Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Zeta logarithmic derivative zero bound

Statement

Write s=σ+it and let ρ range over nontrivial zeta zeros with multiplicity. With the Hadamard constant B, ζζ(s)=B+ρ(1sρ+1ρ)1s1+logπ2Γ(1+s/2)2Γ(1+s/2). This is a meromorphic identity, using convergent genus-one terms. Uniformly for 1σ2, t3 and ζ(s)0, Reζζ(s)=ρRe1sρ12logt+O(1), and this real series is absolutely convergent.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

The Riemann xi function has its genus-one Hadamard product over the nontrivial zeros of zeta: There exist constants A,BC such that ξ(s)=eA+BsρE1(s/ρ), where the product runs over the nontrivial zeros ρ of ζ, counted with multiplicity, and E1(w)=(1w)ew. The product converges in the genus-one canonical sense.

[F2]

The Riemann xi function ξ(s)=12s(s1)Λ(s): The completed function extends meromorphically with simple poles at 0 and 1 by thm-completed-riemann-zeta-functional-equation. The Riemann xi function is defined on C by ξ(s):=12s(s1)Λ(s)=12s(s1)πs/2Γ(s/2)ζ(s). The role of the factor 12s(s1) is to cancel the two simple poles of the completed function Λ. Thus ξ is the entire completion, while Λ remains meromorphic.

[F3]

Stirling's formula for Gamma: Fix δ with 0<δ<π. On the closed sector argzπδ, using the principal logarithm in zz1/2:=exp((z1/2)Logz), one has Γ(z)=2πzz1/2ez(1+Oδ(z1)) as z.

[F4]

All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle: Let f be holomorphic on D(a,R), let 0<r<R, and let γ(t)=a+rexp(it) for 0t2π. Define f(0)=f and, whenever it exists, f(n+1)=(f(n)). Then every f(n) exists on D(a,r) and, for every zD(a,r) and nN, f(n)(z)=n!2πiγf(ζ)(ζz)n+1dζ. In particular, every holomorphic function has complex derivatives of all orders locally.

[F5]

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 T0.

[F6]

The only zeros of zeta on the nonpositive real axis are the negative even integers, and every other zero lies in the open critical strip: For each integer m1, ζ(2m)=0. These are the only zeros of ζ on the nonpositive real axis. Every other zero ρ of ζ satisfies 0<Reρ<1. Moreover, if ρ is a nontrivial zero, then so are 1ρ and ρ.

Proof

1.1

For bounded s away from zeros, the terms 1/(sρ)+1/ρ are Os(ρ2). The unit-interval count, reflected using conjugate zeros, makes their tails normally convergent. Logarithmically differentiating the canonical product therefore gives ξ/ξ=B+ρ(1/(sρ)+1/ρ).

F1F5F6
1.2

Put z=1+s/2. In a wider fixed sector containing these high-height points, Stirling gives Γ(z)=2πexp((z1/2)Logzz)(1+r(z)) with r(z)=O(z1). On discs of radius ϵz in that sector, Cauchy gives r(z)=O(z2). Thus Γ/Γ(z)=Logz1/(2z)+O(z2), whose real part is logtlog2+O(1/t). Remaining bounded heights are compact.

F3F4
2.1

In the defining formula for ξ, use Γ(1+s/2)=(s/2)Γ(s/2) to write ξ=(s1)πs/2Γ(1+s/2)ζ(s). The Gamma recurrence follows by integration by parts on its defining integral and meromorphic continuation. Differentiating this equality proves the first formula wherever its factors are nonzero, hence meromorphically.

F2step 1.1algebra
3.1

For fixed s the real summands have tails Os(Imρ2), since 0<Reρ<1. The same estimate applies to Re(1/ρ). Their absolute convergence follows from the unit-band count. Absorb the constant ReB+Re(1/ρ) and the bounded pole term into O(1), obtaining the second formula.

F5F6step 2.1step 1.2

Depends on

Used by

Dependency tree · two levels

25 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