The multiple Dirichlet product and the multiple Dirichlet series 6 1 0 TOMOKAZU ONOZUKA 2 n a J 2 Abstract 2 First, we define the multiple Dirichlet product and study the ] T propertiesofit.Fromthoseproperties,weobtainazero-freeregionof N a multiple Dirichlet series and a multiple Dirichlet series expression . of the reciprocal of a multiple Dirichlet series. h t a m 1 Introduction [ 1 The Euler-Zagier multiple zeta function ζ (s ,...,s ), the multiple zeta v EZ,k 1 k 4 star function ζ∗(s ,...,s ), and the Mordell-Tornheim multiple zeta func- 2 k 1 k 9 tion ζ (s ,...,s ;s ) are defined by MT,k 1 k k+1 5 0 1 1. ζEZ,k(s1,...,sk) := ms1ms2 ···msk, (1.1) 0 0<m1<Xm2<···<mk 1 2 k 6 1 1 ζk∗(s1,...,sk) := ms1ms2···msk, (1.2) v: 0<m1≤Xm2≤···≤mk 1 2 k i ∞ X 1 ζ (s ,...,s ;s ) := , (1.3) ar MT,k 1 k k+1 m1,.X..,mk=1 ms11 ···mskk(m1 +···+mk)sk+1 respectively, where s (i = 1,...,k +1) are complex variables. Matsumoto i [6] proved that the series (1.1) and (1.2) are absolutely convergent in (s ,...,s ) ∈ Ck | ℜ(s (k −l+1)) > l (l = 1,...,k) (1.4) 1 k k where s(cid:8) (n) = s + s + ··· + s (n = 1,...,k). The s(cid:9)eries (1.3) is k n n+1 k absolutely convergent in (s ,...,s ;s ) ∈ Ck+1 | ℜ(s ) > 1 (l = 1,...,k), ℜ(s ) > 0 . 1 k k+1 l k+1 2(cid:8)010 Mathematics Subject Classification: Primary 11M32;Secondary 11A25.(cid:9) Key words and phrases: Multiple zeta function, Multiple zeta star function, Multi- ple Dirichlet series, Multiple Dirichlet product, Multiple Dirichlet convolution,Zero-free region. The author was supported by JSPS KAKENHI Grant Number 13J00312. 1 The multiple Dirichlet product 2 Akiyama, Egami and Tanigawa [1] and Zhao [11], independently of each other, proved the meromorphic continuation of the series (1.1) to the whole space. Akiyama, Egami and Tanigawa used the Euler-Maclaurin summa- tion formula, while Zhao used generalized functions to prove the analytic continuation. Matsumoto [7] proved the meromorphic continuation of the series (1.3) to Ck+1. The series (1.2) can be expressed by the sum of Euler- Zagier multiple zeta functions and the Riemann zeta function. (Note that the Riemann zeta function is one of Euler-Zagier multiple zeta functions.) For example, ζ∗ and ζ∗ can be expressed as a sum of Euler-Zagier multiple 2 3 zeta functions as follows: ζ∗(s ,s ) = ζ (s ,s )+ζ(s +s ), 2 1 2 EZ,2 1 2 1 2 ζ∗(s ,s ,s ) = ζ (s ,s ,s )+ζ (s +s ,s )+ζ (s ,s +s ) 3 1 2 3 EZ,3 1 2 3 EZ,2 1 2 3 EZ,2 1 2 3 +ζ(s +s +s ). 1 2 3 Wecanobtaintheaboveexpressionbyseparatingthesumoftheseries(1.2). From the above expression and the meromorphic continuation of Euler- Zagiermultiplezetafunctions,itfollowsthatthemultiplezetastarfunctions can be continued meromorphically to the whole space. Inthispaper,asageneralizationoftheclassical notionofDirichlet series, we define multiple Dirichlet series by ∞ f(m ,...,m ) 1 k F(s ,...,s ;f) := , (1.5) 1 k ms1···msk m1,.X..,mk=1 1 k where f : Nk → C and (s ,...,s ) is in the region of absolute conver- 1 k gence for F(s ,...,s ;f). The series (1.5) is a generalization of the series 1 k (1.1), (1.2) and (1.3). De la Bret´eche [4] treated function (1.5) in the case f(m ,...,m ) > 0. 