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The Riemann--von Mangoldt zero count

Statement

Assume countable choice.

For T≥2, N(T)=T2πlog⁡T2π−T2π+O(log⁡T).

Facts & Assumptions

[L1]

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

Proof

Given: Countable choice and T≥2. We first treat sufficiently large T not equal to a zero ordinate.

1.1givenalgebra

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/(s−ρ)+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 ∑ρ∣ρ∣−2<∞, the sum of ℜ(1/ρ)=β/∣ρ∣2 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(log⁡T): the zeta term is bounded by ∑n≥2(log⁡n)n−2, and the Gamma term is O(log⁡T). 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+(T−γ)2≤∑ρ2−β(2−β)2+(T−γ)2=O(log⁡T). In particular there are O(log⁡T) zeros with ∣T−γ∣<1, without using the present theorem or its unit-interval corollary.

2.1step 1.1algebra

On the horizontal segment s=σ+iT, 1/2≤σ≤2, subtract the product formula at s0. For ∣T−γ∣≥1, ∣1s−ρ−1s0−ρ∣≤3/2(T−γ)2≤15/24+(T−γ)2. Thus step 1.1 bounds the nonlocal difference sum by O(log⁡T), uniformly in σ. The local subtracted terms 1/(s0−ρ) also total O(log⁡T). The integral of the imaginary part of each remaining local term 1/(σ+iT−ρ) is the argument change of a horizontal segment missing zero, of absolute value at most π. Their number is O(log⁡T). 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(log⁡T).

3.1step 2.1algebra

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 s−1 in ξ(s)=12s(s−1)π−s/2Γ(s/2)ζ(s) have bounded argument changes. The zeta factor also has bounded argument change: ∣ζ(2+it)−1∣≤∑n≥2n−2<1, so it remains in a fixed right half-plane. Stirling with a continuous logarithm in the right half-plane gives ℑlog⁡Γ(1+iT/2)=T2log⁡T2−T2+O(1). Consequently the argument change along this vertical segment is T2log⁡T2π−T2+O(1).

4.1L1step 2.1step 3.1algebra

By [L1], ξ(s)=ξ(1−s), and conjugation symmetry gives ξ(1−sˉ)=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 ℜs=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+it0→2+iTarg⁡ξ+Δ2+iT→1/2+iTarg⁡ξ)+O(1). Steps 2.1 and 3.1 yield the claimed main term and O(log⁡T) error.

5.1step 4.1algebra∎

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<γ≤T, 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 2≤T≤T0 are covered by enlarging the constant.

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