!(−1)m−1
m−1
X
(−1)i
i=m−sj
m−1
= (m − 1)!(−1)
cs1 ,··· ,sk −(m−ik ),··· ,sn
i!
cs1 ,··· ,sj −(m−i),··· ,sn
i!
((−1)m−sj cs1 ,··· ,sj−1 ,0,sj+1··· ,sn + (−1)m−sj +1 cs1 ,··· ,sj−1 ,1,sj+1 ,··· ,sn )
= (m − 1)!(−1)m−1 ((−1)m−sj + (−1)m−sj +1 )
=0
If 0 ≤ sk ≤ 1 for each 1 ≤ k ≤ n, then
cs1 ,··· ,sn = (m − 1)!(−1)m−1
n
m−1
XX
(−1)i
i=1 k=1
= (m − 1)!(−1)m−1
m−1
n
X
X
cs1 ,··· ,sk −(m−i),··· ,sn
i!
(−1)ik
k=1ik =m−sk
m−1
= (m − 1)!(−1)
m−1
X
X
(−1)ik
1≤k≤nik =m−1
sk =1
= (m − 1)!(−1)m−1
X
(−1)m−1
1≤k≤n
cs1 ,··· ,sk −(m−ik ),··· ,sn
ik !
cs1 ,··· ,1−(m−ik ),··· ,sn
ik !
cs1 ,··· ,sk−1 ,0,sk+1,··· ,sn
(m − 1)!
sk =1
= (m − 1)!
X
1
1≤k≤n
sk =1
= m!
Since cs1 ,··· ,sk−1,0,sk+1 ,··· ,sn
n
X
= ( sk − 1)! = (m − 1)! by the inductive assumption. So the
k=1
conclusion holds by induction.
n
X
xm
Let Xn (m) =
i in Theorem 5.2, we have
i=1
X 3 (1) − 3Xn (1) Xn (2) + 2Xn (3)
Xn2 (1) − Xn (2)
, B̄3 (n) = n
,
2!
3!
X 4 (1) − 6Xn2 (1) Xn (2) + 8Xn (1) Xn (3) + 3Xn2 (2) − 6Xn (4)
,
B̄4 (n) = n
4!
5
1 Xn (1) − 10Xn3 (1) Xn (2) + 20Xn2 (1) Xn (3) + 15Xn (1) Xn2 (2)
B̄5 (n) =
,
5! −30Xn (1) Xn (4) − 20Xn (2) Xn (3) + 24Xn (5)
B̄1 (n) = Xn (1) , B̄2 (n) =
m−1
(−1)m−1 X
(−1)i B̄i (n) Xn (m − i).
B̄m (n) =
m
i=0
28
(5.5)
Putting xi =
1
, p > 1 (i = 1, · · · , n) and n → ∞ in Theorem 5.2, we obtain
ip
m−1
(−1)m−1 X
(−1)i ζ ({p}i ) ζ (pm − pi).
m
ζ ({p}m ) =
(5.6)
i=0
Taking xk =
1
(p > 1) in (5.3) and (5.5), letting n → ∞, we can obtain the results
(k + a)p
m−1
(−1)m−1 X
ζ ({p}m ; a + 1) =
(−1)i ζ ({p}i ; a + 1) ζ (pm − pi, a + 1),
m
(5.7)
i=0
ζ ⋆ ({p}m ; a + 1) =
m−1
1 X ⋆
ζ ({p}i ; a + 1) ζ (pm − pi, a + 1).
m
(5.8)
i=0
By using (5.7) and (5.8), we can obtain the following Theorems.
Theorem 5.3 For integers p > 1, m ∈ N and a, b ∈ N0 , a 6= −1, −2, · · · , we have the recurrence
formulas
H (a, b; m, p) =(−1)m−1 ζ (pm + 1) +
m−1
(−1)m−1 X
(−1)i ζ (pm − pi) H (a, b; i, p)
m
i=1
+
m−1 m−1
X
(−1)
m
(−1)i (m − i) ζ (pm − pi + 1) ζ ({p}i ),
(5.9)
i=1
H ⋆ (a, b; m, p) =ζ (pm + 1) +
m−1
1 X
ζ (pm − pi) H ⋆ (a, b; i, p)
m
i=1
+
Proof.
