Special values of generalized multiple Hurwitz zeta function at non-positive integers
aa r X i v : . [ m a t h . N T ] F e b Special values of generalized multiple Hurwitz zetafunction at non-positive integers
Boualem SADAOUI ∗ February 20, 2019
Abstract
In this paper, we provide an alternative method to calculate the values of generalizedmultiple Hurwitz zeta function at non-positive integers by means of
Raabe ’s formula andthe
Bernoulli numbers.
Mathematics Subject Classifications: 11M32; 11M41.Key words: Generalized multiple Hurwitz zeta function; integral representation;special values; Bernoulli numbers; Raabe’s formula.
Introduction and notations
The multiple Hurwitz zeta function is defined by ζ n ( α ; s , . . . , s n ) = X m =( m ,...,m n ) ∈ N n m + α ) s . . . ( m + · · · + m n + α ) s n (0.1)where α = 0 , − , − , ..., and ( s , . . . , s ) s ∈ C n , which introduced by Akiyama and Ishikawa andproved by Akiyama and Ishikawa [3]. Matsumoto and Tanigawa proved the analytic continuationof wide class of multiple Dirichlet series and multiple Hurwitz zeta functions in [12] and [13], andthe analytic continuation of the series (0.1) is a special case of [12, Theorem 1].Our main result in this work is the values at non positive integers of the following series ζ n ( α ; s ) = X m =( m ,...,m n ) ∈ N n n Y i =1 m + · · · + m i + α i ) s i (0.2)where, α = ( α , . . . , α n ) ∈ R n verified some conditions, this series is called the generalizedmultiple Hurwitz zeta function.The key of this study is the use of the Raabe formula [7] which expresses the integral in termsof the sum.In what follows, for any elements x = ( x , . . . , x n ) and y = ( y , . . . , y n ) of C n and s =( s , ..., s n ) denote a vector in C n . ∗ Université de Khemis Miliana, Laboratoire LESI, 44225, Khemis Miliana, Algérie.
E-mail: [email protected] Main results
For real numbers α = ( α , . . . , α n ) ∈ R n , such that, for all ≤ i ≤ n : α i = 0 , − , − , ..., So, for a complex n − tuples s = ( s , . . . , s n ) ∈ C n , we define the generalized multiple Hurwitzzeta function by ζ n ( α ; s ) := ζ ( α , . . . , α n ; s , . . . , s n ) (1.1) = X m =( m ,...,m n ) ∈ N n n Y i =1 m + · · · + m i + α i ) s i (1.2) = X m > ··· >m n ≥ n Y i =1 m i + α i ) s i (1.3)and the corresponding integral function associated to the generalized multiple Hurwitz zetafunction by Y n ( α ; s ) = Z [0 , + ∞ [ n n Y i =1 x + · · · + x i + α i ) s i dx. (1.4) Remark 1.1.
We remark that: • If α = ( α, α, . . . , α ) , then the series (1.1) corresponding to the classical multiple Hurwitzzeta function. • If α = (1 , , . . . , , then the series (1.1) corresponding to the multiple zeta function. For the meromorphic continuation of the integral (1.4) and the series (1.1), we refer the readerto the work [12].We first give well-known elementary result for the integral function.
Lemma 1.1.
Let N = ( N , . . . , N n ) be a point of N n ,(1) The point ( s = − N ) is a polar divisor for the function Y n ( α ; s ) if and only if there existsa k = ( k , ..., k n ) ∈ N n − such that ( s n − s n + s n − − k n ) ... n X i =1 s i − n + n X i =2 k i ! = n Y j =1 n X i = j s i − n + j − n X i = j +1 k i = 0 . (1.5) (2) If ( s = − N ) is not a polar divisor for the integral function, then the value of this functionat this point exists and is given by Y n ( α ; s ) = ( − n P k =( k ,...,k n ) ∈ N n − ( Nn +1 kn ) (cid:18) Nn + Nn − − knkn − (cid:19) ... (cid:16) P ni =2 Ni + n − P ni =3 kik (cid:17) α ( − P ni =1 si + n − P ni =2 ki ) Q nj =1 ( P ni = j N i + n − j +1 − P ni = j +1 k i ) Q nj =2 α k j j with T ( N ) := k = ( k , ..., k n ) ∈ N n − : 0 ≤ k j ≤ n X i = j N i + n − j + 1 − n X i = j +1 k i , ∀ ≤ j ≤ n .
