Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-09-24 (gpt-6-sol)
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Residues in the von Mangoldt contour shift

Statement

Assume countable choice.

For x>1, shifting −ζ′(s)xs/(ζ(s)s) left crosses residues x(s=1),−xρρ(s=ρ),−12log⁡(1−x−2)(s=−2,−4,…),−ζ′(0)ζ(0)(s=0). Zeros are counted with multiplicity and xρ=exp⁡(ρlog⁡x) uses real log⁡x.

Facts & Assumptions

[L1]

The zeros of zeta in Re⁡s≤0 occur exactly at the negative even integers; in particular, s=0 is not a zero (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).

[L2]

Zeta satisfies ζ(s)=2sπs−1sin⁡(πs/2)Γ(1−s)ζ(1−s) as an identity of meromorphic functions (The Riemann zeta function satisfies the classical sine-gamma functional equation).

[L3]

Gamma has poles only at the nonpositive integers and has no zeros, while zeta has no zeros on Re⁡s≥1 (Meromorphic continuation of Gamma, Gamma has no zeros, The Riemann zeta function has no zeros on the closed half-plane Re⁡s≥1, except for its pole at 1).

Proof

Given: Countable choice, x>1 and the meromorphic continuation of zeta.

1.1givenalgebra

A simple pole of zeta at 1 makes −ζ′/ζ have residue 1; a zero ρ of multiplicity m makes it have residue −m. Multiplication by xs/s gives the first two entries.

2.1L1L2L3step 1.1algebra∎

By [L1], the remaining zeros crossed on the nonpositive real axis occur at −2k. At s=−2k, the sine in [L2] has a simple zero, while all its other factors are finite and nonzero by [L3]; hence these zeros are simple. Their residues sum to ∑k≥1x−2k/(2k)=−12log⁡(1−x−2). Also by [L1], zeta is nonzero at 0, so the pole of 1/s gives the final entry.

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Sources