1 k In this paper, we study the multiple Dirichlet series defined by (1.5). In Section 2, we consider a ring of multiple arithmetic functions with the usual addition + and the multiple Dirichlet product ∗. In the case of one variable, Cashwell and Everett [3] proved that the ring is a UFD. In a way similar to that in[3], we find that the ring is also a UFD inthe multivariable case. In Section 3, we treat the multiple Dirichlet series (1.5) and prove the main theorem (Theorem 3.5). The main theorem has two statements. In the first statement, we state about a zero-free region of F(s ,...,s ;f). In 1 k the second statement, we state that the reciprocal of F(s ,...,s ;f) has a 1 k multiple Dirichlet series expression F(s ,...,s ;f−1). Finally, by using the 1 k The multiple Dirichlet product 3 main theorem, we get a multiple Dirichlet series expression of the reciprocal of ζ∗(s ,...,s ). k 1 k The author would like to express his thanks to Professor Kohji Mat- sumoto, Professor La´szlo´ T´oth, and Miss Ade Irma Suriajaya for their valu- able advice and comments. 2 The ring of multiple arithmetic functions We call f : Nk −→ C a multiple (k-tuple) arithmetic function, andwe define the set of k-tuple arithmetic functions by Ω = Ω := {f | f : Nk −→ C}. k We define the set U by; U = U := {f ∈ Ω | f(1,...,1) 6= 0}. k We use bold letters to express k-tuple of integers like a = (a ,...,a ). 1 k In particular, 1 means (1,...,1). Moreover we define the product of two k-tuple of integers a·b by (a b ,...,a b ). 1 1 k k Definition 2.1. For f,g ∈ Ω and n ∈ Nk, we define the multiple Dirichlet product ∗ by (f ∗g)(n) = f(a)g(b). a·b=n aX,b∈Nk If k = 1, the multiple Dirichlet product is the well-known Dirichlet prod- uct. Hence the above product is a generalization of the Dirichlet product. In the present section, our purpose is to study the properties of the multiple Dirichlet product. The properties were studied by some mathematicians. In [10], T´oth described the details of such studies. In particular, if we define an addition as (f +g)(n) := f(n)+g(n), he mentioned that (Ω ,+,∗) is k an integral domain with the identity function I which is defined by 1 (n = 1), I(n) := 0 (otherwise), (cid:26) and its unit group is U. For f ∈ U, f−1 denotes the inverse function of f with respect to the multiple Dirichlet product ∗. Then f−1 is given by the following lemma. The multiple Dirichlet product 4 Lemma 2.2. For each f ∈ U, f−1 ∈ U can be constructed recursively as follows; 1 f−1(1) = , f(1) 1 f−1(n) = − f(a)f−1(b) (n 6= 1). f(1) a·b=n Xb6=n Proof. In the case n +···+n = k, i.e. n = 1, we have 1 k f ∗f−1 (1) = f(1)f−1(1) = 1. (cid:0) (cid:1) Next,letd > k,andassumethatf−1(n)aredeterminedforalln ∈ Nk which satisfies n +···+n < d. Then for n ∈ Nk which satisfies n +···+n = d, 1 k 1 k we have f ∗f−1 (n) = f(a)f−1(b) a·b=n (cid:0) (cid:1) X = f(1)f−1(n)+ f(a)f−1(b) = 0. a·b=n Xb6=n In this section, let us consider (Ω ,+,∗) and prove that (Ω ,+,∗) is a k k UFD. First, we define a norm N : Ω −→ Z by ≥0 0 (f = 0), N(f) = min {n ···n | f(n) 6= 0} (f 6= 0). ( 1 k n∈Nk By the definition of the norm, we can show that N(f) = 1 if and only if f ∈ U holds. Theorem 2.3. For f,g ∈ Ω, we have N(f ∗g) = N(f)N(g). Proof. In the case f = 0 or g = 0, since f ∗ g = 0, we have N(f ∗ g) = N(f)N(g) = 0. Hence we consider the case f,g 6= 0. If n ···n < 1 k