[{"type":"thinking","thinking":"Continue with piece 6 — the digamma ψ-difference bound. \n\nGoal: bound ½‖ψ(s/2) − ψ(s₀/2)‖ for s = σ+it, σ ∈ [−5/2, 3], s₀ = 2+it, t ≥ 10. Let w = s/2, w₀ = s₀/2. Both have im = t/2 ≥ 5.\n\nPlan (from the budget analysis): \nψ(w) − ψ(w₀) = [log w − log w₀] − (1/2)[1/w − 1/w₀] + [r(w) − r(w₀)]\nwhere ψ(w) = log w − 1/(2w) + r(w) with |r(w)| ≤ 1.2/|w|² (Lemma D).\n\nBounds:\n- |w − w₀| = |s − s₀|/2 ≤ 4.5/2 = 2.25.\n- |w₀| ≥ |im w₀| = t/2 ≥ 5; |w| ≥ t/2 ≥ 5.\n- |log w − log w₀|: w/w₀ = 1 + (w−w₀)/w₀, |u| := |w−w₀|/|w₀| ≤ 2.25/5 = 0.45 ≤ 1/2. |log(1+u)| ≤ 2|u| for |u| ≤ 1/2 (Mathlib: Complex.norm_log_one_add_le? or abs_log_one_add_le?). Need the exact lemma. There might be `Complex.norm_log_one_add_le (h : ‖z‖ ≤ r) (hr : r < 1)? : ‖log (1 + z)‖ ≤ ...`. Let me search. Or `Complex.abs_log_one_add_le (hz : ‖z‖ < 1) : ‖log (1+z)‖ ≤ -log(1−‖z‖)`? and −log(1−x) ≤ 2x for x ≤ 1/2. Then |log diff| ≤ 2·0.45 = 0.9 at t=10. Hmm — ½·0.9 = 0.45. Plus the other pieces: ½·|(1/2)(1/w − 1/w₀)| ≤ ½·(1/2)·|w−w₀|/(|w||w₀|) ≤ ¼·2.25/25 = 0.0225. Plus ½·(|r(w)| + |r(w₀)|) ≤ ½·2·1.2/25 = 0.048. Total: 0.45 + 0.0225 + 0.048 = 0.52. Budget 2.1 ✓ HUGE margin. Actually even the crudest |log| bound works.\n\nWait — but the digamma difference is (ψ(s/2) − ψ(s₀/2))/2 in the decomposition (Rexpr_decomp has `− (Complex.digamma (s/2) − Complex.digamma (s₀/2))/2`). So piece 6's target: ‖(ψ(s/2) − ψ(s₀/2))/2‖ ≤ something ≤ 2.1. With the above: ≤ 0.52. Let me set the target: ‖(digamma (s/2) − digamma (s₀/2))/2‖ ≤ 1 (round, comfortable). Actually let me compute a cleaner number: ≤ 0.9/2 + 0.0225 + 0.048... the /2 applies to the whole difference: ‖(ψs − ψs₀)/2‖ = ½‖ψs − ψs₀‖ ≤ ½(0.9 + 0.045 + 0.096) = ½·1.041 = 0.5205. So ≤ 0.53 with room; claim ≤ 1 for comfort.\n\nNow the Lean proof structure:\n1. **The sharp digamma remainder** (Lemma D form): for w with |im w| ≥ 5 (re arbitrary? or re w ∈ some range?): ψ(w) = log w − 1/(2w) + r(w), ‖r(w)‖ ≤ 1.2/‖w‖². \n\nHow to prove in Lean via the level-2 EM? What's available: LiRBound has the level-2 Euler–Maclaurin for ζ. For ψ/digamma, what exists? From 5d: `norm_digamma_sub_log_le_of_two_le_abs_im` : ‖ψ(w) − log w‖ ≤ 3 for 2 ≤ |im w| (digamma_add_natCast shift + first-order EM + arctan-FTC). The first-order EM gave ψ(w) − log w = −1/(2w) − ∫ B̃₁(x)/(x+w)² dx-ish. For the sharp remainder I need one more EM level: ψ(w) = log w − 1/(2w) − ∫₀^∞ B̃₂(x)/(x+w)³ dx·(factor).\n\nHmm — actually, what does the digamma EM look like exactly? The digamma series: ψ(w) = log w − 1/(2w) − Σ... The classical Binet: ψ(w) = log w − 1/(2w) − 2∫₀^∞ t/((t²+w²)(e^{2πt}−1)) dt (Binet's second formula for ψ). D-056 §1 used Binet. But in Lean, we don't have Binet. The EM route: how did 5d's norm_digamma_sub_log_le work? Let me look at its proof structure (LiReflect.lean) — it built ψ(w) − log w via the shift + first-order EM `sum_eq_integral_add_integral_deriv` + arctan-FTC. \n\nFor the sharp version, one more EM level on the digamma series: ψ(w) = −γ − 1/w + Σ_{n≥1} (1/n − 1/(w+n))·... hmm. Actually the cleanest EM form for ψ: from the series ψ(w) = −γ + Σ_{n≥0} (1/(n+1) − 1/(w+n)): and Σ_{n≥0} 1/(w+n) vs ∫₀^∞ dx/(w+x) + (1/2)(1/w) + ...: EM: Σ_{n≥0} f(n) = ∫₀^∞ f + f(0)/2 + ∫ B̃₁ f' — divergent for f = 1/(w+x). The convergent form uses the difference series. Hmm.\n\nLet me look at how 5d's norm_digamma_sub_log_le_of_two_le_abs_im was actually proved — that's the pattern to extend. Let me read LiReflect.lean's relevant section.","signature":""},{"type":"text","text":"Continuing with piece 6. First, studying how 5d's digamma bound was built (the EM pattern to extend)."},{"type":"tool_use","id":"Bash_212","name":"Bash","input":{"command":"cd D:/Python/Quantum/research/lean/NguyenRH1/NguyenRH1 && grep -n \"norm_digamma_sub_log_le_of_two_le_abs_im\\|theorem\\|lemma\" LiReflect.lean | grep -n \"digamma\\|Digamma\" | head -20","description":"Find the digamma bound structure in LiReflect"},"extra_content":null}]