Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedaudited 2026-09-07
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The Riemann--von Mangoldt zero count

Statement

For T2, N(T)=T2πlogT2πT2π+O(logT).

Facts & Assumptions

[L1]

The completed zeta function satisfies Λ(s)=Λ(1s) (The completed zeta function satisfies Λ(s)=Λ(1s)).

Proof

Given: T2. We first treat sufficiently large T not equal to a zero ordinate.

1.1

The Hadamard product The Riemann xi function has its genus-one Hadamard product over the nontrivial zeros of zeta gives ξ/ξ(s)=B+ρ(1/(sρ)+1/ρ). Write ρ=β+iγ, with 0<β<1 by 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. Since ρρ2<, the sum of (1/ρ)=β/ρ2 converges. At s0=2+iT, the xi identity, Stirling's formula and The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1 give ξ/ξ(s0)=O(logT): the zeta term is bounded by n2(logn)n2, and the Gamma term is O(logT). Differentiating Stirling here is justified by Cauchy's estimate for its analytic remainder on disks of radius proportional to s0 in a larger sector. Taking real parts of the product formula, all variable summands are positive and ρ14+(Tγ)2ρ2β(2β)2+(Tγ)2=O(logT). In particular there are O(logT) zeros with Tγ<1, without using the present theorem or its unit-interval corollary.

givenalgebra
2.1

On the horizontal segment s=σ+iT, 1/2σ2, subtract the product formula at s0. For Tγ1, 1sρ1s0ρ3/2(Tγ)215/24+(Tγ)2. Thus step 1.1 bounds the nonlocal difference sum by O(logT), uniformly in σ. The local subtracted terms 1/(s0ρ) also total O(logT). The integral of the imaginary part of each remaining local term 1/(σ+iTρ) is the argument change of a horizontal segment missing zero, of absolute value at most π. Their number is O(logT). Including the reference value ξ/ξ(s0), the total argument change of ξ on the top segment from 2+iT to 1/2+iT is therefore O(logT).

step 1.1algebra
3.1

Fix a height t0(0,2) not equal to any zero ordinate. On the vertical segment from 2+it0 to 2+iT, the factors s and s1 in ξ(s)=12s(s1)πs/2Γ(s/2)ζ(s) have bounded argument changes. The zeta factor also has bounded argument change: ζ(2+it)1n2n2<1, so it remains in a fixed right half-plane. Stirling with a continuous logarithm in the right half-plane gives logΓ(1+iT/2)=T2logT2T2+O(1). Consequently the argument change along this vertical segment is T2logT2πT2+O(1).

step 2.1algebra
4.1

By [L1], ξ(s)=ξ(1s), and conjugation symmetry gives ξ(1sˉ)=ξ(s). Apply the argument principle to the rectangle with real sides 1,2 and heights t0,T. Its zeros are precisely the nontrivial zeta zeros in that height range. Reflection in s=1/2 pairs the two vertical edges and the two halves of each horizontal edge, doubling the argument change on the right vertical edge followed by the right half of the top; the bottom contributes a constant independent of T. Thus N(T)=1π(Δ2+it02+iTargξ+Δ2+iT1/2+iTargξ)+O(1). Steps 2.1 and 3.1 yield the claimed main term and O(logT) error.

L1step 2.1step 3.1algebra
5.1

If T is a zero ordinate, take nonzero-ordinate heights decreasing to T. Discreteness of zeros in the bounded strip makes their counts eventually equal to the convention 0<γT, with full multiplicities. The preceding error constant is independent of the distance to zero ordinates, so passage to the limit preserves the estimate. Finally compact heights 2TT0 are covered by enlarging the constant.

step 4.1algebra

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