aa r X i v : . [ m a t h . N T ] J a n AN IDENTITY CONCERNING THE RIEMANN-ZETAFUNCTION
DOUGLAS AZEVEDO
Abstract.
For a certain function J ( s ) we prove that the identity ζ (2 s ) ζ ( s ) − (cid:18) s − (cid:19) J ( s ) = ζ (2 s + 1) ζ ( s + 1 / , holds in the half-plane Re( s ) > / Introduction
Let s = σ + it , ζ ( s ), as usual, the Riemann zeta-function and λ ( n ) the Liouvillefunction, that is, a completely multiplicative function with λ (1) = 1 and λ ( n ) = − p .It is well know that λ ( n ) is deeply related to the Riemann hypothesis, or simply,RH and also to the Prime Number Theorem (PNT). For instance, RH is equivalentto X n ≤ x λ ( n ) = O ǫ ( x + ǫ ) , for all ǫ >
0. Whereas the PNT is equivalent to ([1] ) ∞ X n =1 λ ( n ) n = 0 . We also refer to [4, p.179] for more details on the PNT.Any improvements in the zero-free region for ζ ( s ) will immediately imply im-provements in the error term of the prime number theorem. For example, if theRiemann Hypothesis is true then we obtain π ( x ) = Z x du log( u ) + O ( √ x log( x )) , and this last is not only implied by the RH but actually implies the RH itself.The identity(1.1) ζ (2 s ) ζ ( s ) = ∞ X n =1 λ ( n ) n s , Re( s ) > , Mathematics Subject Classification.
Primary 11M26, Secondary 11M41.
Key words and phrases.
Riemann zeta-function, Riemann hypothesis, Liouville function,Prime Number Theorem. was investigated by P´olya [5] and Tur´an [6] aiming information about the RH. P´olyastudied the change of sign of sum P ( x ) = X n ≤ x λ ( n )and its relation to the RH. He remarked that the RH would follow if one couldestablish that P ( x ) eventually has constant sign. The assertion that P ( x ) ≤ x ≥ T ( x ) = X n ≤ x λ ( n ) n . He showed that if there exists a positive constant c such that T ( n ) > − c/ √ n forall sufficiently large n , then the RH would follow. He also reported that severalassistants had verified that T ( n ) > n ≤ P ( x ) and T ( x ). In 1958, Haselgrove [2] proved that both P ( x ) and T ( x ) changesign infinitely often. Consequently one can not rely on this to infer about the RH.These approaches to the RH depend upon an application of a Landau’s resultabout the domain of convergence of Dirichlet integrals ( Lemma 2.1 below). Let ussummarize the idea behind the approach which consists of two main features: thefirst, is to represent the ratio ζ (2 s ) ζ ( s ) by means of a Dirichlet integral, that is, Z ∞ A ( u ) u − s du for all Re( s ) >
1. The second, is that A ( u ) does not change sign for all u large. Bypartial summation it easy to see that ζ (2 s ) ζ ( s ) = s Z ∞ P ( u ) u u − s du and ζ (2 s ) ζ ( s ) = ( s − Z ∞ T ( u ) u − s du, Re( s ) >
1. Therefore, if P ( x ) or T ( x ) do not change sign for all x large (which isnot the case), by Landau’s result, since ζ (2 s ) ζ ( s ) has real singularity at s = , then ζ (2 s ) ζ ( s ) is analytic in the half-plane Re( s ) > . Hence, ζ ( s ) = 0 in this half-plane. Inother words, we would have the validity of the RH.In this paper we follow a similar idea to investigate the RH. Precisely, we provethat there exist certain functions F ( x ) and J ( s ), with lim x →∞ F ( x ) = − (cid:18) s − (cid:19) − / (cid:18) ζ (2 s ) ζ ( s ) − (cid:19) + J ( s ) = Z ∞ F ( u ) u − s − / du, holds in the half plane Re( s ) >
1. Hence, since, F ( x ) < x large and theintegral on the right hand side converges absolutely for s > / s = 1 /
2, as a consequence of Landau’s theorem for Dirichlet’s integrals (Lemma2.1) we can conclude that both sides of the identity are analytic in the half-planeRe( s ) > / µ ( n ) instead λ ( n ), withsome adaptations. N IDENTITY CONCERNING THE RIEMANN-ZETA FUNCTION 3 Auxiliary results
Let us present some results that will be needed throughout this paper.For the next result, which is an analogue of Landau’s theorem concerning Dirich-let series with non-negative coefficients, we refer [4, Lemma 15.1]. This result is themain tool used to obtain the central results of this paper.
