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On complete monotonicity of the Riemann zeta function

Abstract

Under the assumption of the Riemann hypothesis for the Riemann zeta function and some Dirichlet L-series we demonstrate that certain products of the corresponding zeta functions are completely monotonic. This may provide a method to disprove a certain Riemann hypothesis numerically.

MSC:30E15, 33D45.

1 Introduction

The Riemann zeta function ζ(s) can be defined by

ζ(s)= n = 1 1 n s ,(s)>1,
(1.1)

and on the rest of the complex plane by analytic continuation. It is known that the extended ζ(s) is meromorphic with infinitely many zeros at 2n for nN (a.k.a trivial zeros) and with infinitely many zeros within the vertical strip 0<(s)<1 (nontrivial zeros). The Riemann hypothesis for ζ(s) says that all nontrivial zeros are actually on the critical line (s)= 1 2 .

For any complex number zC, let Γ(z) be Euler’s Gamma function defined by [18]

1 Γ ( z ) =z j = 1 ( 1 + z j ) ( 1 + 1 j ) z .
(1.2)

Then, the Riemann Ξ(z) function [17]

Ξ(z)= 1 + 4 z 2 8 π 1 + 2 i z 4 Γ ( 1 + 2 i z 4 ) ζ ( 1 + 2 i z 2 )
(1.3)

is an even entire function of order 1. The celebrated Riemann hypothesis is equivalent to the statement that Ξ(z) has only real zeros.

Let χ(n) be a real primitive character with modulus m; the function L(s,χ) is defined by [3, 8]

L(s,χ)= n = 1 χ ( n ) n s ,(s)>1.
(1.4)

Let

α={ 0 , χ ( 1 ) = 1 , 1 , χ ( 1 ) = 1 ,
(1.5)

then

Ξ(z,χ)= ( π m ) ( 1 + 2 α + 2 i z ) / 4 Γ ( 1 + 2 α + 2 i z 4 ) L ( 1 + 2 i z 2 , χ )
(1.6)

is an even entire function of order 1. The Riemann hypothesis for L(s,χ) is equivalent to Ξ(z,χ) having only real zeros.

Given real numbers a, b with a<b and an indefinite differentiable real valued function f(x) on (a,b), f(x) is called completely monotonic on (a,b) if ( 1 ) m f ( m ) (x)0 for all x(a,b) and m=0,1, . In this work, under the assumptions of the Riemann hypothesis for the Riemann zeta function and certain L-series, we apply the ideas from [8, 9] to prove that some products of these zeta functions are completely monotonic. This complete monotonicity may provide a method to disprove a certain Riemann hypothesis via numerical methods.

2 Main results

Lemma 1 Given a non-increasing sequence of positive numbers such that

n = 1 | λ n |<,
(2.1)

then, the entire function

f(x)= n = 1 (1x λ n )
(2.2)

is completely monotonic on (, λ 1 1 ).

Proof It is a direct consequence of Theorem 1 of [8]. □

Assuming the Riemann hypothesis is true, we list all positive zeros of Ξ(z) as

z 1 z 2 z n ,
(2.3)

and z 1 is approximately 14.1347. Then,

Ξ(z)=Ξ(0) n = 1 ( 1 z 2 z n 2 ) .
(2.4)

Thus,

n = 1 ( 1 z z n 2 ) = Ξ ( z ) Ξ ( 0 ) ,
(2.5)
Ξ ( z 1 4 ) Ξ ( i z 1 4 ) Ξ 2 ( 0 ) = n = 1 ( 1 z z n 4 ) ,
(2.6)

and

Ξ ( z 1 6 ) Ξ ( ρ z 1 6 ) Ξ ( ρ 2 z 1 6 ) Ξ 3 ( 0 ) = n = 1 ( 1 z z n 6 )
(2.7)

for 0arg(z)<2π, where ρ= e 2 π i 3 . In fact, for any positive integer >1 and assume that ρ is a primitive th root of unity; then we have

j = 1 Ξ ( ρ j z 1 2 ) Ξ ( 0 ) = n = 1 ( 1 z z n 2 ) .
(2.8)

Corollary 2 Under the Riemann hypothesis, let z 1 be the least positive zeros of Ξ(z); then the function Ξ( z ) is completely monotonic for z(, z 1 2 ), Ξ( z 1 4 )Ξ(i z 1 4 ) is completely monotonic for z(, z 1 4 ), and Ξ( z 1 6 )Ξ(ρ z 1 6 )Ξ( ρ 2 z 1 6 ) is completely monotonic for z(, z 1 6 ). Let ρ be a primitive ℓth root of unity for some positive integer ; then j = 1 Ξ( ρ j z 1 2 ) is completely monotonic for z(, z 1 2 ).

Proof Notice that Ξ(0) is a positive constant, and the claims are obtained by applying Corollary 1 to equations (2.5)-(2.8). □

Assuming the Riemann hypothesis for L(s,χ), we list all the positive zeros for Ξ(z,χ) as [8]

z 1 (χ) z 2 (χ) z n (χ).
(2.9)

Then

Ξ(z,χ)=Ξ(0,χ) n = 1 ( 1 z 2 z n ( χ ) 2 ) .
(2.10)

Evidently,

Ξ(0,χ)0,
(2.11)

otherwise Ξ(z,χ)0, which is clearly false. Thus,

n = 1 ( 1 z z n ( χ ) 2 ) = Ξ ( z , χ ) Ξ ( 0 , χ )
(2.12)

for 0arg(z)<2π. Furthermore,

Ξ ( z 1 4 , χ ) Ξ ( i z 1 4 , χ ) Ξ 2 ( 0 ) = n = 1 ( 1 z z n 4 ( χ ) )
(2.13)

and

Ξ ( z 1 6 , χ ) Ξ ( ρ z 1 6 , χ ) Ξ ( ρ 2 z 1 6 , χ ) Ξ 3 ( 0 ) = n = 1 ( 1 z z n 6 ( χ ) )
(2.14)

for 0arg(z)<2π, where ρ= e 2 π i 3 . Let ρ be a primitive th root of unity for some positive integer ; then we have

j = 1 Ξ ( ρ j z 1 2 , χ ) Ξ ( 0 , χ ) = n = 1 ( 1 z z n 2 ( χ ) ) .
(2.15)

Corollary 3 Assume that the Riemann hypothesis is true for L(s,χ) and z 1 (χ) is the least positive zero of Ξ(z,χ); then the function Ξ ( z , χ ) Ξ ( 0 , χ ) is completely monotonic for z(, z 1 2 (χ)), Ξ ( z 1 4 , χ ) Ξ ( i z 1 4 , χ ) Ξ 2 ( 0 ) is completely monotonic for z(, z 1 4 (χ)), and Ξ ( z 1 6 , χ ) Ξ ( ρ z 1 6 , χ ) Ξ ( ρ 2 z 1 6 , χ ) Ξ 3 ( 0 ) is completely monotonic for z(, z 1 6 (χ)). Let ρ be a primitive ℓth root of unity for some positive integer , then j = 1 Ξ( ρ j z 1 2 ,χ) is completely monotonic for z(, z 1 2 (χ)).

Proof These are consequences of Lemma 1 and equations (2.12)-(2.15). □

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Acknowledgements

This research is partially supported by National Natural Science Foundation of China, grant No. 11371294.

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Correspondence to Ruiming Zhang.

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Zhang, R. On complete monotonicity of the Riemann zeta function. J Inequal Appl 2014, 15 (2014). https://doi.org/10.1186/1029-242X-2014-15

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Keywords

  • Riemann zeta function
  • Dirichlet series
  • Riemann hypothesis
  • complete monotonic functions