[{"type":"thinking","thinking":"I left sorries again for §5 — I need to think through these carefully before writing.\n\n**norm_b2Integral_le**: \n‖∫₁^∞ ↑b2Fun·x^(−s−2)‖ ≤ ∫₁^∞ ‖↑b2Fun x·x^(−s−2)‖ = ∫₁^∞ |b2Fun x|·x^(−σ−2) ≤ ∫₁^∞ (1/8)x^(−σ−2) = (1/8)/(σ+1).\n\nComponents:\n- h1: `norm_integral_le_integral_norm`? For set-integral: ‖∫ x in Ioi 1, f x‖ ≤ ∫ x in Ioi 1, ‖f x‖: the lemma `norm_integral_le_integral_norm (f)` works for any measure ✓ (set integral is restrict-integral).\n- h2: pointwise ≤ + integral monotonicity: `integral_mono_ae (hf : Integrable f) (hg : Integrable g) (h : f ≤ᵐ[μ] g)`? — for the set-integral: both integrable on Ioi 1: ‖f‖ side: IntegrableOn (fun x => ‖↑b2Fun x·↑x^(−s−2)‖) (Ioi 1) — from hbound in integrableOn_b2_cpow's proof — that bound said ‖·‖ ≤ x^(−σ−2) — so ‖f‖ integrable via mono' with the rpow. The RHS (1/8)x^(−σ−2): integrable via integrableOn_Ioi_rpow_of_lt.const_mul.\n  integral_mono_ae signature: `integral_mono_ae (hf : IntegrableOn f s μ)? (hg) (h : ∀ᵐ x ∂(μ.restrict s), f x ≤ g x)`? — there's `setIntegral_mono_ae`? or `integral_mono_ae (hf : Integrable f) (hg : Integrable g) (h : f ≤ᵐ[μ] g) : ∫ f ≤ ∫ g` ✓ exists. With restrict measure forms.\n- h3: ∫ x in Ioi 1, (1/8)·x^(−σ−2) = (1/8)·∫ x in Ioi 1, x^(−σ−2) = (1/8)·(1/(σ+1)): the rpow integral value: `integral_rpow_of_lt`? Let me check: `integral_Ioi_rpow_of_lt {a : ℝ} (ha : a < -1) {c : ℝ} (hc : 0 < c) : ∫ x in Ioi c, x ^ a = - (c ^ (a + 1)) / (a + 1)`? something like that. Let me grep.\n\n**norm_riemannZeta_le_disc**: the assembly. Steps:\n1. Geometry: from hs (closedBall): ‖s − (2+tI)‖ ≤ 5/2:\n   - re s ≥ −1/2: |re s − 2| = |(s − (2+tI)).re| ≤ ‖s − (2+tI)‖ ≤ 5/2 ⟹ re s ∈ [−1/2, 9/2] ✓.\n   - |im s − t| ≤ 5/2 ⟹ |im s| ≥ |t| − 5/2 ≥ 15/2: so sign-split: t ≥ 10 ⟹ im s ≥ 15/2 > 5; t ≤ −10 ⟹ im s ≤ −15/2 < −5 ✓.\n2. s ≠ 1 ✓ (|im s| ≥ 7.5).\n3. Apply the identity (of_im_gt or of_im_lt by cases).\n4. Norm the RHS: ‖1/2 + 1/(s−1) − s(s+1)J‖ ≤ 1/2 + 1/‖s−1‖ + ‖s‖‖s+1‖‖J‖:\n   - ‖s−1‖ ≥ |im(s−1)| = |im s| ≥ 15/2 ⟹ 1/‖s−1‖ ≤ 2/15.\n   - ‖s‖ ≤ ‖2+tI‖ + 5/2 ≤ (2 + |t|) + 5/2 = |t| + 9/2 ✓ (norm_le_abs_re_add_abs_im for ‖2+tI‖ ≤ |2|+|t|).\n   - ‖s+1‖ ≤ ‖s‖ + 1 ≤ |t| + 11/2.\n   - ‖J‖ ≤ (1/8)/(re s + 1) ≤ (1/8)/(1/2) = 1/4.\n   Total: ‖ζ‖ ≤ 1/2 + 2/15 + (|t|+9/2)(|t|+11/2)/4.\n   For |t| ≥ 10: (|t|+9/2)(|t|+11/2)/4 = (t² + 10|t| + 99/4)/4 ≤ (t² + t² + 99/4)/4 hmm — 10|t| ≤ t² for |t| ≥ 10 ✓; 99/4 ≤ 0.2475 t²; so ≤ (1 + 1 + 0.2475)t²/4 = 0.561875 t². Plus 1/2+2/15 = 19/30 ≈ 0.633 ≤ 0.633·(t²/100) = 0.00633 t². Total ≤ 0.5682 t² ≤ 0.75 t² ✓ slack.\n   The final nlinarith: give it: h1 : (|t|+9/2)(|t|+11/2)/4 ≤ 0.57·|t|²? — do the full: want 19/30 + X ≤ 3/4·T² with X = (T+9/2)(T+11/2)/4, T = |t| ≥ 10: ⟺ 19/30 + (T²+10T+99/4)/4 ≤ 3T²/4 ⟺ 19/30 + T²/4 + 5T/2 + 99/16 ≤ 3T²/4 ⟺ 19/30 + 99/16 + 5T/2 ≤ T²/2 ⟸ T ≥ 10: T²/2 ≥ 50; LHS ≤ 19/30 + 99/16 + 25 ≈ 0.633+6.19+25 = 31.8 ≤ 50 ✓. nlinarith [sq_nonneg..., T ≥ 10, T² ≥ 100]: should close with the right hints (T² ≥ 10T from T ≥ 10 etc.).\n\nLet me check integral_Ioi_rpow_of_lt's name first, then write §5's two theorems completely.","signature":""},{"type":"text","text":"Let me check the rpow integral evaluation lemma name before writing §5 properly."},{"type":"tool_use","id":"Bash_299","name":"Bash","input":{"command":"cd D:/Python/Quantum/zeta-23-lean/.lake/packages/mathlib && grep -rn \"theorem integral_Ioi_rpow_of_lt\\|lemma integral_Ioi_rpow_of_lt\\|theorem integral_rpow_of_lt\" Mathlib/Analysis/SpecialFunctions/ImproperIntegrals.lean Mathlib/Analysis/SpecialFunctions/JapaneseBracket.lean 2>/dev/null | head -4","description":"Find the rpow integral evaluation"},"extra_content":null}]