1
m
m−1
X
(m − i) ζ (pm − pi + 1) ζ ⋆ ({p}i ).
(5.10)
i=1
In (5.6) and (5.7), taking the derivative with respect to a gives
H (a, b; m, p) = −
1 ∂
(ζ ({p}m ; a + 1))
p ∂a
=−
m−1 m−1
X
(−1)
pm
i=0
(−1)i
a=0
∂
(ζ ({p}i ; a + 1) ζ (pm − pi, a + 1))
∂a
1 ∂
(ζ ⋆ ({p}m ; a + 1))
H ⋆ (a, b; m, p) = −
p ∂a
=−
1
pm
m−1
X
i=0
a=0
∂
(ζ ⋆ ({p}i ; a + 1) ζ (pm − pi, a + 1))
∂a
By simple caluculation, we obtain the result.
From (5.3), (5.6), (5.9) and (5.10), we can give the following corollary.
29
,
a=0
.
a=0
Corollary 5.4 For integers p ∈ N \ {1}, m ∈ N and a, b ∈ N0 , then the sums
X
ζ ({p}a , p + 1, {p}b )
a+b=m−1
and
X
ζ ⋆ ({p}a , p + 1, {p}b )
a+b=m−1
can be expressed as a rational linear combination of products of Riemann zeta values.
Acknowledgments. The authors would like to thank the anonymous referee for his/her helpful
comments, which improve the presentation of the paper.
References
[1] David H. Bailey, Jonathan M. Borwein and Roland Girgensohn. Experimental evaluation of Euler
sums. Experimental Mathematics., 1994, 3(1): 17-30.
[2] J. Blumlein, S. Kurth. Harmonic sums and Mellin transforms up to two loop order. Phys. Rev. D.,
1999, 60: 14-18.
[3] David Borwein, Jonathan M. Borwein and Roland Girgensohn. Explicit evaluation of Euler sums.
Proc. Edinburgh Math., 1995, 38: 277-294.
[4] J. Borwein, P. Borwein, R. Girgensohn, S. Parnes. Making sense of experimental mathematics.
Mathematical Intelligencer., 1996, 18(4): 12-18.
[5] Jonathan M. Borwein, David M. Bradley, David J. Broadhurst. Evaluations of k-fold Euler/Zagier
sums: a compendium of results for arbitrary k. Electronic Journal of Combinatorics., 1997, 4(2):
1-21.
[6] J. M. Borwein, D.M. Bradley. Multiple polylogarithms: a brief survey. Contemp. Math., 2001, 291:
71-92.
[7] Jonathan M. Borwein, David M. Bradley, David J. Broadhurst, Petr. Lisonk. Special values of
multiple polylogarithms. Trans. Amer. Math. Soc., 2001, 353(3): 907-941.
[8] J. M. Borwein, I. J. Zucker, J. Boersma. The evaluation of character Euler double sums. Ramanujan
J., 2008, 15 (3): 377-405.
[9] J. M. Borwein, R. Girgensohn. Evaluation of triple Euler sums, Electron. J. Combin., 1996: 2-7.
[10] Kwang-Wu Chen, Chan-Liang Chung, Minking Eie. Sum formulas and duality theorems of multiple
zeta values. J. Number Theory., 2016, 158: 33-53.
[11] Chan-Liang Chung, Minking Eie, Tung-Yang Lee. Another expression of the restricted sum formula
of multiple zeta values. J. Number Theory., 2016, 166: 452-472.
[12] Ayhan Dil, Khristo N. Boyadzhiev. Euler sums of hyperharmonic numbers. J. Number Theory., 2015,
147: 490-498.
[13] V.G. Drinfeld. On quasitriangular quasi-Hopf algebras and on a group that is closely connected with
Gal(Q/Q). Leningrad Math. J., 1991, 2(4): 829-860.
[14] M. Eie, W.C. Liaw, Y.L. Ong. A restricted sum formula among multiple zeta values. J. Number
Theory., 2009, 129: 908-921.
