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ζ\zetaζ has no zeros in the rectangle 0<Re⁡s<10<\operatorname{Re} s<10<Res<1, ∣Im⁡s∣≤5|\operatorname{Im} s|\le 5∣Ims∣≤5

Proved
zeta_ne_zero_of_mem_strip_of_abs_im_le_five

by Gabewhigham · Sep 6, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

analytic-number-theorycomplex-analysisnumber-theoryzeta-functions

Let ζ\zetaζ denote the Riemann zeta function. The assertion is an unconditional, completely explicit zero-free region for the low-lying part of the critical strip, of height 555:

0<Re⁡s<1  and  ∣Im⁡s∣≤5 ⟹ ζ(s)≠0.0<\operatorname{Re} s<1 \ \text{ and } \ |\operatorname{Im} s|\le 5 \ \Longrightarrow\ \zeta(s)\neq 0.0<Res<1  and  ∣Ims∣≤5 ⟹ ζ(s)=0.

This sharpens the height-222 statement zeta_ne_zero_of_mem_strip_of_abs_im_le_two. The first nontrivial zero occurs at height ≈14.13\approx 14.13≈14.13, so the region is still a proper truncation of the search region for zeros, but a larger one.

The proof is a quantitative estimate on the completed zeta function. Write Λ(s)=π−s/2Γ(s/2)ζ(s)\Lambda(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)Λ(s)=π−s/2Γ(s/2)ζ(s) and Λ0(s)=Λ(s)+1s+11−s\Lambda_0(s)=\Lambda(s)+\frac1s+\frac1{1-s}Λ0​(s)=Λ(s)+s1​+1−s1​ for its entire part, which is the Mellin transform

Λ0(s)=12∫0∞f(t) ts/2−1 dt\Lambda_0(s)=\tfrac12\int_0^\infty f(t)\,t^{s/2-1}\,dtΛ0​(s)=21​∫0∞​f(t)ts/2−1dt

of the modified Jacobi theta kernel fff, where f(t)=θ(t)−1f(t)=\theta(t)-1f(t)=θ(t)−1 for t>1t>1t>1 and f(t)=t−1/2(θ(1/t)−1)f(t)=t^{-1/2}\bigl(\theta(1/t)-1\bigr)f(t)=t−1/2(θ(1/t)−1) for 0<t<10<t<10<t<1.

Comparison with a geometric series gives ∣θ(t)−1∣≤52e−πt|\theta(t)-1|\le \tfrac52 e^{-\pi t}∣θ(t)−1∣≤25​e−πt for t≥1t\ge 1t≥1, so for 0<Re⁡s<10<\operatorname{Re} s<10<Res<1 the integrand is dominated by 52e−πt\tfrac52 e^{-\pi t}25​e−πt on (1,∞)(1,\infty)(1,∞) and by 52t−2e−π/t\tfrac52 t^{-2}e^{-\pi/t}25​t−2e−π/t on (0,1](0,1](0,1]. The substitution t↦1/tt\mapsto 1/tt↦1/t identifies the second integral with the first, so

∥Λ0(s)∥≤12⋅2⋅52⋅e−ππ≤128\|\Lambda_0(s)\|\le \tfrac12\cdot 2\cdot \tfrac52\cdot\frac{e^{-\pi}}{\pi}\le \frac1{28}∥Λ0​(s)∥≤21​⋅2⋅25​⋅πe−π​≤281​

using e−π≤1/23e^{-\pi}\le 1/23e−π≤1/23. On the other hand ∥1s+11−s∥=1∥s∥ ∥1−s∥≥126\left\|\frac1s+\frac1{1-s}\right\|=\frac1{\|s\|\,\|1-s\|}\ge \frac1{26}​s1​+1−s1​​=∥s∥∥1−s∥1​≥261​ in the rectangle 0<Re⁡s<10<\operatorname{Re} s<10<Res<1, ∣Im⁡s∣≤5|\operatorname{Im} s|\le 5∣Ims∣≤5. Hence Λ(s)=Λ0(s)−1s−11−s\Lambda(s)=\Lambda_0(s)-\frac1s-\frac1{1-s}Λ(s)=Λ0​(s)−s1​−1−s1​ cannot vanish there, and since π−s/2Γ(s/2)≠0\pi^{-s/2}\Gamma(s/2)\neq0π−s/2Γ(s/2)=0 for Re⁡s>0\operatorname{Re} s>0Res>0, neither can ζ(s)\zeta(s)ζ(s).

Formalization note. riemannZeta is Mathlib's zeta function; |s.im| is the absolute value of the imaginary part of sss.

Preamble
import Mathlib.NumberTheory.LSeries.RiemannZeta
import Mathlib.NumberTheory.LSeries.Nonvanishing

open Complex
Formal statement
theorem zeta_ne_zero_of_mem_strip_of_abs_im_le_five (s : ℂ) (h0 : 0 < s.re) (h1 : s.re < 1)
    (him : |s.im| ≤ 5) : riemannZeta s ≠ 0 := by sorry
Source
Standard theory of the completed Riemann zeta function; the quantitative form given here (bounding the entire part of the completed zeta function against its pole terms, with the substitution t -> 1/t on (0,1)) is elementary. See e.g. Titchmarsh, The Theory of the Riemann Zeta-Function, 2nd ed., Ch. II.

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