math.NT — Number Theory
Contributed by Pedro Caceres
Let X_n(s) = Σ_{k≤n} k⁻ˢ and let C₂(n,s) = |X_n(s)|² − |X_{n−1}(s)|² − n⁻²σ be the cross-energy of consecutive partial sums. At every point of the critical strip the limit of C₂(n,s) is 1/(¼ + t²), 0 or +∞ according as σ = ½, σ > ½ or σ < ½; this trichotomy is the trace of the pole at s = 1. On the critical line the associated helix has a natural orientation whose signed radius is Hardy’s Z(t), and the zeros of the truncated functions spiral into the zeros of ζ at the rate n⁻ᵝ. We prove that the mean energy of the normalized error E(u) in the prime number theorem equals Σ_ρ lim_{n→∞} C₂(n,ρ) in [0,∞]; it is at least 2 + γ₀ − log(4π), with equality if and only if the Riemann hypothesis (RH) holds and all zeros are simple, and an unconditional local version expresses the energy in a Gaussian window as a double sum over zeros. We then study the pointwise inequality S(σ,t) = ∂σ log|ξ(σ + it)| > 0, which is equivalent to RH: an off-line pair lowers S exactly inside a disc, the failure set at abscissa σ up to height T has measure at most N(σ,T + 1), and S is Weil’s functional at a Poisson kernel, whose truncation is an exact finite expression in the primes up to eᴸ. Every criterion is tested against the Davenport–Heilbronn function, which has zeros off the critical line, and against its Euler-product twin L(s,χ) mod 5: the former violates each criterion in the predicted way, the latter behaves like ζ. For ζ, the primes up to 10⁸ give mean energy 0.0453 against 0.0462. On the circle of the Euler factor of a prime p the zeros lean toward the angle π, and toward π + argχ(p) for L(s,χ). We do not prove the Riemann hypothesis.
A mix of human-written and AI-generated textHuman understanding: all partsC2 FunctionDirichlet SumsRiemann Zeta Function