Lemma 2.1.
Suppose that G ( u ) is bounded Riemann-integrable function on everycompact interval [1 , a ] , and that G ( u ) ≥ for all u > M or G ( u ) ≤ for all u > M .Let σ c denote the infimum of those σ for which R ∞ M G ( u ) u − σ du converges. Thenthe function ϕ ( s ) = Z ∞ G ( u ) u − s du is analytic in the half-plane Re ( s ) > σ c but not at s = σ c . Our first result is just an observation derived from the identity (1.1) and anapplication of the PNT.
Lemma 2.2.
Let λ ( n ) be the Liouville function and a ( n ) be the arithmetic functiondefined as a ( n ) = ( , n = 1 λ ( n ) , n ≥ . We have that, ζ (2 s ) ζ ( s ) − ∞ X n =1 a ( n ) n s , Re ( s ) > . In particular, ∞ X n =1 a ( n ) n = − . Proof.
Since λ (1) = 1, it is immediate that ζ (2 s ) ζ ( s ) = ∞ X n =1 λ ( n ) n s = 1 + ∞ X n =2 λ ( n ) n s = 1 + ∞ X n =1 a ( n ) n s , Re( s ) > . The conclusion of the proof follows the definition of a ( n ) and from the fact thatthe PNT is equivalent to ∞ X n =1 λ ( n ) n = 0 . (cid:3) Let x ≥ b ( n ) an arithmetical function. Consider the Dirichlet polynomials(2.1) F x ( α ) = X n ≤ x b ( n ) n − α , α ∈ R . A first consequence that can be extracted from Lemma 2.2 is an alternative prooffor the well-known fact that ζ ( s ) = 0 for Re( s ) >
1. Indeed, by partial summation(2.2) (cid:18) − ζ ( s ) (cid:19) ( s − − = Z ∞ F u (1) u − s du, Re( s ) > . DOUGLAS AZEVEDO
Since lim x →∞ F u (1) = −
1, it is clear that F u (1) < x large. Also note that,since the inequalities(2.3) 1 σ − < ζ ( σ ) < σσ − σ > σ = 1 but isanalytic on the real line for σ >
1. Hence, by an application of Lemma 2.1, (2.2)holds for Re( s ) > ζ ( s ) = 0 for Re( s ) > Lemma 2.3.
Let β > α and b ( n ) an arithmetic function. There exists a sequence ξ = ( ξ ( n )) ⊂ ] α, β [ such that F x ( β ) − F x ( α ) = − ( β − α ) X n ≤ x b ( n ) log( n ) n ξ ( n ) , for all x ≥ . Moreover, ξ decreases and lim n →∞ ξ ( n ) = α. Proof.
Let n > β > α . By the Mean Value Theorem n − β − n − α = − ( β − α ) log( n ) n − ξ ( n ) , for some ξ ( n ) ∈ ] α, β [. That is, for each n ≥
1, there exists ξ ( n ) ∈ ] α, β [ for which F x ( β ) − F x ( α ) = − ( β − α ) X n ≤ x b ( n ) log( n ) n ξ ( n ) , for all x ≥ ξ follows from the equality ξ ( n ) = log (cid:16) ( β − α ) log( n ) n α + β n β − n α (cid:17) log( n ) = log(log( n ))log( n ) + log (cid:16) ( β − α )1 − n α − β (cid:17) log( n ) + α, for n > (cid:3) Main results
Let a ( n ) be as in Lemma 2.2 and we now fix the notation F x ( α ) = X n ≤ x a ( n ) n α , with α > x ≥ Lemma 3.1.
Let Re ( s ) > . The following identity holds true (cid:18) s − (cid:19) − (cid:18) ζ (2 s ) ζ ( s ) − (cid:19) = Z ∞ F u (1 / u − s − / du. N IDENTITY CONCERNING THE RIEMANN-ZETA FUNCTION 5
Proof.
Let x ≥
1. By partial summation we obtain that X n ≤ x a ( n ) n s = F x (1 / x − s +1 / + (cid:18) s − (cid:19) Z x F u (1 / u − s − / du, for Re( s ) >
1. Since lim x →∞ F x (1 / x − s +1 / = 0 , whenever Re( s ) >
1, we obtain that ∞ X n =1 a ( n ) n s = (cid:18) s − (cid:19) Z ∞ F u (1 / u − s − / du, holds in the half-plane Re( s ) >
1. By Lemma 2.2 we have that ζ (2 s ) ζ ( s ) − (cid:18) s − (cid:19) Z ∞ F u (1 / u − s − du, holds in the half-plane Re( s ) > (cid:3) In the following result we prove an identity that concerning the analiticity of ζ (2 s ) ζ ( s ) in the half-plane Re( s ) > / Theorem 3.2.
