MULTIPLE ZETA VALUES, POLY-BERNOULLI NUMBERS, AND RELATED ZETA...
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T. Arakawa and M. Kaneko Nagoya Math. J. Vol. 153 (1999), 189–209
MULTIPLE ZETA VALUES,
POLY-BERNOULLI NUMBERS,
AND RELATED ZETA FUNCTIONS
TSUNEO ARAKAWA and MASANOBU KANEKO
Abstract. We study the function
ζ(k1, . . . , kn−1; s) = ∑
0
190 T. ARAKAWA AND M. KANEKO
probably because “nice properties like the functional equation of the Rie-
mann zeta function do not seem to extend” to ζ(s1, s2, . . . , sn) for n ≥ 2.
In this paper, we study the single variable function
ζ(k1, . . . , kn−1; s) = ∑
0
MULTIPLE ZETA VALUES 191
ζ(k1, k2, . . . , kn−1; s) := ∑
0
192 T. ARAKAWA AND M. KANEKO
Proof. (i) is straightforward, and (ii) is derived from (i) and Li1(z) =
− log(1 − z) by integrating repeatedly.
Proposition 2. (i) For Re(s) > 1, we have
ζ(k1, k2, . . . , kn−1; s) = 1
Γ(s)
∫ ∞
0
ts−1
et − 1 · Lik1,k2,...,kn−1(e
−t)dt.(1)
(ii) For Re(s) > 0 and integers n ≥ 2, j ≥ 0, we have
∫ ∞
0 ts+j−1Lik1,k2,...,kn−1(e
−t)dt = Γ(s + j)ζ(k1, . . . , kn−2; s + j + kn−1).
Proof. To obtain (i), write the sum in the definition of ζ(k1, k2, . . . ,
kn−1; s) as
∑
0
MULTIPLE ZETA VALUES 193
Proof. Writing the sum in the definition of ζ(k1, k2, . . . , kn−1; s) as ∞∑
m1=1
∞∑
m2=m1+1
· · · ∞∑
mn=mn−1+1
, we can prove (i) by the repetition of the same
calculation as in the proof of Proposition 2 (i). We prove (ii) by induction
on n. The case of n = 1 is the analytic continuation of the Riemann zeta
function. Let N be an arbitrary non-negative integer. Write
1
et − 1 =
N∑
j=0
Bj j!
tj−1 + RN (t)
with
RN (t) =
∞∑
j=N+1
Bj j!
tj−1,
where Bj are the classical Bernoulli numbers defined by this expansion. By
Proposition 2 (ii), we have
∫ ∞
0 ts−1Lik1,k2,...,kn−1(e
−t)
N∑
j=0
Bj j!
tj−1dt
=
N∑
j=0
Bj j!
Γ(s + j − 1)ζ(k1, . . . , kn−2; s + j + kn−1 − 1),
and thus
Γ(s)ζ(k1, k2, . . . , kn−1; s) =
∫ ∞
0 ts−1Lik1,k2,...,kn−1(e
−t)RN (t)dt(2)
+
N∑
j=0
Bj j!
Γ(s + j − 1)ζ(k1, . . . , kn−2; s + j + kn−1 − 1).
The second term in the right-hand side of (2) can be analytically continued
to a meromorphic function in the whole s-plane by the induction assump-
tion. Write the integral in (2) as a sum of two integrals:
∫ 1
0 ts−1Lik1,k2,...,kn−1(e
−t)RN (t)dt +
∫ ∞
1 ts−1Lik1,k2,...,kn−1(e
−t)RN (t)dt.
194 T. ARAKAWA AND M. KANEKO
The second integral converges absolutely for any s ∈ C and represents an
entire function of s. Using the estimates
Lik1,k2,...,kn−1(e −t) ≤
Lik1,k2,...,kn−1(1) = ζ(k1, k2, . . . , kn−1),
for kn−1 > 1,
Li1,1,...,1 ︸ ︷︷ ︸
n−1
(e−t) = 1
(n − 1)! (− log(1 − e−t))n−1
for kn−1 = 1,
and
RN (t) = O(|t| N ) (|t| < 1),
we see that the first integral converges absolutely for Re(s + N) > 0 and
hence holomorphic as a function of s in the same region. Since we can choose
N arbitrary large, this shows via (2) that the function ζ(k1, k2, . . . , kn−1; s)
has a meromorphic continuation to the whole s-plane.
By means of the inductive relation (2), we can, in principle, detect the
location and the order of poles of ζ(k1, k2, . . . , kn−1; s), which are in general
quite complicated to describe. Possible poles lie at the integers ≤ 1. Here
we only state results at s = 1 as a proposition.
Proposition 4. (i) Let the index set (k1, k2, . . . , kn−1) be of the form
(k1, . . . , ki, 1, 1, . . . , 1 ︸ ︷︷ ︸
m
) with m ≥ 0, i ≥ 1, and ki > 1. Then, the order of
pole of ζ(k1, k2, . . . , kn−1; s) at s = 1 is m + 1 and the leading coefficient is
ζ(k1, k2, . . . , ki) (i = n − m − 1), i.e.
ζ(k1, k2, . . . , kn−1; s) = ζ(k1, k2, . . . , ki)
(s − 1)m+1 + O((s − 1)−m).
