New Equivalent of the Riemann Hypothesis

Preprint

Article

New Equivalent of the Riemann Hypothesis

Altmetrics

Downloads

Views

Comments

Leonardo De Lima^*

This version is not peer-reviewed

Submitted:

11 December 2023

Posted:

28 December 2023

You are already at the latest version

Alerts

Abstract

In this article, it is demonstrated that if the zeta function does not have a sequence of zeros whose real part converges to 1, then it cannot have any zeros in the critical strip, showing that the Riemann Hypothesis is false.

Keywords:

Subject: Computer Science and Mathematics - Mathematics

1. Introduction

The distribution of zeros of the Riemann zeta function is one of the great mysteries of modern mathematics. Its significance is related to the estimation of the error in the prime number theorem and a set of equivalent results (see [1,2,3,4]).

One of the fundamental issues in the theory of the zeta function is the delimitation of a zero-free region in the critical strip. The first promising result in this direction was established by Vallée Poussin (see [5]), but this and recent findings have been unable to demarcate a zero-free region of the form

ℜ (s) > 1 - ϵ

(see [6]).

In this work, I will demonstrate that the absence of a set of zeros in the zeta function, whose real part converges to 1, implies the absence of zeros along the critical strip.

2. Fundamental Theorems

In this section, some theorems used throughout the article will be listed.

Theorem 1.

φ (s)

is analytic in the strip

a < ℜ (s) < b

, and if it tends to zero uniformly as

I m (s) \to \pm \infty

for any real value c between a and b, with its integral along such a line converging absolutely, then if

f (x) = M^{- 1} φ = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} x^{- s} φ (s) d s,

we have that

φ (s) = M f = \int_{0}^{\infty} x^{s - 1} f (x) d x .

Conversely, suppose

f (x)

is piecewise continuous on the positive real numbers, taking a value halfway between the limit values at any jump discontinuities, and suppose the integral

φ (s) = \int_{0}^{\infty} x^{s - 1} f (x) d x

is absolutely convergent when

a < ℜ (s) < b

. Then f is recoverable via the inverse Mellin transform from its Mellin transform φ.

Proof.

[7] □

Theorem 2.

ℜ (s) > 1

, we have:

\frac{1}{ζ (s)} = \sum_{n = 1}^{\infty} \frac{μ (n)}{n^{s}}

If the zeta function has no zeros in the region

ℜ (s) > ρ

, we can extend the equality above to such a region.

Proof.

[8] □

Theorem 3.

\sum_{n = 1}^{\infty} \frac{μ (n)}{n} = 0

If the zeta function is free of zeros in

ℜ (s) > ρ

, we have:

\sum_{n = 1}^{x} \frac{μ (n)}{n} = O (\frac{1}{x^{1 - ρ - ϵ}})

for all

ϵ > 0

Proof.

[8] □

Theorem 4.

0 < ℜ (s) < 1

, we have:

- \frac{ζ (s)}{s} = \int_{0}^{\infty} \frac{{x}}{x^{s + 1}} d x

Proof.

[5] □

Theorem 5.

For any natural number

n > 1

, the sum of the values of the Möbius function

μ (d)

over all positive divisors of n is given by:

\sum_{d ∣ n} μ (d) = \{\begin{matrix} 1 & if n = 1, \\ 0 & if n > 1 . \end{matrix}

Proof.

[9] □

Theorem 6.

It is stated that

θ_{n} (x) = \int_{0}^{x} ϕ_{n} (u) d u,

where

ϕ_{n} (x) = \int_{0}^{x} {n u} \frac{μ (n)}{n} d u .

Then:

\sum_{n = 1}^{\infty} θ_{n} (x) = \frac{1}{2 π^{2}} (\frac{sin (2 π x)}{2 π} - x)

and convergence occurs uniformly.

Proof.

Using the fact that, for every x, we have

{x} = \frac{1}{2} - \frac{1}{π} \sum_{n = 1}^{\infty} \frac{sin (2 π n x)}{n} .