2e give now a similar result for the generalized multiple Hurwitz zeta function.
Theorem 1.
Let N = ( N , . . . , N n ) a point of N n , if the point ( s = − N ) is not a polar divisor for theintegral function Y n ( α ; s ) , then the value of the generalized multiple Hurwitz zeta function ζ n ( α ; s ) at the point ( s = − N ) exists and is given by ζ n ( α ; − N ) = ( − n P k =( k ,...,k n ) ∈ N n − P v =( v ,...,vn ) ∈ N nvj ≤ kj ∀ ≤ j ≤ n ; v ≤ ( P ni =1 Ni + n − P ni =2 ki ) A ( − N ) B v Q nj =1 1 ( P ni = j N i + n − j +1 − P ni = j +1 k i ) (1.6) with A ( − N ) = (cid:18) P ni =1 N i + n − P ni =2 k i v (cid:19) α ( P ni =1 N i + n − P ni =2 v i ) n Y j =2 (cid:18) P ni = j N i + n − j + 1 − P ni = j +1 k i k j (cid:19) (cid:18) k j v j (cid:19) . (1.7) and T ( N ) := k = ( k , ..., k n ) ∈ N n − : 0 ≤ k j ≤ n X i = j N i + n − j + 1 − n X i = j +1 k i , ∀ ≤ j ≤ n . and B v = n Y j =1 B v j where B v j is the v j − th Bernoulli number.
Let the integral function Y n ( α ; s ) = Z [0 , + ∞ [ n n Y i =1 ( x + · · · + x i + α i ) − s i dx. (2.1)If we use the following change of variables: y i = x + · · · + x i + α i (2.2)for all ≤ i ≤ n , we find Y n ( α ; s ) = Z Q ni =1 [ α i , + ∞ [ n n Y i =1 ( y + · · · + y i ) − s i dx. (2.3)Now, using the following change of variables: z i = y + · · · + y i − i X j =2 α j (2.4)for all ≤ i ≤ n . This change gives (cid:26) y = z y i = z i − z i − + α i , ∀ ≤ i ≤ n (2.5)3ince y = ( y , ..., y n ) ∈ Q ni =1 [ α i , + ∞ [ , this gives z ∈ V n = { z ∈ R n : α ≤ z ≤ z ≤ · · · ≤ z n } (2.6)and, we find Y n ( α ; s ) = Z V n n Y i =1 ( z i + i X j =2 α j ) − s i dz. (2.7)This integral can be rewritten as follows. Y n ( α ; s ) = Z V n − n − Y i =1 ( z i + i X j =2 α j ) − s i Z + ∞ z n − ( z n + n X j =2 α j ) − s n dz n dz ...dz n − (2.8)with R + ∞ z n − ( z n + P nj =2 α j ) − s n dz n = ( z n − + P n − j =2 α j ) − sn +1 s n − (cid:18) α n z n − + P n − j =2 α j (cid:19) − s n +1 = P k n ∈ N (cid:16) − s n +1 k n (cid:17) ( z n − + P n − j =2 α j ) − sn +1 − kn s n − α k n n (2.9)if and only if ℜ ( s n ) − > .Inductively on n , we find Y n ( α ; s ) = X k =( k ,...,k n ) ∈ N n − (cid:16) − s n +1 k n (cid:17) (cid:16) − s n − s n − +2 − k n k n − (cid:17) ... (cid:16) − P ni =2 s i + n − P ni =3 k i k (cid:17) α ( − P ni =1 s i + n − P ni =2 k i ) ( s n − s n + s n − − k n ) ... ( P ni =1 s i − n + P ni =2 k i ) n Y j =2 α k j j if and only if for all ≤ i ≤ n − ℜ n X i =1 s i ! − n + j − n X i =2 k i > (2.10)and ℜ ( s n ) − > . (2.11)Therefore, for any point N = ( N , . . . , N n ) ∈ N n
