[{"type":"thinking","thinking":"I have the exact patterns:\n- `Complex.hasDerivAt_log hslit` with `hslit : z ∈ Complex.slitPlane` (via `Complex.mem_slitPlane_iff` — the right case: im ≠ 0).\n- `(Complex.hasDerivAt_log hslit).comp (x:ℂ) ((hasDerivAt_id' (x:ℂ)).const_add w)` for the affine precomposition, then `.comp_ofReal`.\n- `intervalIntegral.integral_eq_sub_of_hasDerivAt hanti hconti` : ∫ x in a..b, f' = f b − f a (with hanti : ∀ x ∈ uIcc, HasDerivAt f (f' x) x and hconti : IntervalIntegrable f').\n- The norm bound for the integral: I'll need `intervalIntegral.norm_integral_le_of_norm_le_const` or similar — let me just bound via: ‖∫₀¹ δ·z(y)⁻¹ dy‖ ≤ ∫₀¹ ‖δ·z(y)⁻¹‖ dy (intervalIntegral.norm_integral_le_integral_norm or intervalIntegral.norm_integral_le_of_norm_le_const). The latter: `intervalIntegral.norm_integral_le_of_norm_le_const (h : ∀ x ∈ uIoc a b, ‖f x‖ ≤ C) : ‖∫ x in a..b, f x‖ ≤ C * |b − a|`? — I think it exists. Or the norm-of-integral ≤ integral-of-norm + the constant bound. Let me plan with norm_integral_le_of_norm_le_const (check at build).\n\nFor piece 6's log difference, I need:\ng(y) := Complex.log (w₀ + (y:ℂ)·δ) where δ := (w+2) − w₀.\n- z(y) := w₀ + (y:ℂ)·δ has re ≥ 3/4 (so z(y) ≠ 0 and in slitPlane — via re > 0: mem_slitPlane_iff's LEFT case: re ≠ 0? slitPlane = ℂ ∖ (−∞, 0]: mem_slitPlane_iff : z ∈ slitPlane ↔ re z ≠ 0 ∨ im z ≠ 0? no: slitPlane = {z | re z > 0} ∪ {im ≠ 0}? Complex.mem_slitPlane_iff : z ∈ slitPlane ↔ 0 < z.re ∨ z.im ≠ 0 — I think it's (0 < re ∨ im ≠ 0). re ≥ 3/4 > 0 ✓ left case.\n- HasDerivAt g (δ·z(y)⁻¹) y: via (Complex.hasDerivAt_log hslit).comp ((hasDerivAt_id' _).const_add ... hmm the composition: z(y) = w₀ + y·δ: as a function of the real y: first build the ℂ-version: HasDerivAt (fun z : ℂ => Complex.log (w₀ + z·δ)) (Complex.log' ... ·δ)?? — the chain: (fun z => w₀ + z·δ) has deriv δ; log has deriv z⁻¹; composite: δ·z⁻¹ or z⁻¹·δ — HasDerivAt.comp: (hasDerivAt_log hslit).comp z₀ (h_affine): h_affine : HasDerivAt (fun z => w₀ + z·δ) δ z₀: (hasDerivAt_id' z₀).const_mul δ |>.const_add w₀? HasDerivAt.const_mul : HasDerivAt f f' x → HasDerivAt (fun x => f x * c) (f' * c) x ✓ then const_add. Then comp gives HasDerivAt (fun z => log (w₀ + z·δ)) (z₀-related value ·) — the comp value: (log' at w₀+z₀·δ) * δ = (w₀+z₀δ)⁻¹ * δ. Then .comp_ofReal : HasDerivAt (fun y : ℝ => log (w₀ + (y:ℂ)·δ)) (((w₀+(x:ℂ)δ)⁻¹·δ)) x — then the value form (w₀+xδ)⁻¹·δ vs δ·(w₀+xδ)⁻¹ — mul_comm ✓.\n- FTC: ∫₀¹ (fun y => δ·z(y)⁻¹) = g 1 − g 0 = log(w₀ + δ) − log w₀ = log(w+2) − log w₀ ✓ (w₀ + 1·δ = w₀ + δ = w+2 ✓ by ring-ish: δ := w+2−w₀: w₀ + δ = w+2 ✓).\n- norm bound: ‖∫₀¹ δ·z⁻¹‖ ≤ ∫₀¹ ‖δ·z⁻¹‖ ≤ (5/2)·(2/t)·(1−0) = 5/t: via intervalIntegral.norm_integral_le_of_norm_le_const (h : ∀ y ∈ uIoc 0 1, ‖δ·z(y)⁻¹‖ ≤ 5/t): ‖δ·z⁻¹‖ = ‖δ‖·‖z‖⁻¹ ≤ (5/2)·(2/t) = 5/t ✓ (‖z‖ ≥ t/2 from |im z| = t/2). \n\nThen the final algebra. Let me now write the whole piece 6. Structure:\n\n```lean\n/-! ## 7. Piece 6: the digamma difference -/\n\n/-- The log difference along the constant-imaginary segment:\nfor w, w₀ with the same imaginary part t/2 ≥ 5 and real parts in\n[3/4, 7/2]... ