TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Zeta bounds in classical zero free region

Statement

There are $0 < c_{2} < c_{1} < c_{0}$ and $C > 0$ such that for $∣ t ∣ \geq 3$ and $σ \geq 1 - c_{1} / \log (∣ t ∣ + 2)$ , $∣ ζ^{'} / ζ (σ + i t) ∣ \leq C \log (∣ t ∣ + 2) \leq C \log^{2} (∣ t ∣ + 2) .$ In the narrower $c_{2}$ region, $∣ 1 / ζ (s) ∣ \leq C \log (∣ t ∣ + 2)$ . For $∣ t ∣ \leq 3$ and $1 - c_{2} / \log (∣ t ∣ + 2) \leq σ \leq 2$ , $ζ^{'} / ζ (s) + 1 / (s - 1) = O (1), 1 / ζ (s) = O (∣ s - 1 ∣),$ with removable interpretations at one.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Zeta horizontal logarithmic derivative comparison: There are absolute $d > 0, C > 0$ , with $d < c_{0}$ , such that for $∣ t ∣ \geq 3$ and $σ \geq 1 - d / \log (∣ t ∣ + 2)$ , $∣ \frac{ζ^{'} (σ + i t)}{ζ (σ + i t)} ∣ \leq C \log (∣ t ∣ + 2) .$

[F2]

Riemann zeta classical zero free region: 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.

[F3]

The Riemann zeta function has its Euler product on the half-plane $Re s > 1$ : For every $s \in C$ with $Re s > 1$ , $ζ (s) = \prod_{p} \frac{1}{1 - p^{- s}},$ where the product ranges over the primes and converges absolutely and locally uniformly on $Re s > 1$ .

[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

1.1

Choose $c_{1}$ smaller than the constant in the horizontal comparison. This gives the stated derivative bound throughout the high-height region.

2.1

At $s_{1} = 1 + 1 / L + i t$ , $L = \log (∣ t ∣ + 2)$ , the Euler logarithm satisfies $∣ \log ζ (s_{1}) ∣ \leq \log ζ (1 + 1 / L) \leq \log (1 + L)$ , by comparing the positive real zeta series to its integral. Integrate $ζ^{'} / ζ$ from $s_{1}$ horizontally to $s$ for $1 - c_{2} / L \leq σ \leq 1 + 1 / L$ . The length is $O (1 / L)$ and the integrand is $O (L)$ , so the change in the continued logarithm is $O (1)$ . Exponentiating its negative real part gives $∣ 1 / ζ (s) ∣ = O (L)$ . For larger sigma the Euler logarithm already gives that bound.

F3step 1.1

3.1

On the compact low-height portion choose $c_{2} < c_{1}$ sufficiently small that $h (s) = (s - 1) ζ (s)$ is holomorphic and nonvanishing on a neighborhood, including $h (1) = 1$ . Then $h^{'} / h$ and $1 / h$ are bounded there. The identities $ζ^{'} / ζ + 1 / (s - 1) = h^{'} / h$ and $1 / ζ = (s - 1) / h$ prove both low-height estimates and their removable interpretations.

F2F4∎

Depends on

Used by

Chebyshev psi prime number theorem error Theorem

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theorem 6.7, pp.174–175 (standard reference, not scraped)