[{"type":"thinking","thinking":"§3 green — `hasDerivAt_b2Integral` proved. Now §4 (the extended identity via identity principle) and §5 (the disc bound).\n\n§4:\n```lean\n/-! ## 4. The level-2 representation of ζ on re s > −1 -/\n\n/-- The level-2 Euler–Maclaurin representation: for re s > −1, s ≠ 1,\n  ζ(s) = 1/2 + 1/(s−1) − s(s+1)·∫₁^∞ B̃₂(x)·x^{−s−2} dx. -/\ntheorem riemannZeta_eq_level2 {s : ℂ} (hs : -1 < s.re) (hs1 : s ≠ 1) :\n    riemannZeta s = 1 / 2 + 1 / (s - 1)\n      - s * (s + 1) * ∫ x in Set.Ioi (1:ℝ), (b2Fun x : ℂ) * (x : ℂ) ^ (-s - 2) := by\n  -- both sides analytic on U := {s | −1 < re s ∧ s ≠ 1}, agree on {re > 0}\n  ...\n```\n\nDetails:\n- U := {s : ℂ | −1 < s.re ∧ s ≠ 1} — preconnected: copy Zeta23's isPathConnected_aux (they did {z ≠ 1 ∧ 0 < z.re}): I need {-1 < re ∧ ≠ 1} — adapt their proof (read it first).\n- LHS analytic on U: HolomorphicOn_riemannZeta.mono.\n- RHS analytic on U: pieces: 1/2 + 1/(s−1) analytic off 1 ✓; s(s+1)·∫: the integral has hasDerivAt at every point of {re > −1} (§3) ⟹ DifferentiableOn ⟹ analyticOnNhd via DifferentiableOn.analyticOnNhd (open set {re > −1}). Then the product/sum.\n- Agree on {re > 0, ≠ 1}: via ζ₀(1) — Zeta0EqZeta (N_pos := 1, re > 0, ≠ 1): ζ₀ 1 s = ζ s; and ζ₀ 1 s = 1/2 + 1/(s−1) − s∫B̃₁·x^{−s−1} (massage riemannZeta0 1 s) = 1/2 + 1/(s−1) − s(s+1)∫B̃₂... (via integral_b1_cpow_eq ✓).\n- Then eqOn_of_preconnected_of_eventuallyEq at 2.\n\nLet me read Zeta23's isPathConnected_aux to clone for the shifted half-plane. And check riemannZeta0 1 s's exact simplification (the Finset sum at N=1: range 2 = {0, 1}: 1/(0:ℂ)^s + 1/(1:ℂ)^s — 0^s: for s ≠ 0: zero_cpow ✓ = 0; 1/0 = 0; 1/1^s = 1. So sum = 1 ✓ for s ≠ 0. And (−1^(1−s))/(1−s) = −1/(1−s) = 1/(s−1) (one_cpow ✓); (−1^(−s))/2 = −1/2 ✓.\n\nHmm wait — ζ₀'s sum at N = 1: `∑ n ∈ Finset.range (N + 1), 1 / (n : ℂ) ^ s` with N = 1: range 2: {0, 1}: 1/(0:ℂ)^s + 1/(1:ℂ)^s. 0^s: cpow with 0 base: `Complex.zero_cpow (hs : s ≠ 0) : (0:ℂ)^s = 0` ✓ so 1/0^s = 1/0 = 0 ✓ (div_zero). So sum = 0 + 1 = 1 for s ≠ 0 ✓ (s = 0 has re 0 — in V re > 0 — s ≠ 0? re s > 0 ⟹ s ≠ 0 ✓).\n\nSo on V: riemannZeta0 1 s = 1 + (−(1:ℂ)^(1−s))/(1−s) + (−(1:ℂ)^(−s))/2 + s·∫₁^∞(⌊x⌋+1/2−x)·x^(−(s+1)) = 1 + 1/(s−1) − 1/2 − s∫B̃₁x^{−s−1}·... \n\nwait — the integral in riemannZeta0: `s * ∫ x in Ioi (N:ℝ), (⌊x⌋ + 1/2 - x) / (x:ℂ)^(s+1)` — the div form: /x^(s+1) vs my ∫ B̃₁·x^(−s−1): (⌊x⌋+1/2−x)/x^(s+1) = −B̃₁x·x^(−(s+1)) ✓ = −B̃₁·x^(−s−1) (exponent assoc — ring_nf handles? −(s+1) = −s−1 by ring ✓).\n\nSo ζ s = 1/2 + 1/(s−1) − s·∫B̃₁·x^(−s−1) on V — then integral_b1_cpow_eq gives −s·∫B̃₁ = −s(s+1)∫B̃₂ — wait: ∫B̃₁x^{−s−1} = (s+1)∫B̃₂x^{−s−2} so −s·∫B̃₁ = −s(s+1)∫B̃₂·... ✓ ζ = 1/2 + 1/(s−1) − s(s+1)∫B̃₂x^{−s−2} ✓ matches my statement (with the − sign) ✓.\n\nLet me read isPathConnected_aux fully to clone.","signature":""},{"type":"text","text":"§3 green. Reading Zeta23's path-connectedness proof to clone for §4's identity-principle step."},{"type":"tool_use","id":"Bash_274","name":"Bash","input":{"command":"cd D:/Python/Quantum/zeta-23-lean && sed -n '1092,1125p' Zeta23/FromPNTPlus/ZetaBounds.lean","description":"Read isPathConnected_aux proof"},"extra_content":null}]