(1)

It concludes that

ϕ_{n} (x) = \frac{x}{2} \frac{μ (n)}{n} + \frac{μ (n)}{2 n^{2} π^{2}} \sum_{k = 1}^{\infty} \frac{cos (2 π n k x) - 1}{k^{2}} .

(2)

Performing another integration we obtain the following expression:

θ_{n} (x) = \frac{x^{2}}{4} \frac{μ (n)}{n} + \frac{μ (n)}{4 n^{3} π^{3}} \sum_{k = 1}^{\infty} \frac{sin (2 π n k x)}{k^{3}} - \frac{x}{2 π^{2}} \frac{μ (n)}{n^{2}} \sum_{k = 1}^{\infty} \frac{1}{k^{2}} .

(3)

Thus

\sum_{n = 1}^{\infty} θ_{n} (x) = \frac{x^{2}}{4} \sum_{n = 1}^{\infty} \frac{μ (n)}{n} - \frac{x}{2 π^{2}} \sum_{n = 1}^{\infty} \frac{μ (n)}{n^{2}} \sum_{k = 1}^{\infty} \frac{1}{k^{2}} + \frac{1}{4 π^{3}} \sum_{n, k = 1}^{\infty} \frac{sin (2 π n k x) μ (n)}{n^{3} k^{3}} .

(4)

Rearranging the series, it follows that

\sum_{n, k = 1}^{\infty} \frac{sin (2 π n k x) μ (n)}{n^{3} k^{3}} = \sum_{l = 1}^{\infty} \frac{sin (2 π l x)}{l^{3}} \sum_{n | l} μ (n) = sin (2 π x) .

(5)

(The rearrangement of the summations is justified by the uniform convergence of the series).

Finally, it can be observed that:

\sum_{n = 1}^{\infty} \frac{μ (n)}{n} = 0

(6)

\sum_{n = 1}^{\infty} \frac{μ (n)}{n^{3}} \sum_{k = 1}^{\infty} \frac{1}{k^{3}} = 1,

(7)

concluding the proof. □

Theorem 7.

Suppose the Riemann zeta function has no zeros in the region

1 > R e (s) > ρ

. In this case, for every

0 < γ < 1 - ρ

, we have

ψ_{γ} (M) = max {∥ \frac{1}{(γ + 2 + i t)} \sum_{n = M}^{\infty} \frac{μ (n)}{n^{1 - γ - i t}} ∥, t \in R};

lim_{M \to \infty} ψ_{γ} (M) = 0 .

Proof.

Since

\sum_{n = k}^{\infty} \frac{μ (n)}{n^{s}} = \frac{M (k)}{k^{s}} - s \int_{k}^{\infty} \frac{M (x)}{x^{s + 1}} d x

(8)

where

M (x) = \sum_{n = 1}^{x} μ (n) .

(9)

The fact that the zeta function has no zeros in

1 > ℜ (s) > ρ

is equivalent to

M (x) = O (x^{ρ})

, therefore, we conclude that:

lim_{M \to \infty} ψ_{γ} (M) = 0

(10)

0 < γ < 1 - ρ

. □

3. Proof

Theorem 8.

If the zeta function does not exhibit a sequence of zeros whose real part converges to 1, then the zeta function has no zeros along the critical strip.

Proof.

By Theorem (2.4), we have:

- \frac{ζ (s)}{s} = \int_{0}^{\infty} \frac{{x}}{x^{s + 1}} d x

(11)

Making the variable change

x = n \cdot y

in (3.1), we obtain:

- \frac{ζ (s)}{s} n^{s} = \int_{0}^{\infty} \frac{{n x}}{x^{s + 1}} d x

(12)

We then conclude:

- \frac{ζ (s)}{s} \sum_{n = 1}^{M} \frac{μ (n)}{n^{1 - s}} = \int_{0}^{\infty} \sum_{n = 1}^{M} \frac{μ (n)}{n} \frac{{n x}}{x^{s + 1}} d x

(13)

Performing two integrations by parts in (3.3), we obtain:

- \frac{ζ (s)}{s (s + 1) (s + 2)} \sum_{n = 1}^{M} \frac{μ (n)}{n^{1 - s}} = \int_{0}^{\infty} \sum_{n = 1}^{M} \frac{θ_{n} (x)}{x^{s + 3}} d x,

(14)

where

θ_{n} (x) = \int_{0}^{x} ϕ_{n} (u) d u,

(15)

and

ϕ_{n} (x) = \int_{0}^{x} {n \cdot u} \frac{μ (n)}{n} d u .