N(f)N(g), then we have (f ∗g)(n) = f(a)g(b) = 0, a·b=n X The multiple Dirichlet product 5 since a ···a < N(f) or b ···b < N(g). Therefore it is sufficient to 1 k 1 k find n such that (f ∗ g)(n) 6= 0 and n ···n = N(f)N(g). We define 1 k (M ,...,M ),(L ,...,L ) ∈ Nk as follows; 1 k 1 k M := min{m | f(m ,...,m ) 6= 0, m ···m = N(f)}, 1 1 1 k 1 k M := min{m | f(M ,m ,...,m ) 6= 0, M m ···m = N(f)}, 2 2 1 2 k 1 2 k . . . M := min{m | f(M ,...,M ,m ) 6= 0, M ···M m = N(f)}, k k 1 k−1 k 1 k−1 k L := min{l | g(l ,...,l ) 6= 0, l ···l = N(g)}, 1 1 1 k 1 k L := min{l | g(L ,l ,...,l ) 6= 0, L l ···l = N(g)}, 2 2 1 2 k 1 2 k . . . L := min{l | g(L ,...,L ,l ) 6= 0, L ···L l = N(g)}. k k 1 k−1 k 1 k−1 k Next we prove (f ∗g)(M ·L) = f(a)g(b) a·b=M·L X = f(M)g(L) 6= 0. (2.1) For a 6= M with a ···a = N(f), there exists an index i such that a 6= M 1 k i i and a = M for all j < i. If a < M , by the definition of M , f(a) = 0. j j i i i If a > M , then b < L and b = L for j < i, so g(B) = 0. Hence (2.1) i i i i j j holds. Hence we have N(f ∗g) = N(f)N(g). Next, we define an equivalence relation among functions belonging to Ω. For f,g ∈ Ω, we write f ∼ g and say f is equivalent to g if there exists a function ε ∈ U such that f = ε∗g. Also we write f|g if g = f ∗h for some h ∈ Ω. Theorem 2.4. Suppose f,g ∈ Ω. Then f is equivalent to g, if and only if f|g and g|f. Proof. Assume f ∼ g. Then there exists a function ε ∈ U such that f = ε∗g, that is, g|f. Since ε ∈ U, ε−1 ∗f = g holds. This means f|g. Next,weassumef|g andg|f.Iff = 0,theng isalso0.Thereforef = ε∗g holds for any ε ∈ U. Hence we are only left to consider the case f 6= 0. We seethatf|g andg|f ifandonlyifthereexistα,β ∈ Ωsuchthatf = α∗g and g = β ∗f, and if so, then N(f) = N(α)N(g) and N(g) = N(β)N(f). From the above equations, it follows that N(α)N(β) = 1. It implies α,β ∈ U. Hence f is equivalent to g. The multiple Dirichlet product 6 For p ∈ Ω \U, we call p prime if p = f ∗g implies f ∈ U or g ∈ U. We k define P = P to be the set of all prime functions. If N(f) is prime in N, k then f is prime in Ω. Hence we can find infinitely many primes. A multiple arithmetic function f ∈ Ω is said to be composite if f satisfies f 6= 0, f ∈/ U, and f ∈/ P. If f ∼ g, then f ∈ P implies g ∈ P. This property indicates that the equivalence relation ∼ preserves primitivity. The same property holds for 0, U, and composite functions, respectively. Next, we show that each non-zero function f ∈ Ω\U can be decomposed into a finite product of prime functions. In the case f ∈ P, the function f itself isthefiniteproduct ofprimefunctions. Hence weconsider thecasef ∈/ P. By the assumption, there exist multiple arithmetic functions g,h ∈ Ω\U such that f = g ∗h. Then the inequalities 1 < N(g),N(h) < N(f) hold. If both of g and h are primes, then g∗h is the prime factorization of f. If g is a composite function, then there exist multiple arithmetic functions g ,g ∈ 1 2 Ω\U such that g = g ∗g , and the inequalities 1 < N(g ),N(g ) < N(g) 1 2 1 2 hold. Repeating the above algorithm, we can obtain the prime factorization of f. Because the norm is a non-negative integer, the product is finite. Let PN be the set of all prime numbers in N. We define a set Pk = P by Pk = P := {(1,...,1,p,1,...,1) ∈ Nk | p ∈ PN, 1 ≤ j ≤ k}. j We take a bijection S : N −→ P. Since P is a countably infinite set, we can take the bijection S. We put p := S(j). Then each m ∈ Nk can be j decomposedintoafiniteproductofP.Wedefinemapsα (j ∈ N)asfollows; j m =: pα1(m) ·pα2(m) ···· . 