Open Access

On complete monotonicity of the Riemann zeta function

Journal of Inequalities and Applications20142014:15

https://doi.org/10.1186/1029-242X-2014-15

Received: 29 September 2013

Accepted: 13 December 2013

Published: 9 January 2014

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.

Keywords

Riemann zeta functionDirichlet seriesRiemann hypothesiscomplete monotonic functions

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 2 n for n N (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 z C , 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 ( 1 x 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 0 arg ( z ) < 2 π , 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 0 arg ( z ) < 2 π . 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 0 arg ( z ) < 2 π , 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). □

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
College of Science, Northwest A&F University

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Copyright

© Zhang; licensee Springer. 2014

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