1) The point ( s = − N ) is a polar divisor for the function Y n ( α ; s ) if there exists a k =( k , ..., k n ) ∈ N n − such that ( s n − s n + s n − − k n ) ... n X i =1 s i − n + n X i =2 k i ! = n Y j =1 n X i = j s i − n + j − n X i = j +1 k i = 0 . (2.12)2) If ( s = − N ) is not a polar divisor we get (cid:18) N n + 1 k n (cid:19) ... (cid:18) P ni =2 N i + n − P ni =3 k i k (cid:19) = n Y j =2 (cid:18) P ni = j N i + n − j + 1 − P ni = j +1 k i k j (cid:19) = 0 (2.13)4f and only if there exists an k = ( k , ..., k n ) ∈ N n − and ≤ j ≤ n , such that k j > n X i = j N i + n − j + 1 − n X i = j +1 k i . Let T ( N ) := k = ( k , ..., k n ) ∈ N n − : 0 ≤ k j ≤ n X i = j N i + n − j + 1 − n X i = j +1 k i , ∀ ≤ j ≤ n (2.14)which is finite, then Y n ( α ; s ) = ( − n P k =( k ,...,k n ) ∈ N n − ( Nn +1 kn ) (cid:18) Nn + Nn − − knkn − (cid:19) ... (cid:16) P ni =2 Ni + n − P ni =3 kik (cid:17) α ( − P ni =1 si + n − P ni =2 ki ) Q nj =1 ( P ni = j N i + n − j +1 − P ni = j +1 k i ) Q nj =2 α k j j For a = ( a , ..., a n ) ∈ R n + and s = ( s , ..., s n ) ∈ C n , we define the function Y n,a ( α ; s ) = Z Q ni =1 [ α i , + ∞ [ n n Y i =1 ( x + ... + x i + a + ... + a i ) − s i dx. (3.1)We prove the following useful result. Proposition 3.1.
Let N = ( N , ..., N n ) a point of N n , then we have for a ∈ R + Y n,a ( α ; − N ) = ( − n P k =( k ,...,k n ) ∈ N n − P v =( v ,...,vn ) ∈ N nvj ≤ kj ∀ ≤ j ≤ n ; v ≤ ( − P ni =1 si + n − P ni =2 ki ) A ( − N ) a v ( P ni =1 N i + n − P ni =2 k i ) Q nj =2 a vjj ( P ni = j N i + n − j +1 − P ni = j +1 k i ) . (3.2) with A ( − N ) = (cid:18) P ni =1 N i + n − P ni =2 k i v (cid:19) α ( P ni =1 N i + n − P ni =2 v i ) n Y j =2 (cid:18) P ni = j N i + n − j + 1 − P ni = j +1 k i k j (cid:19) (cid:18) k j v j (cid:19) α k j − v j j . (3.3) and T ( N ) := k = ( k , ..., k n ) ∈ N n − : 0 ≤ k j ≤ n X i = j N i + n − j + 1 − n X i = j +1 k i , ∀ ≤ j ≤ n (3.4) Proof.
