# 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 $\zeta \left(s\right)$ can be defined by

$\zeta \left(s\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{s}},\phantom{\rule{1em}{0ex}}\mathrm{\Re }\left(s\right)>1,$
(1.1)

and on the rest of the complex plane by analytic continuation. It is known that the extended $\zeta \left(s\right)$ is meromorphic with infinitely many zeros at $-2n$ for $n\in \mathbb{N}$ (a.k.a trivial zeros) and with infinitely many zeros within the vertical strip $0<\mathrm{\Re }\left(s\right)<1$ (nontrivial zeros). The Riemann hypothesis for $\zeta \left(s\right)$ says that all nontrivial zeros are actually on the critical line $\mathrm{\Re }\left(s\right)=\frac{1}{2}$.

For any complex number $z\in \mathbb{C}$, let $\mathrm{\Gamma }\left(z\right)$ be Euler’s Gamma function defined by [18]

$\frac{1}{\mathrm{\Gamma }\left(z\right)}=z\prod _{j=1}^{\mathrm{\infty }}\left(1+\frac{z}{j}\right){\left(1+\frac{1}{j}\right)}^{-z}.$
(1.2)

Then, the Riemann $\mathrm{\Xi }\left(z\right)$ function [17]

$\mathrm{\Xi }\left(z\right)=-\frac{1+4{z}^{2}}{8}{\pi }^{-\frac{1+2iz}{4}}\mathrm{\Gamma }\left(\frac{1+2iz}{4}\right)\zeta \left(\frac{1+2iz}{2}\right)$
(1.3)

is an even entire function of order 1. The celebrated Riemann hypothesis is equivalent to the statement that $\mathrm{\Xi }\left(z\right)$ has only real zeros.

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

$L\left(s,\chi \right)=\sum _{n=1}^{\mathrm{\infty }}\frac{\chi \left(n\right)}{{n}^{s}},\phantom{\rule{1em}{0ex}}\mathrm{\Re }\left(s\right)>1.$
(1.4)

Let

$\alpha =\left\{\begin{array}{cc}0,\hfill & \chi \left(-1\right)=1,\hfill \\ 1,\hfill & \chi \left(-1\right)=-1,\hfill \end{array}$
(1.5)

then

$\mathrm{\Xi }\left(z,\chi \right)={\left(\frac{\pi }{m}\right)}^{-\left(1+2\alpha +2iz\right)/4}\mathrm{\Gamma }\left(\frac{1+2\alpha +2iz}{4}\right)L\left(\frac{1+2iz}{2},\chi \right)$
(1.6)

is an even entire function of order 1. The Riemann hypothesis for $L\left(s,\chi \right)$ is equivalent to $\mathrm{\Xi }\left(z,\chi \right)$ having only real zeros.

Given real numbers a, b with $a and an indefinite differentiable real valued function $f\left(x\right)$ on $\left(a,b\right)$, $f\left(x\right)$ is called completely monotonic on $\left(a,b\right)$ if ${\left(-1\right)}^{m}{f}^{\left(m\right)}\left(x\right)\ge 0$ for all $x\in \left(a,b\right)$ and $m=0,1,\dots$ . 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

$\sum _{n=1}^{\mathrm{\infty }}|{\lambda }_{n}|<\mathrm{\infty },$
(2.1)

then, the entire function

$f\left(x\right)=\prod _{n=1}^{\mathrm{\infty }}\left(1-x{\lambda }_{n}\right)$
(2.2)

is completely monotonic on $\left(-\mathrm{\infty },{\lambda }_{1}^{-1}\right)$.

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

Assuming the Riemann hypothesis is true, we list all positive zeros of $\mathrm{\Xi }\left(z\right)$ as

${z}_{1}\le {z}_{2}\le \cdots \le {z}_{n}\le \cdots ,$
(2.3)

and ${z}_{1}$ is approximately 14.1347. Then,

$\mathrm{\Xi }\left(z\right)=\mathrm{\Xi }\left(0\right)\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{{z}^{2}}{{z}_{n}^{2}}\right).$
(2.4)

Thus,

$\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{z}{{z}_{n}^{2}}\right)=\frac{\mathrm{\Xi }\left(\sqrt{z}\right)}{\mathrm{\Xi }\left(0\right)},$
(2.5)
$\frac{\mathrm{\Xi }\left({z}^{\frac{1}{4}}\right)\mathrm{\Xi }\left(i{z}^{\frac{1}{4}}\right)}{{\mathrm{\Xi }}^{2}\left(0\right)}=\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{z}{{z}_{n}^{4}}\right),$
(2.6)

and

$\frac{\mathrm{\Xi }\left({z}^{\frac{1}{6}}\right)\mathrm{\Xi }\left(\rho {z}^{\frac{1}{6}}\right)\mathrm{\Xi }\left({\rho }^{2}{z}^{\frac{1}{6}}\right)}{{\mathrm{\Xi }}^{3}\left(0\right)}=\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{z}{{z}_{n}^{6}}\right)$
(2.7)

for $0\le arg\left(z\right)<2\pi$, where $\rho ={e}^{\frac{2\pi i}{3}}$. In fact, for any positive integer $\ell >1$ and assume that ${\rho }_{\ell }$ is a primitive th root of unity; then we have

$\frac{{\prod }_{j=1}^{\ell }\mathrm{\Xi }\left({\rho }_{\ell }^{j}{z}^{\frac{1}{2\ell }}\right)}{{\mathrm{\Xi }}^{\ell }\left(0\right)}=\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{z}{{z}_{n}^{2\ell }}\right).$
(2.8)

