# 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

$ζ(s)= ∑ n = 1 ∞ 1 n s ,ℜ(s)>1,$
(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 $n∈N$ (a.k.a trivial zeros) and with infinitely many zeros within the vertical strip $0<ℜ(s)<1$ (nontrivial zeros). The Riemann hypothesis for $ζ(s)$ says that all nontrivial zeros are actually on the critical line $ℜ(s)= 1 2$.

For any complex number $z∈C$, let $Γ(z)$ be Euler’s Gamma function defined by 

$1 Γ ( z ) =z ∏ j = 1 ∞ ( 1 + z j ) ( 1 + 1 j ) − z .$
(1.2)

Then, the Riemann $Ξ(z)$ function 

$Ξ(z)=− 1 + 4 z 2 8 π − 1 + 2 i z 4 Γ ( 1 + 2 i z 4 ) ζ ( 1 + 2 i z 2 )$
(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]

$L(s,χ)= ∑ n = 1 ∞ χ ( n ) n s ,ℜ(s)>1.$
(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 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 . □

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

$∏ j = 1 ℓ Ξ ( ρ ℓ j z 1 2 ℓ ) Ξ ℓ ( 0 ) = ∏ n = 1 ∞ ( 1 − z z n 2 ℓ ) .$
(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 $ρ ℓ$ be a primitive ℓth root of unity for some positive integer ; then $∏ j = 1 ℓ Ξ( ρ ℓ j z 1 2 ℓ )$ is completely monotonic for $z∈(−∞, z 1 2 ℓ )$.

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 

$z 1 (χ)≤ z 2 (χ)≤⋯≤ z n (χ)≤⋯.$
(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 $0≤arg(z)<2π$. 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 $0≤arg(z)<2π$, where $ρ= e 2 π i 3$. Let $ρ ℓ$ be a primitive th root of unity for some positive integer ; then we have

$∏ j = 1 ℓ Ξ ( ρ ℓ j z 1 2 ℓ , χ ) Ξ ℓ ( 0 , χ ) = ∏ n = 1 ∞ ( 1 − z z n 2 ℓ ( χ ) ) .$
(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 $ρ ℓ$ be a primitive ℓth root of unity for some positive integer , then $∏ j = 1 ℓ Ξ( ρ ℓ j z 1 2 ℓ ,χ)$ is completely monotonic for $z∈(−∞, z 1 2 ℓ (χ))$.

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.

## Author information

Authors

### Corresponding author

Correspondence to Ruiming Zhang.

### Competing interests

The author declares that they have no competing interests.

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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 