(16)

Using the inverse Mellin transform in (3.4), we have:

\sum_{n = 1}^{M} θ_{n} (x) = - \int_{σ - i \infty}^{σ + i \infty} x^{s + 2} \frac{ζ (s)}{s (s + 1) (s + 2)} \sum_{n = 1}^{M} \frac{μ (n)}{n^{1 - s}} d s .

(17)

Similarly, we can conclude:

\sum_{n = M}^{M + P} θ_{n} (x) = - \int_{σ - i \infty}^{σ + i \infty} x^{s + 2} \frac{ζ (s)}{s (s + 1) (s + 2)} \sum_{n = M}^{M + P} \frac{μ (n)}{n^{1 - s}} d s .

(18)

Note that, by the Cauchy theorem, the integral is independent of the value of

σ

, provided that

σ \in (0, 1)

We will now make the following hypothesis:

Hypothesis 1.Suppose there exists a ρ such that

1 - ϵ > ℜ (ρ) > ϵ > 0

, where

ϵ < \frac{1}{2}

, and

ζ (ρ) = 0

. In other words, let’s assume that the zeta function has a zero within the critical strip.

In this case, by equation (3.4), we have:

\int_{0}^{\infty} \sum_{n = 1}^{M} \frac{θ_{n} (x)}{x^{ρ + 3}} d x = 0 .

(19)

Dividing this integral into two parts, we obtain:

\int_{0}^{1} \sum_{n = 1}^{M} \frac{θ_{n} (x)}{x^{ρ + 3}} d x + \int_{1}^{\infty} \sum_{n = 1}^{M} \frac{θ_{n} (x)}{x^{ρ + 3}} d x = 0 .

(20)

By Theorem (2.6), we have:

\int_{1}^{\infty} \sum_{n = 1}^{M} \frac{θ_{n} (x)}{x^{ρ + 3}} d x \to \int_{1}^{\infty} \frac{1}{x^{ρ + 3}} \frac{1}{2 π^{2}} (\frac{sin (2 π x)}{2 π} - x) d x,

(21)

M \to \infty

.(See lemma (1))

For the first integral, it is initially observed that:

\int_{0}^{1} \sum_{n = M}^{M + P} \frac{θ_{n} (x)}{x^{ρ + 3}} d x = - \int_{0}^{1} (\int_{σ - i \infty}^{σ + i \infty} x^{s - ρ - 1} \frac{ζ (s)}{s (s + 1) (s + 2)} \sum_{n = M}^{M + P} \frac{μ (n)}{n^{1 - s}} d s) d x .

(22)

Choosing

1 > σ > ℜ (ρ)

, we get:

\int_{0}^{1} \sum_{n = M}^{M + P} \frac{θ_{n} (x)}{x^{ρ + 3}} d x = - \int_{σ - i \infty}^{σ + i \infty} \frac{ζ (s)}{s (s - ρ) (s + 1) (s + 2)} \sum_{n = M}^{M + P} \frac{μ (n)}{n^{1 - s}} d s .

(23)

(The permutation of integrals is allowed by the uniform convergence of the line integral).

It is noted that, because

ζ (ρ) = 0

, the integrand in (3.13) has no poles in

0 < σ < 1

. Thus, by the Cauchy theorem, we can take

σ

as close to zero as desired.