Let a ∈ R n + , such that for all x = ( x , ..., x n ) ∈ [ α, + ∞ [ n and for all ≤ i ≤ nα i + a i x + ... + x i − + a + ... + a i − < , (3.5)we have Y n,a ( α ; s ) = Z Q ni =1 [ α i , + ∞ [ n n Y i =1 ( x + ... + x i + a + ... + a i ) − s i dx. (3.6)5his integral can be written as follows Y n,a ( α ; s ) = R Q n − i =1 [ α i , + ∞ [ n − Q n − i =1 ( x + ... + x i + a + ... + a i ) − s i (cid:16)R + ∞ α n ( x + ... + x n + a + ... + a n ) − s n dx n (cid:17) dx ...dx n − Since for ℜ ( s n ) > we have Z + ∞ α n ( x + ... + x n + a + ... + a n ) − s n dx n = ( x + ... + x n − + a + ...a n − + α n + a n ) − s n +1 s n − (3.7)condition (3.5) yields Z + ∞ α n ( x + ... + x n + a + ... + a n ) − s n dx n = X k n ∈ N (cid:18) − s n + 1 k n (cid:19) ( α n + a n ) k n s n − x + ... + x n − + a + ...a n − ) − s n +1 − k n . (3.8)If for ≤ j ≤ n − n X i = j ℜ ( s i ) − n + j − n X i = j +1 k i > (3.9)then inductively we find Y n,a ( α ; s ) = ( − n P k =( k ,...,k n ) ∈ N n − ( α + a ) − P ni =1 si + n − P ni =2 ki ( − P ni =1 s i + n − P ni =2 k i ) Q nj =2 (cid:16) − P ni = j s i + n − j +1 − P ni = j +1 k i k j (cid:17) ( α j + a j ) kj ( − P ni = j s i + n − j +1 − P ni = j +1 k i ) . (3.10)But, for all ≤ j ≤ n we have ( α j + a j ) k j = X vj ∈ N vj ≤ kj (cid:18) k j v j (cid:19) α k j − v j j a v j j (3.11)and ( α + a ) − P ni =1 s i + n − P ni =2 k i = X v ∈ N v ≤ ( − P ni =1 si + n − P ni =2 ki ) (cid:18) − P ni =1 s i + n − P ni =2 k i v (cid:19) α ( − P ni =1 s i + n − P ni =2 k i ) a v (3.12)which yields Y n,a ( α ; s ) = ( − n P k =( k ,...,k n ) ∈ N n − P v =( v ,...,vn ) ∈ N nvj ≤ kj ∀ ≤ j ≤ n ; v ≤ ( − P ni =1 si + n − P ni =2 ki ) A ( s ) a v ( − P ni =1 s i + n − P ni =2 k i ) Q nj =2 a vjj ( − P ni = j s i + n − j +1 − P ni = j +1 k i ) . (3.13)with A ( s ) = (cid:18) − P ni =1 s i + n − P ni =2 k i v (cid:19) α ( − P ni =1 s i + n − P ni =2 v i ) n Y j =2 (cid:18) − P ni = j s i + n − j + 1 − P ni = j +1 k i k j (cid:19) (cid:18) k j v j (cid:19) α k j − v j j . (3.14)Setting s = − N = − ( N , ..., N n ) ∈ N n yields (3.2) and ends the proof of Proposition 3.1.6 Proof of Theorem 1
The proof relies on the
Raabe formula [7], which expresses the integral in terms of the sum.
Proposition 4.1. (1) Raabe formula:for all s ∈ C n , outside the possible polar divisors of Y n ( α ; s ) , we have: Y n ( α ; s ) = Z t ∈ [0 , n ζ n, t ( α ; s ) d t (4.1) where: ζ n, t ( α ; s ) = X m ∈ N n n Y i =1 m + t ) + · · · + ( m i + t i + α i )) s i and d t is the Lebesgue measure on R n .(2) For a fixed point N = ( N , ..., N n ) in N n the maps a Y n,a ( α ; − N ) and a ζ n, t ( α ; − N ) are polynomials in a = ( a , ..., a n ) ∈ R n + .Proof. (1) Let s ∈ C n be chosen in such a way that the integral function and the generalized multipleHurwitz zeta function are absolutely convergent.Thus, for t ∈ R n + , we have: Z [0 , n ζ n, t ( α ; s ) d t = Z [0 , n X m ∈ N n n Y i =1 ( t + · · · + t i + m + · · · + m i + α i ) − s i d t = X m =( m ,...,m n ) ∈ N n Z Q ni =1 [ m i ,m i +1] n Y i =1 ( t + · · · + t i + m + · · · + m i + α i ) − s i d t = Z [0 , + ∞ [ n n Y i =1 ( x + · · · + x i + α i ) − s i d x = Y n ( α ; s ) . This last equality which is verified for all s ∈ C n follows by analytic continuation outsidethe polar divisors.(2) follows from (3.2) combined with the Raabe formula.