There exists a decreasing sequence ξ = ( ξ ( n )) ⊂ ]1 / , such that lim n →∞ ξ ( n ) = 1 / for which (3.1) (cid:18) s − (cid:19) − (cid:18) ζ (2 s ) ζ ( s ) − (cid:19) − J ξ ( s ) = Z ∞ F u (1) u − s − / du, holds for Re ( s ) > / and both sides of the equality are analytic in this half-plane.In the equality above we have that J ξ ( s ) = Z ∞ L u ( ξ ) u − s − / du, Re ( s ) > and L x ( ξ ) = 12 X n ≤ x a ( n ) log( n ) n ξ ( n ) , x ≥ . Proof.
From the previous lemma we have that(3.2) (cid:18) s − (cid:19) − (cid:18) ζ (2 s ) ζ ( s ) − (cid:19) = Z ∞ F u (1 / u − s − / du, in the half-plane Re( s ) >
1. For β = 1 and α = 1 / ξ = ( ξ ( n )) ⊂ ]1 / ,
1[ such thatlim n →∞ ξ ( n ) = 1 / F u (1 /
2) = F u (1) + L u ( ξ ) , for all u ≥
1, where L u ( ξ ) = 12 X n ≤ u a ( n ) log( n ) n ξ ( n ) , u ≥ . DOUGLAS AZEVEDO
Hence, we can rewrite (3.2) as(3.3) (cid:18) s − (cid:19) − (cid:18) ζ (2 s ) ζ ( s ) − (cid:19) − J ξ ( s ) = Z ∞ F u (1) u − s − / du, for Re( s ) >
1, for some sequence ξ = ( ξ ( n )) ⊂ ]1 / , J ξ ( s ) = Z ∞ L u ( ξ ) u − s − / du. In order to show that the equality (3.3) extends to the half-plane Re( s ) > / u →∞ F u (1) = −
1. Thus there exists M ≥ F u (1) < u > M . Moreover, clearlythe integral Z ∞ F u (1) u − σ − / du converges (absolutely) at every σ > / σ = 1 /
2. By Lemma 2.1,this implies that the function Z ∞ F u (1) u − s − / du is analytic for Re( s ) > /
2. Therefore, Lemma 2.1 implies that (3.3) holds forRe( s ) > / (cid:3) Note that by Lemma 2.1, if F x (1 / ≤ x large, then by Lemma 3.1 ζ (2 s ) ζ ( s ) is analytic in the half-plane Re( s ) > /
2, which implies the truth of RH. In thefollowing result it is provided another condition to obtain the RH.
Theorem 3.3.
If there exists r > such that L x ( ξ ) ≤ − r for all x large, then ζ (2 s ) ζ ( s ) is analytic in Re ( s ) > / .Proof. From Lemma 2.3, we have that F x (1 /
2) = F x (1) + L x ( ξ ), for all x ≥ ǫ > F x (1) < − ǫ , for all x large. Hence, if L x ( ξ ) ≤ − r , for 0 < ǫ ≤ r sufficiently small F x (1 /
2) = F x (1) + L x ( ξ ) < ǫ − r ≤ x ≥
1. By Lemma 2.1 and Lemma 3.1 this implies the analyticity of ζ (2 s ) ζ ( s ) inthe half-plane Re( s ) > / (cid:3) Now note that by partial summation, it follows that ∞ X n =1 a ( n ) n s +1 / = ( s − / Z ∞ F u (1) u − s − / du for Re( s ) > /
2. Hence, as a consequence of Lemma 3.2 and Lemma 2.2, equation(3.1) can be writen as(3.4) ζ (2 s ) ζ ( s ) − (cid:18) s − (cid:19) J ξ ( s ) = ∞ X n =1 λ ( n ) n s +1 / , Re( s ) > /
2, and both sides are analytic in this half-plane. Moreover, from (1.1),this previous conclusions imply that
Theorem 3.4. (3.5) ζ (2 s ) ζ ( s ) − (cid:18) s − (cid:19) J ξ ( s ) = ζ (2 s + 1) ζ ( s + 1 / , Re ( s ) > / and both sides are analytic in this half-plane. N IDENTITY CONCERNING THE RIEMANN-ZETA FUNCTION 7
References
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Federal Technological University of Parana
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