We will now make another hypothesis:

Hypothesis 2.Suppose there exists an

1 > ϵ > 0

, such that the Riemann zeta function has no zeros for

1 > ℜ (s) > 1 - ϵ

In this case, by taking

σ

sufficiently close to 0 and

σ > 0

in (3.13), we can conclude by Theorem (2.7) :

\int_{0}^{1} \sum_{n = M}^{\infty} \frac{θ_{n} (x)}{x^{ρ + 3}} d x \to 0,

(24)

M \to \infty

.(See lemma (2))

In summary, we have proven:

\int_{0}^{\infty} \frac{1}{x^{ρ + 3}} \frac{1}{2 π^{2}} (\frac{sin (2 π x)}{2 π} - x) d x = 0,

(25)

or equivalently:

\int_{0}^{\infty} \frac{1}{x^{ρ + 1}} \frac{sin (2 π x)}{π} d x = 0 .

(26)

But this cannot happen, because:

\int_{0}^{\infty} \frac{1}{x^{ρ + 1}} \frac{sin (2 π x)}{π} d x = π^{\frac{1}{2} - ρ} \frac{Γ (\frac{1 - ρ}{2})}{Γ (\frac{ρ}{2})} = \frac{ζ (ρ)}{ζ (1 - ρ) ρ} .

(27)

Therefore, we conclude that if hypothesis 2 is true, hypothesis 1 cannot occur, which means that the Riemann hypothesis is false as the zeta function has an infinite set of zeros along the critical line. □

Corollary 1.

The Riemann Hypothesis is false.

Lemma 1.

0 < ℜ (ρ) < 1

, then:

\int_{1}^{\infty} \sum_{n = 1}^{M} \frac{θ_{n} (x)}{x^{ρ + 3}} d x \to \int_{1}^{\infty} \frac{1}{x^{ρ + 3}} \frac{1}{2 π^{2}} (\frac{sin (2 π x)}{2 π} - x) d x,

(28)

M \to \infty

Proof.

By Theorem (2.6), we have:

\sum_{n = 1}^{\infty} θ_{n} (x) = \frac{1}{2 π^{2}} (\frac{sin (2 π x)}{2 π} - x)

(29)

and

\sum_{n = M}^{\infty} θ_{n} (x) = \frac{x^{2}}{4} \sum_{n = M}^{\infty} \frac{μ (n)}{n} - \frac{x}{2 π^{2}} \sum_{n = M}^{\infty} \frac{μ (n)}{n^{2}} \sum_{k = 1}^{\infty} \frac{1}{k^{2}} + \frac{1}{4 π^{3}} \sum_{n = M, k = 1}^{\infty} \frac{sin (2 π n k x) μ (n)}{n^{3} k^{3}} .

(30)

Note that the last two terms of (3.20) go to zero uniformly as

M \to \infty

. Moreover, they are bounded by

O (x)

, so it follows that the integral of this term divided by

x^{ρ + 3}

over the interval

[1, \infty)

tends to zero uniformly as

M \to \infty

For the first term, it suffices to observe that its integral is (in case

ℜ (ρ) > 0

\int_{1}^{\infty} \frac{x^{- 1 - ρ}}{4} \sum_{n = M}^{\infty} \frac{μ (n)}{n} = \frac{1}{4 ρ} \sum_{n = M}^{\infty} \frac{μ (n)}{n} = O (\frac{1}{ln M})

(31)

□

Lemma 2.

ζ (ρ) = 0

and there exists an

1 > ϵ > 0

such that for

1 > ℜ (s) > 1 - ϵ

, the zeta function has no zeros, then:

\int_{0}^{1} \sum_{n = M}^{\infty} \frac{θ_{n} (x)}{x^{ρ + 3}} d x \to 0,

(32)

M \to \infty

This implies that:

\int_{0}^{1} \sum_{n = 1}^{M} \frac{θ_{n} (x)}{x^{ρ + 3}} d x \to \int_{0}^{1} \frac{1}{x^{ρ + 3}} \frac{1}{2 π^{2}} (\frac{sin (2 π x)}{2 π} - x) d x,

(33)

M \to \infty

Proof.