1 2 The definition of α implies the equation α (a · b) = α (a) + α (b). By j j j j using α , we define a map R : Ω −→ C := C{x ,x ,...} by R(f) = j ω 1 2 ∞ f(m)xα1(m)xα2(m)··· for f ∈ Ω, where C is the set of the m1,...,mk=1 1 2 ω infinite series of the above form, where each term has only a finite number P of x . Note that although the series R(f) contains k, the set C does not j ω depend on k. Lemma 2.5. The map R is an isomorphism. Proof. The equation R(f + g) = R(f) + R(g) is trivial. In addition, we The multiple Dirichlet product 7 have ∞ R(f ∗g) = f(a)g(b)xα1(a)+α1(b)xα2(a)+α2(b)··· 1 2 m1,.X..,mk=1aX·b=m = f(a)xα1(a)xα2(a)··· g(b)xα1(b)xα2(b)··· 1 2 1 2 a ! b ! X X = R(f)R(g). Hence R is homomorphism. Next, we show that R isa bijection. Trivially f 6= g implies R(f) 6= R(g). For any A ∈ C , we can constitute a multiple arithmetic function f ∈ Ω ω A by using coefficients of A such that R(f ) = A. A In the case of one variable, Cashwell and Everett [3] proved that the ring (Ω ,+,∗) is a UFD by first showing that Ω is isomorphic to C and 1 1 ω then showing that (C ,+,×) is a UFD. Here, for the multivariable case, we ω use Lemma 2.5 and the result of Cashwell and Everett that (C ,+,×) is a ω UFD, and we obtain the following theorem. Theorem 2.6. (Ω ,+,∗) is a UFD. k In addition to the above theorem, Lemma 2.5 implies an another theo- rem. Lemma 2.5 states that Ω is isomorphic to C for all k ∈ N. This fact k ω means that Ω is isomorphic to Ω for k,l ∈ N. k l Theorem 2.7. All of Ω are isomorphic, i.e. Ω ∼= Ω holds for all k,l ∈ N. k k l 3 The multiple Dirichlet series In this section, we consider the zero-free region of the multiple Dirichlet series and the reciprocal of the multiple Dirichlet series by using the notion of multiple Dirichlet product. The following theorem is a basic property of the multiple Dirichlet series. Theorem 3.1. ([10], Proposition 10) For f,g ∈ Ω , we have k F(s ,...,s ;f)+F(s ,...,s ;g) = F(s ,...,s ;f +g), 1 k 1 k 1 k F(s ,...,s ;f)F(s ,...,s ;g) = F(s ,...,s ;f ∗g), 1 k 1 k 1 k where (s ,...,s ) lies in the region of absolute convergence for the series 1 k F(s ,...,s ;f) and F(s ,...,s ;g). 1 k 1 k The multiple Dirichlet product 8 Corollary 3.2. Let f ∈ U. If F(s ,...,s ;f) and F(s ,...,s ;f−1) are 1 k 1 k absolutely convergent on R ⊂ Ck, then F(s ,...,s ;f) has no zeros on R. 1 k Proof. Let (s ,...,s ) ∈ R. Then by applying Theorem 3.1, we have 1 k F(s ,...,s ;f)F(s ,...,s ;f−1) = F(s ,...,s ;I) = 1. 1 k 1 k 1 k To find the zero-free region of the series F(s ,...,s ;f), we have to find 1 k the region of absolute convergence R ⊂ Ck, and to find the region R, we have to evaluate f−1. For this purpose, we prepare the following lemma. Lemma 3.3. For α > 1, we have dα ≤ ζ(α)nα. d|n X Proof. We have α d 1 dα = nα = nα ≤ ζ(α)nα. n dα d|n d|n (cid:18) (cid:19) d|n X X X By the above lemma, we can evaluate the function f−1. Theorem 3.4. Let f ∈ U satisfy the condition that there exist constants C > 0 and r ,...,r ∈ R such that |f(n)| ≤ Cnr1nr2···nrk for n 6= 1. We 1 k 1 2 k put α > 1+r (j = 1,...,k) satisfying ζ(α −r )ζ(α −r )···ζ(α −r ) ≤ j j 1 1 2 2 k k 1+|f(1)|/C. Then we have nα1nα2 ···nαk |f−1(n)| ≤ 1 2 k . |f(1)| Proof. We use induction on n +···+n . In the case n + ···+n = k, 1 k 1 k i.e. n = 1, then f−1(1) = 1/f(1), so we have 1 |f−1(1)| = . |f(1)| Next, let d > k, and assume that |f−1(n)| ≤ nα1nα2 ···nαk/|f(1)| for all 1 2 k n ∈ Nk which satisfies n +···+n < d. Then for n ∈ Nk which satisfies 1 k The multiple Dirichlet product 9 n +···+n = d, by using Lemma 3.3, we have 1 k 1 |f−1(n)| ≤ |f(a)||f−1(b)| f(1) (cid:12) (cid:12) a·b=n (cid:12) (cid:12)aX,b∈Nk (cid:12) (cid:12) b6=n (cid:12) (cid:12) C ≤ ar1bα1 ···arkbαk |f(1)|2 1 1 k k a·b=n Xb6=n C = nr1 ···nrk bα1−r1 ··· bαk−rk −nα1 ···nαk |f(1)|2  1 k  1   k  1 k   bX1|n1 bXk|nk  C     ≤ |f(1)|2 (ζ(α1 −r1)nα11 ···ζ(αk −rk)nαkk −nα11nα22 ···nαkk)  nα1nα2 ···nαk ≤ 1 2 k . |f(1)| Finally, we obtain the main theorem. Theorem 3.5. Let f and α ,...,α satisfy the conditions in Theorem 3.4. 1 k Then F(s ,...,s ;f) and F(s ,...,s ;f−1) have no zeros in the region 1 k 1 k (s ,...,s ) ∈ Ck | ℜ(s ) > 1+α (j = 1,··· ,k) . 1 k j j (cid:8) (cid:9) Moreover, in the same region F(s ,...,s ;f) and F(s ,...,s ;f−1) satisfy 1 k 1 k the relation (F(s ,...,s ;f))−1 = F(s ,...,s ;f−1). 1 k 1 k Proof. Since f(n) ≪ nr1nr2···nrk, F(s ,...,s ;f) is convergent abso- 1 2 k 1 k lutely in (s ,...,s ) ∈ Ck | ℜ(s ) > 1+r (j = 1,...,k) . 1 k j j (cid:8) (cid:9) Since f−1(n) ≪ nα1 ···nαk by Theorem 3.4, F(s ,...,s ;f−1) is convergent 1 k 1 k absolutely in (s ,...,s ) ∈ Ck | ℜ(s ) > 1+α (j = 1,...,k) . 1 k j j (cid:8) (cid:9) Hence we obtain the theorem by using Theorem 3.2. The multiple Dirichlet product 10 Next, we consider a restricted multiple Dirichlet series. To express the series (1.1), (1.2) and (1.3), we define three functions as follows; 1 (n < n < ··· < n ), u (n) := 1 2 k EZ (0 (otherwise), 1 (n ≤ n ≤ ··· ≤ n ), u∗(n) := 1 2 k (0 (otherwise), 1 (n = n +n +···+n ), u (n) := k+1 1 2 k MT (0 (otherwise). Then the series (1.1), (1.2) and (1.3) can be expressed by ζ (s ,...,s ) = F(s ,...,s ;u ), EZ,k 1 k 1 k EZ ζ∗(s ,...,s ) = F(s ,...,s ;u∗), k 1 k 1 k ζ (s ,...,s ;s ) = F(s ,...,s ;u ). MT,k 1 k k+1 1 k+1 MT In [2], Akiyama and Ishikawa treated the multiple L-function which is defined by χ (m )···χ (m ) 1 1 k k L (s ,...,s |χ ,...,χ ) := , k 1 k 1 k ms1ms2···msk m1<mX2<···<mk 1 2 k where χ are Dirichlet characters. This is a generalization of the series (1.1). j Moreover in [8], Matsumoto and Tanigawa treated the series ∞ a (m )a (m )···a (m ) 1 1 2 2 k k , m1,.X..,mk=1 ms11(m1 +m2)s2···(m1 +···+mk)sk where series ∞ a (m)/ms (1 ≤ j ≤ k) satisfy good conditions. This is m=1 j also a generalization of the series (1.1). P To consider such series, we generalize the series (1.1) as follows; f(m ,...,m ) 1 k . ms1ms2···msk 0<m1<Xm2<···<mk 1 2 k This series is a restriction of the multiple Dirichlet series (1.5). To treat the restricted multiple Dirichlet series, we define three sets by Ω := {f ∈ Ω | f(n) = 0 for n which does not satisfy n < ··· < n }, EZ 1 k Ω∗ := {f ∈ Ω | f(n) = 0 for n which does not satisfy n ≤ ··· ≤ n }, 1 k Ω := {f ∈ Ω | f(n) = 0 for n which satisfies n < n +n +···+n }. MT k 1 2 k−1