Lemma 4.1 ([6]) . Let P and Q to be two polynomials in n variables linked by P ( a ) = Z t ∈ [0 , n Q ( a + t ) d t . (4.2) Write out P ( a ) = P ( a , ..., a n ) = X L h L n Y i =1 a L i i (4.3)7 here h L ∈ C and L = ( L , ..., L n ) ∈ N n ranges over a finite set of multi-index. Then Q ( a ) = Q ( a , ..., a n ) = X L h L n Y i =1 B L i ( a i ) (4.4) where the B L i ( a i ) are the Bernoulli polynomials [4].Conversely, if Q is given by (4.4), then the relations (4.2) and (4.3) yield equivalent formulasfor the polynomial P . Proposition 4.2.
If we write out the polynomial Y a ( α ; − N ) as a sum of monomials, Y a ( α ; − N ) = X L C L a L with a L = n Y i =1 a L i i and C L = C L ( N ) ∈ C .Then ζ n ( α ; − N ) = X L C L B L where B L = n Y i =1 B L i is a product of Bernoulli numbers.More generally, for a = ( a , ..., a n ) ∈ R n + , we have: ζ n, a ( α ; − N ) = X L C L B L ( a ) where B L ( a ) = n Y i =1 B L i ( a i ) is a product of Bernoulli numbers.Proof. It follows from the above lemma, with P ( a ) = Y n,a ( α ; − N ) and Q ( a ) = ζ n,a ( α ; − N ) . Relation (3.2) shows that for all a ∈ R n + Y n,a ( α ; − N ) = ( − n P k =( k ,...,k n ) ∈ N n − P v =( v ,...,vn ) ∈ N nvj ≤ kj ∀ ≤ j ≤ n ; v ≤ ( − P ni =1 si + n − P ni =2 ki ) A ( − N ) a v ( P ni =1 N i + n − P ni =2 k i ) Q nj =2 a vjj ( P ni = j N i + n − j +1 − P ni = j +1 k i ) . (4.5)with A ( − N ) = (cid:18) P ni =1 N i + n − P ni =2 k i v (cid:19) α ( P ni =1 N i + n − P ni =2 v i ) n Y j =2 (cid:18) P ni = j N i + n − j + 1 − P ni = j +1 k i k j (cid:19) (cid:18) k j v j (cid:19) α k j − v j j . (4.6)and T ( N ) := k = ( k , ..., k n ) ∈ N n − : 0 ≤ k j ≤ n X i = j N i + n − j + 1 − n X i = j +1 k i , ∀ ≤ j ≤ n . (4.7)8etting, a v = n Y j =1 a v j j (4.8)this gives Y n,a ( α ; − N ) = ( − n P k =( k ,...,k n ) ∈ N n − P v =( v ,...,vn ) ∈ N nvj ≤ kj ∀ ≤ j ≤ n ; v ≤ ( P ni =1 Ni + n − P ni =2 ki ) A ( − N ) a v Q nj =1 1 ( P ni = j N i + n − j +1 − P ni = j +1 k i ) . (4.9)It follows from Proposition 4.2 that ζ n ( α ; − N ) = ( − n P k =( k ,...,k n ) ∈ N n − P v =( v ,...,vn ) ∈ N nvj ≤ kj ∀ ≤ j ≤ n ; v ≤ ( P ni =1 Ni + n − P ni =2 ki ) A ( − N ) B v Q nj =1 1 ( P ni = j N i + n − j +1 − P ni = j +1 k i ) (4.10)with B v = n Y j =1 B v j and B v j is the v j − th Bernoulli number, which ends the proof of Theorem 1.
