TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-24 (gpt-6-sol)

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Riemann zeta classical zero free region

Statement

Assume countable choice.

There is an absolute $c_{0} > 0$ such that $ζ$ has no zeros in $σ \geq 1 - c_{0} / \log (∣ t ∣ + 2)$ . The pole at $s = 1$ is not a zero.

Facts & Assumptions

Given: Countable choice and the data and hypotheses of the statement.

[F1]

Zeta logarithmic derivative zero bound: Write $s = σ + i t$ and let $ρ$ range over nontrivial zeta zeros with multiplicity. With the Hadamard constant $B$ , $\frac{ζ^{'}}{ζ} (s) = B + \sum_{ρ} (\frac{1}{s - ρ} + \frac{1}{ρ}) - \frac{1}{s - 1} + \frac{\log π}{2} - \frac{Γ^{'} (1 + s / 2)}{2 Γ (1 + s / 2)} .$ This is a meromorphic identity, using convergent genus-one terms. Uniformly for $- 1 \leq σ \leq 2$ , $∣ t ∣ \geq 3$ and $ζ (s) \neq 0$ , $Re \frac{ζ^{'}}{ζ} (s) = \sum_{ρ} Re \frac{1}{s - ρ} - \frac{1}{2} \log ∣ t ∣ + O (1),$ and this real series is absolutely convergent.

[F2]

Zeta three four one logarithmic derivative inequality: For $σ > 1$ and $t \in R$ , $- 3 \frac{ζ^{'} (σ)}{ζ (σ)} - 4 Re \frac{ζ^{'} (σ + i t)}{ζ (σ + i t)} - Re \frac{ζ^{'} (σ + 2 i t)}{ζ (σ + 2 i t)} \geq 0.$

[F3]

The Riemann zeta function has no zeros on the closed half-plane $Re s \geq 1$ , except for its pole at $1$ : The meromorphic continuation of $ζ$ has no zeros on the closed half-plane $Re s \geq 1$ . Its only singularity there is the simple pole at $s = 1$ .

[F4]

For $Re s > 0$ , zeta admits the fractional-part integral formula with a simple residue-one pole at $1$ : For every complex number $s$ with $Re s > 0$ and $s \neq 1$ , $ζ (s) = \frac{s}{s - 1} - s \int_{1}^{\infty} {x} x^{- s - 1} d x,$ where ${x} = x - ⌊ x ⌋$ is the fractional part. The integral defines a holomorphic function on $Re s > 0$ , so the right-hand side is meromorphic there with a single simple pole at $s = 1$ of residue $1$ .

Proof

1.1F1F4

Under countable choice, the supplied zero-bound identity applies. For $1 < σ \leq 2$ , the simple pole gives $- ζ^{'} / ζ (σ) = 1 / (σ - 1) + O (1)$ . If $ρ_{0} = β + i γ$ is a zero with $∣ γ ∣ \geq 3$ , positivity of the real zero summands gives $- Re (ζ^{'} / ζ) (σ + i γ) \leq 1 ⁣ \cdot C_{1} \log (∣ γ ∣ + 2) - 1 / (σ - β)$ and $- Re (ζ^{'} / ζ) (σ + 2 i γ) \leq C_{1} \log (∣ γ ∣ + 2)$ .

2.1F2step 1.1

Insert these bounds in the three-four-one inequality. For an absolute $C \geq 1$ , put $L = \log (∣ γ ∣ + 2)$ ; then $4 / (σ - β) \leq 3 / (σ - 1) + C L$ . Taking $σ - 1 = 1 / (2 C L)$ yields $1 - β \geq 1 / (14 C L)$ . Choosing a strictly smaller constant excludes even the closed boundary of the claimed high-height region.

3.1F3F4step 2.1∎

The function $h (s) = (s - 1) ζ (s)$ is holomorphic near the compact segment ${1 + i t : ∣ t ∣ \leq 3}$ , nonzero there, and $h (1) = 1$ . Finitely many nonvanishing neighborhoods cover this segment and contain a uniform thin rectangle about it. Shrink $c_{0}$ so the proposed bounded-height region to the left of one lies in that rectangle. To the right use the already proved zero-free half-plane. This proves the claim at every height, including zero.

Depends on

Used by

Zeta zero count near the one line Corollary
Zero free region parameter balance Example
Zeta explicit formula zero free error balance Lemma
Zeta horizontal logarithmic derivative comparison Lemma
The classical zeta region is not a uniform dirichlet l region Remark
Zeta bounds in classical zero free region Theorem

Dependency tree · two levels

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Sources

Theorem 6.6, pp.172–173 (standard reference, not scraped)