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

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Zeta zero count near the one line

Statement

Assume countable choice.

For $t \in R$ and $0 < r \leq 3 / 4$ , let $n (r; t)$ count nontrivial zeros with $∣ ρ - (1 + i t) ∣ \leq r$ , including multiplicity. Then $n (r; t) = O (r \log (∣ t ∣ + 2))$ , uniformly.

Facts & Assumptions

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

[F1]

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.

[F2]

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.

[F3]

The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1: For $ℜ s > 1$ , if $ζ (s) : = \sum_{n \geq 1} n^{- s},$ then $- \frac{ζ^{'} (s)}{ζ (s)} = \sum_{n \geq 1} Λ (n) n^{- s} .$

[F4]

A unit-interval bound for zeta zeros: The number of nontrivial zeta zeros, with multiplicity, whose ordinates lie in $[T, T + 1]$ is $O (\log (T + 2))$ for $T \geq 0$ .

Proof

1.1F1

Under countable choice, the supplied zero-free region applies. Write $L = \log (∣ t ∣ + 2)$ . For a sufficiently small absolute $a > 0$ , $r < a / L$ makes the disc zero-free: within it $\log (∣ Im ρ ∣ + 2) \leq K L$ , whereas $1 - Re ρ \leq r$ . This contradicts the region bound if a zero occurs.

2.1F2F3step 1.1

For $∣ t ∣ \geq 3$ and $a / L \leq r \leq 1 / 6$ , evaluate at $s_{1} = 1 + r + i t$ . The Euler series gives $∣ ζ^{'} / ζ (s_{1}) ∣ = O (1 / r)$ , using its simple-pole expansion on the real axis. The positive real zero sum is thus $O (1 / r + L)$ . Every counted zero contributes at least $r / (4 r^{2} + r^{2}) = 1 / (5 r)$ , since its real separation is between r and 2r and its imaginary separation at most r. Hence $n (r; t) = O (1 + r L) = O (r L)$ .

3.1F4step 1.1∎

If $r \geq 1 / 6$ and $∣ t ∣ \geq 3$ , finitely many adjacent unit ordinate bands give $n (r; t) = O (L) = O (r L)$ . Negative bands have the same count by conjugation of zeta. For $∣ t ∣ \leq 3$ , all counted zeros lie in one compact rectangle and are finite in number, while a nonempty disc must have $r \geq a / L \geq a / \log 5$ . Enlarging the constant handles these remaining cases.

Depends on

Used by

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Sources

Theorem 6.8, p.175 (standard reference, not scraped)