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### Transcript of MULTIPLE ZETA VALUES, POLY-BERNOULLI NUMBERS, AND RELATED ZETA...

• 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.