(ii) For (k1, k2, . . . , kn−1) = (1, 1, . . . , 1), we have
Γ(s)ζ(1, 1, . . . , 1 ︸ ︷︷ ︸
n−1
; s) = 1
(s − 1)n + O(s − 1).(3)
Proof. (i) The integral in (2) is holomorphic at s = 1 even when N = 0,
in which case the relation (2) divided by Γ(s) reads
ζ(k1, k2, . . . , kn−1; s) = 1
Γ(s)
∫ ∞
0 ts−1Lik1,k2,...,kn−1(e
−t)R0(t)dt
+ 1
s − 1 ζ(k1, . . . , kn−2; s + kn−1 − 1).
MULTIPLE ZETA VALUES 195
By our definition of m, the function ζ(k1, . . . , kn−m−2; s + kn−m−1 − 1) is
holomorphic at s = 1 with the value ζ(k1, k2, . . . , kn−m−1). From this the
case of m = 0 follows, and we can easily obtain the proposition by induction
on the number of indices ki, starting with
ζ(s) = 1
s − 1 + O(1).
(ii) By Proposition 2 (i) and Lemma 1 (ii), we have
Γ(s)ζ(1, 1, . . . , 1 ︸ ︷︷ ︸
n−1
; s) = (−1)n−1
(n − 1)!
∫ ∞
0
ts−1
et − 1
( log
( 1 − e−t
))n−1 dt.
Let ζ∗(1, 1, . . . , 1 ︸ ︷︷ ︸
n−1
; s) be defined by
Γ(s)ζ∗(1, 1, . . . , 1 ︸ ︷︷ ︸
n−1
; s) = (−1)n−1
(n − 1)!
∫ ∞
0
ts−1
et − 1
(
log
( 1 − e−t
t
))n−1
dt.
By the standard method already used to prove the analytic continuation of
ζ(k1, k2, . . . , kn−1; s), we see that the function ζ ∗(1, 1, . . . , 1
︸ ︷︷ ︸
n−1
; s) is continued
to an entire function in the whole s-plane. Now we claim that the following
relation holds:
Γ(s)ζ(1, . . . , 1 ︸ ︷︷ ︸
n−1
; s) = (−1)n−1
(n − 1)! ·
dn−1
dsn−1 (Γ(s)ζ(s))(4)
+
n−2∑
j=0
(−1)j
j! ·
dj
dsj
Γ(s)ζ
∗(1, . . . , 1 ︸ ︷︷ ︸
n−1−j
; s)
.
This can be seen by using
( log(1 − e−t)
)n−1 =
n−1∑
j=0
( n − 1
j
)
(log t)j (
log
( 1 − e−t
t
))n−1−j
and ∫
∞
0
ts−1
et − 1 (log t)j
(
log
( 1 − e−t
t
))n−1−j
dt
= (−1)n−1−j(n − 1 − j)! dj
dsj
{
Γ(s)ζ∗(1, . . . , 1 ︸ ︷︷ ︸
n−1−j
; s) }
.
196 T. ARAKAWA AND M. KANEKO
Using (4) and the formula
Γ(s)ζ(s) = 1
s − 1 + O(s − 1)
which is equivalent to the more familiar (“Kronecker limit formula for ζ(s)”,
see e.g. Siegel [S])
ζ(s) = 1
s − 1 + γ + O(s − 1) (γ = Euler’s constant),
we conclude that
Γ(s)ζ(1, . . . , 1 ︸ ︷︷ ︸
n−1
; s) = 1
(s − 1)n + O(1).
Let c (n) 0 be the constant term on the right-hand side. From the relation
(obtained from (2) by putting N = 0 and ∀ki = 1)
ζ(1, . . . , 1 ︸ ︷︷ ︸
n−1
; s)(5)
= 1
Γ(s)
∫ ∞
0 ts−1Li1,1,...,1
︸ ︷︷ ︸
n−1
(e−t)R0(t)dt + 1
s − 1 ζ(1, . . . , 1
︸ ︷︷ ︸
n−2
; s),
we obtain
1
(s − 1)n + c
(n) 0 + O(s − 1) =
1
(s − 1)n +
c (n−1) 0
s − 1 + O(1),
hence we necessarily have c (n−1) 0 = 0. Since n is arbitrary, this completes
the proof of the Proposition.
Remarks. 1. The values of the holomorphic function ζ∗(1, 1, . . . , 1 ︸ ︷︷ ︸
r
, s)
at non-positive integers are evaluated in terms of Bernoulli numbers:
ζ∗(1, 1, . . . , 1 ︸ ︷︷ ︸
r
, 1 − m) = (−1)r+m−1
r! · D
(r) m
m (m ≥ 1),
where D (r) m is defined by
1
et − 1 ·
{
log
( 1 − e−t
t
)}r
=
∞∑
m=1
D(r)m · tm−1
m! ,
MULTIPLE ZETA VALUES 197
(note that D (r) m = 0 for 1 ≤ m ≤ r − 1). It is easy to see that (noting
(
log (
1−e−t
t
)) ′
= 1 et−1 −
1 t )
D(1)m =
m−1∑
j=0
( m
j
)
· Bj · Bm−j m − j
and that
D(r+1)m =
m−1∑
l=r
( m
l
)
· D (r) l ·
Bm−l m − l
for r ≥ 1 and m ≥ r + 1. Hence D (r) m is the sum of (r + 1)-fold products of
Bernoulli numbers.