By equation (3.13):

\int_{0}^{1} \sum_{n = M}^{M + P} \frac{θ_{n} (x)}{x^{ρ + 3}} d x = - \int_{σ - i \infty}^{σ + i \infty} \frac{ζ (s)}{s (s - ρ) (s + 1) (s + 2)} \sum_{n = M}^{M + P} \frac{μ (n)}{n^{1 - s}} d s .

(34)

The contour integral (3.24) does not depend on

σ

0 < σ < 1

, because the integrand is analytic, given that by hypothesis

ζ (ρ) = 0

Choosing a

σ

sufficiently close to zero, we can make

\sum_{n = M}^{M + P} \frac{μ (n)}{n^{1 - s}}

(35)

converge, because there exists an

ϵ

such that for

1 > ℜ (s) > 1 - ϵ \to ζ (s) \neq 0

, by hypothesis.

Thus, we have:

\int_{0}^{1} \sum_{n = M}^{\infty} \frac{θ_{n} (x)}{x^{ρ + 3}} d x = - \int_{σ - i \infty}^{σ + i \infty} \frac{ζ (s)}{s (s - ρ) (s + 1) (s + 2)} \sum_{n = M}^{\infty} \frac{μ (n)}{n^{1 - s}} d s .

(36)

And by the notation of Theorem (2.7):

∥ \int_{σ - i \infty}^{σ + i \infty} \frac{ζ (s)}{s (s - ρ) (s + 1) (s + 2)} \sum_{n = M}^{\infty} \frac{μ (n)}{n^{1 - s}} d s ∥ < ψ_{σ} (M) \int_{σ - i \infty}^{σ + i \infty} ∥ \frac{ζ (s)}{s (s - ρ) (s + 1)} ∥ d t

(37)

Hence, it follows that

\int_{0}^{1} \sum_{n = M}^{\infty} \frac{θ_{n} (x)}{x^{ρ + 3}} d x \to 0,

(38)

M \to \infty

For the second statement of the lemma, just note that:

\int_{0}^{1} \sum_{n = 1}^{M} \frac{θ_{n} (x)}{x^{ρ + 3}} d x = \int_{0}^{1} \frac{1}{x^{ρ + 3}} \frac{1}{2 π^{2}} (\frac{sin (2 π x)}{2 π} - x) d x - \int_{0}^{1} \sum_{n = M}^{\infty} \frac{θ_{n} (x)}{x^{ρ + 3}} d x

(39)

and use (3.28). □

References

Conrey, J.B. Riemann’s Hypothesis; American Mathematical Society: Providence, RI, USA, 2003. [Google Scholar]
Choie, Y.; Lichiardopol, N.; Moree, P.; Solé, P. On Robin’s criterion for the Riemann hypothesis. Journal de théorie des nombres de Bordeaux 2007, 19, 357–372. [Google Scholar] [CrossRef]
Borwein, P.; Choi, S.; Rooney, B.; Weirathmueller, A. The Riemann Hypothesis: A Resource for the Afficionado and Virtuoso Alike; CMS Books in Mathematics / Springer: New York, NY, USA, 2008. [Google Scholar]
Schumayer, D.; Hutchinson, D.A.W. Colloquium: Physics of the Riemann hypothesis. Rev. Mod. Phys. 2011, 83, 307–330. [Google Scholar] [CrossRef]
Karatsuba, A.; Voronin, S.; Koblitz, N. The Riemann Zeta-Function; De Gruyter: Berlin, Germany, 2011. [Google Scholar]
Yang, A. Explicit bounds on ζ(s) in the critical strip and a zero-free region. arXiv preprint arXiv:2301.03165, arXiv:2301.03165 2023.
Lang, S. Complex Analysis; Springer New York: New York, NY, 1985. [Google Scholar]
Titchmarsh, E. The Theory of the Riemann Zeta-Function; Oxford University Press: Oxford, UK, 1986. [Google Scholar]
Apostol, T.M. Introduction to Analytic Number Theory; Springer: New York, NY, USA, 1976. [Google Scholar]

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

© 2020 by the author. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https://creativecommons.org/licenses/by/4.0/).

Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.