In this part, we give two applications of this results:
In this paragraph, we give the values of the classical multiple zeta values at non positive integers.So, for α := ( α, . . . , α ) , where α = 0 , − , − , . . . , , we find ζ n ( α ; s ) = ζ n ( α ; s , . . . , s n ) = X m =( m ,...,m n ) ∈ N n n Y i =1 m + · · · + m i + α ) s i (5.1)which is the multiple Hurwitz zeta function.Thus, if we apply Theorem 1, we find the same result given in [16]. Corollary 5.1.
Let N = ( N , . . . , N n ) a point of N n , if the point ( s = − N ) is not a polar divisor for theintegral function Y n ( α ; s ) , then the value of the multiple Hurwitz zeta function ζ n ( α ; s ) at thepoint ( s = − N ) exists and is given by ζ n ( α ; − N ) = ( − n P k =( k ,...,k n ) ∈ N n − P v =( v ,...,vn ) ∈ N nvj ≤ kj ∀ ≤ j ≤ n ; v ≤ ( P ni =1 Ni + n − P ni =2 ki ) A ( − N ) B v Q nj =1 1 ( P ni = j N i + n − j +1 − P ni = j +1 k i ) (5.2) with A ( − N ) = (cid:18) P ni =1 N i + n − P ni =2 k i v (cid:19) α ( P ni =1 N i + n − P ni =2 v i ) n Y j =2 (cid:18) P ni = j N i + n − j + 1 − P ni = j +1 k i k j (cid:19) (cid:18) k j v j (cid:19) . (5.3)9 nd T ( N ) := k = ( k , ..., k n ) ∈ N n − : 0 ≤ k j ≤ n X i = j N i + n − j + 1 − n X i = j +1 k i , ∀ ≤ j ≤ n . and B v = n Y j =1 B v j where B v j is the v j − th Bernoulli number.
Now, for α := (1 , . . . , , we find ζ n ( α ; s ) = ζ n (1; s , . . . , s n ) = ζ n ( s ) = X m =( m ,...,m n ) ∈ N n n Y i =1 m + · · · + m i + 1) s i (5.4) = X m =( m ,...,m n ) ∈ N ∗ n n Y i =1 m + · · · + m i ) s i (5.5)which is the multiple zeta function.Thus, if we apply Theorem 1, we find the same result given in [14]. Corollary 5.2.
Let N = ( N , . . . , N n ) a point of N n , if the point ( s = − N ) is not a polar divisor for theintegral function Y ( s ) , then the value of the multiple zeta function Z ( s ) at the point ( s = − N ) exists and is given by . ζ n ( − N ) = ( − n P k =( k ,...,k n ) ∈ T ( N ) P v =( v ,...,vn ) ∈ N nvj ≤ kj ∀ ≤ j ≤ n ; v ≤ ( P ni =1 Ni + n − P ni =2 ki ) A ( − N ) B v Q nj =1 1 ( P ni = j N i + n − j +1 − P ni = j +1 k i ) (5.6) with A ( − N ) = (cid:18) P ni =1 N i + n − P ni =2 k i v (cid:19) n Y j =2 (cid:18) P ni = j N i + n − j + 1 − P ni = j +1 k i k j (cid:19) (cid:18) k j v j (cid:19) T ( N ) := k = ( k , ..., k n ) ∈ N n − : 0 ≤ k j ≤ n X i = j N i + n − j + 1 − n X i = j +1 k i , ∀ ≤ j ≤ n . and B v = n Y j =1 B v j where B v j is the v j − th Bernoulli number. This corrects a typo in[14, Theorem 1] cknowledgement We thank the referees for their numerous and very helpful comments and suggestions whichgreatly contributed in improving the final presentation.
References [1] S. Akiyama and S. Egami and Y. Tanigawa, Analytic continuation of multiple zeta-functionsand their values at non-positive integers, Acta Arith., 98,2, 107–116,2001.[2] S. Akiyama and Y. Tanigawa, Multiple zeta values at non-positive integers, Ramanujan J.,5, 327-351 (2002).[3] S. Akiyama and H. Ishikawa, On analytic continuation of multiple L −−