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 $\zeta (s)$ can be defined by

$$\zeta (s)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s},\mathrm{\Re }(s)>1,$$

(1.1)

and on the rest of the complex plane by analytic continuation. It is known that the extended $\zeta (s)$ 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 $0<\mathrm{\Re }(s)<1$ (nontrivial zeros). The Riemann hypothesis for $\zeta (s)$ says that all nontrivial zeros are actually on the critical line $\mathrm{\Re }(s)=\frac{1}{2}$.

For any complex number $z\in \mathbb{C}$, let $\mathrm{\Gamma }(z)$ be Euler’s Gamma function defined by [1–8]

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

(1.2)

Then, the Riemann $\mathrm{\Xi }(z)$ function [1–7]

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

(1.3)

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

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

$$L(s,\chi )=\sum _{n=1}^{\mathrm{\infty }}\frac{\chi (n)}{n^s},\mathrm{\Re }(s)>1.$$

(1.4)

Let

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

(1.5)

then

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

(1.6)

is an even entire function of order 1. The Riemann hypothesis for $L(s,\chi )$ is equivalent to $\mathrm{\Xi }(z,\chi )$ 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 ${(-1)}^m{f}^{(m)}(x)\ge 0$ for all $x\in (a,b)$ 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

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

(2.2)

is completely monotonic on $(-\mathrm{\infty },{\lambda }_1^{-1})$.

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

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

$$z_1\le z_2\le \cdots \le z_n\le \cdots ,$$

(2.3)

and $z_1$ is approximately 14.1347. Then,

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

(2.4)

Thus,

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

(2.5)

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

(2.6)

and

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

(2.7)

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

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

(2.8)

Corollary 2 Under the Riemann hypothesis, let $z_1$ be the least positive zeros of $\mathrm{\Xi }(z)$; then the function $\mathrm{\Xi }(\sqrt{z})$ is completely monotonic for $z\in (-\mathrm{\infty },z_1^2)$, $\mathrm{\Xi }(z^{\frac{1}{4}})\mathrm{\Xi }(i{z}^{\frac{1}{4}})$ is completely monotonic for $z\in (-\mathrm{\infty },z_1^4)$, and $\mathrm{\Xi }(z^{\frac{1}{6}})\mathrm{\Xi }(\rho z^{\frac{1}{6}})\mathrm{\Xi }({\rho }^2{z}^{\frac{1}{6}})$ is completely monotonic for $z\in (-\mathrm{\infty },z_1^6)$. Let ${\rho }_{\ell }$ be a primitive ℓth root of unity for some positive integer ℓ; then ${\prod }_{j=1}^{\ell }\mathrm{\Xi }({\rho }_{\ell }^j{z}^{\frac{1}{2\ell }})$ is completely monotonic for $z\in (-\mathrm{\infty },z_1^{2\ell })$.

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

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

$$z_1(\chi )\le z_2(\chi )\le \cdots \le z_n(\chi )\le \cdots .$$

(2.9)

Then

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

(2.10)

Evidently,

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

(2.11)

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

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

(2.12)

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

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

(2.13)

and

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

(2.14)

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

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

(2.15)

Corollary 3 Assume that the Riemann hypothesis is true for $L(s,\chi )$ and $z_1(\chi )$ is the least positive zero of $\mathrm{\Xi }(z,\chi )$; then the function $\frac{\mathrm{\Xi }(\sqrt{z},\chi )}{\mathrm{\Xi }(0,\chi )}$ is completely monotonic for $z\in (-\mathrm{\infty },z_1^2(\chi ))$, $\frac{\mathrm{\Xi }(z^{\frac{1}{4}},\chi )\mathrm{\Xi }(i{z}^{\frac{1}{4}},\chi )}{{\mathrm{\Xi }}^2(0)}$ is completely monotonic for $z\in (-\mathrm{\infty },z_1^4(\chi ))$, and $\frac{\mathrm{\Xi }(z^{\frac{1}{6}},\chi )\mathrm{\Xi }(\rho z^{\frac{1}{6}},\chi )\mathrm{\Xi }({\rho }^2{z}^{\frac{1}{6}},\chi )}{{\mathrm{\Xi }}^3(0)}$ is completely monotonic for $z\in (-\mathrm{\infty },z_1^6(\chi ))$. Let ${\rho }_{\ell }$ be a primitive ℓth root of unity for some positive integer ℓ, then ${\prod }_{j=1}^{\ell }\mathrm{\Xi }({\rho }_{\ell }^j{z}^{\frac{1}{2\ell }},\chi )$ is completely monotonic for $z\in (-\mathrm{\infty },z_1^{2\ell }(\chi ))$.

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