[15] Minking Eie, Tung-Yang Lee. Identities among restricted sums of multiple zeta values. J. Number
Theory., 2016, 164: 208-222.
30
[16] L. Euler. Meditationes circa singulare serierum genus. Novi Comm. Acad. Sci. Petropol. 177, 520:
140-186. Reprinted in Opera Omnia, ser. I, vol. 15. B. G. Teubner, Berlin, 1927: 217-267.
[17] Philippe Flajolet, Bruno Salvy. Euler sums and contour integral representations. Experimental Mathematics., 1998, 7(1): 15–35.
[18] Pedro Freitas. Integrals of polylogarithmic functions, recurrence relations, and associated Euler sums.
Mathematics of Computation,. 2005, 74(251): 1425-1440.
[19] M. E. Hoffman. Multiple harmonic series. Pacific J. Math., 1992, (152): 275-290.
[20] M. Igarashi. On generalizations of the sum formula for multiple zeta values. arXiv:1110.4875
[21] Comtet L. Advanced combinatorics, Boston: D Reidel Publishing Company, 1974.
[22] Zhonghua Li. Another proof of Zagier’s evaluation formula of the multiple zeta values ζ(2.., 2, 3, 2.., 2).
Mathematical Research Letters,. 2012, 20(5).
[23] Masanobu Kaneko, Yasuo Ohno. On a kind of duality of multiple zeta-star values. International
Journal of Number Theory., 2010, 8(8): 1927-1932.
[24] C. Markett. Triple Sums and the Riemann Zeta Function. J. Number Theory, 1994, 48(2): 113-132.
[25] I. Mezö, A. Dil. Hyperharmonic series involving Hurwitz zeta function. Journal of Number Theory,
2010, 130(2): 360–369.
[26] I. Mezö. Nonlinear Euler sums. Pacific J. Math., 2014, 272: 201-226.
[27] Y. Ohno. A generalization of the duality and sum formulas on the multiple zeta values. J. Number
Theory., 1999, 74(1): 39-43.
[28] Kh. Hessami Pilehrood, T. Hessami Pilehrood, R. Tauraso. New properties of multiple harmonic
sums modulo p and p-analogues of Leshchiner’s series. Trans. Amer. Math. Soc., 2013, 366(6):
3131-3159.
[29] T.M. Rassias, H.M. Srivastava. Some classes of infinite series associated with the Riemann zeta
function and polygamma functions and generalized harmonic numbers. Appl. Math. Comput., 2002,
131: 593–605.
[30] A. Sofo. Integral forms of sums associated with harmonic numbers. Applied Mathematics and Computation., 2009, 207(2): 365-372.
[31] A. Sofo. Quadratic alternating harmonic number sums. J. Number Theory., 2015, 154: 144-159.
[32] A. Sofo, H.M. Srivastava. Identities for the harmonic numbers and binomial coefficients. Ramanujan J., 2011, 25: 93-113.
[33] A. Sofo. Harmonic sums and integral representations. J. Appl. Anal., 2010, 16: 265-277.
[34] Ce Xu, Jinfa Cheng. Some results on Euler sums. Functions et Approximatio., 2016, 54(1): 25-37.
[35] Ce Xu, Yuhuan Yan, Zhijuan Shi. Euler sums and integrals of polylogarithm functions. J. Number
Theory., 2016, 165: 84-108.
[36] Chika Yamazaki. On the duality for multiple zeta-star values of height one. Kyushu Journal of
Mathematics., 2009, 64(1): 145-152.
[37] D. Zagier.Values of zeta functions and their applications. First European Congress of Mathematics,
Volume II, Birkhauser, Boston., 1994, (120): 497-512.
[38] D. Zagier. Evaluation of the multiple zeta values ζ(2, ..., 2, 3, 2, ..., 2). Annals of Mathematics., 2012,
2(2): 977-1000.
[39] J. Zhao. On a conjecture of Borwein, Bradley and Broadhurst. J. reine angew. Math., 2010, (639):
223-233.
31