Corollary 2 Under the Riemann hypothesis, let ${z}_{1}$ be the least positive zeros of $\mathrm{\Xi }\left(z\right)$; then the function $\mathrm{\Xi }\left(\sqrt{z}\right)$ is completely monotonic for $z\in \left(-\mathrm{\infty },{z}_{1}^{2}\right)$, $\mathrm{\Xi }\left({z}^{\frac{1}{4}}\right)\mathrm{\Xi }\left(i{z}^{\frac{1}{4}}\right)$ is completely monotonic for $z\in \left(-\mathrm{\infty },{z}_{1}^{4}\right)$, and $\mathrm{\Xi }\left({z}^{\frac{1}{6}}\right)\mathrm{\Xi }\left(\rho {z}^{\frac{1}{6}}\right)\mathrm{\Xi }\left({\rho }^{2}{z}^{\frac{1}{6}}\right)$ is completely monotonic for $z\in \left(-\mathrm{\infty },{z}_{1}^{6}\right)$. Let ${\rho }_{\ell }$ be a primitive ℓth root of unity for some positive integer ; then ${\prod }_{j=1}^{\ell }\mathrm{\Xi }\left({\rho }_{\ell }^{j}{z}^{\frac{1}{2\ell }}\right)$ is completely monotonic for $z\in \left(-\mathrm{\infty },{z}_{1}^{2\ell }\right)$.

Proof Notice that $\mathrm{\Xi }\left(0\right)$ is a positive constant, and the claims are obtained by applying Corollary 1 to equations (2.5)-(2.8). □

Assuming the Riemann hypothesis for $L\left(s,\chi \right)$, we list all the positive zeros for $\mathrm{\Xi }\left(z,\chi \right)$ as [8]

${z}_{1}\left(\chi \right)\le {z}_{2}\left(\chi \right)\le \cdots \le {z}_{n}\left(\chi \right)\le \cdots .$
(2.9)

Then

$\mathrm{\Xi }\left(z,\chi \right)=\mathrm{\Xi }\left(0,\chi \right)\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{{z}^{2}}{{z}_{n}{\left(\chi \right)}^{2}}\right).$
(2.10)

Evidently,

$\mathrm{\Xi }\left(0,\chi \right)\ne 0,$
(2.11)

otherwise $\mathrm{\Xi }\left(z,\chi \right)\equiv 0$, which is clearly false. Thus,

$\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{z}{{z}_{n}{\left(\chi \right)}^{2}}\right)=\frac{\mathrm{\Xi }\left(\sqrt{z},\chi \right)}{\mathrm{\Xi }\left(0,\chi \right)}$
(2.12)

for $0\le arg\left(z\right)<2\pi$. Furthermore,

$\frac{\mathrm{\Xi }\left({z}^{\frac{1}{4}},\chi \right)\mathrm{\Xi }\left(i{z}^{\frac{1}{4}},\chi \right)}{{\mathrm{\Xi }}^{2}\left(0\right)}=\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{z}{{z}_{n}^{4}\left(\chi \right)}\right)$
(2.13)

and

$\frac{\mathrm{\Xi }\left({z}^{\frac{1}{6}},\chi \right)\mathrm{\Xi }\left(\rho {z}^{\frac{1}{6}},\chi \right)\mathrm{\Xi }\left({\rho }^{2}{z}^{\frac{1}{6}},\chi \right)}{{\mathrm{\Xi }}^{3}\left(0\right)}=\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{z}{{z}_{n}^{6}\left(\chi \right)}\right)$
(2.14)

for $0\le arg\left(z\right)<2\pi$, where $\rho ={e}^{\frac{2\pi i}{3}}$. Let ${\rho }_{\ell }$ be a primitive th root of unity for some positive integer ; then we have

$\frac{{\prod }_{j=1}^{\ell }\mathrm{\Xi }\left({\rho }_{\ell }^{j}{z}^{\frac{1}{2\ell }},\chi \right)}{{\mathrm{\Xi }}^{\ell }\left(0,\chi \right)}=\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{z}{{z}_{n}^{2\ell }\left(\chi \right)}\right).$
(2.15)

Corollary 3 Assume that the Riemann hypothesis is true for $L\left(s,\chi \right)$ and ${z}_{1}\left(\chi \right)$ is the least positive zero of $\mathrm{\Xi }\left(z,\chi \right)$; then the function $\frac{\mathrm{\Xi }\left(\sqrt{z},\chi \right)}{\mathrm{\Xi }\left(0,\chi \right)}$ is completely monotonic for $z\in \left(-\mathrm{\infty },{z}_{1}^{2}\left(\chi \right)\right)$, $\frac{\mathrm{\Xi }\left({z}^{\frac{1}{4}},\chi \right)\mathrm{\Xi }\left(i{z}^{\frac{1}{4}},\chi \right)}{{\mathrm{\Xi }}^{2}\left(0\right)}$ is completely monotonic for $z\in \left(-\mathrm{\infty },{z}_{1}^{4}\left(\chi \right)\right)$, and $\frac{\mathrm{\Xi }\left({z}^{\frac{1}{6}},\chi \right)\mathrm{\Xi }\left(\rho {z}^{\frac{1}{6}},\chi \right)\mathrm{\Xi }\left({\rho }^{2}{z}^{\frac{1}{6}},\chi \right)}{{\mathrm{\Xi }}^{3}\left(0\right)}$ is completely monotonic for $z\in \left(-\mathrm{\infty },{z}_{1}^{6}\left(\chi \right)\right)$. Let ${\rho }_{\ell }$ be a primitive ℓth root of unity for some positive integer , then ${\prod }_{j=1}^{\ell }\mathrm{\Xi }\left({\rho }_{\ell }^{j}{z}^{\frac{1}{2\ell }},\chi \right)$ is completely monotonic for $z\in \left(-\mathrm{\infty },{z}_{1}^{2\ell }\left(\chi \right)\right)$.

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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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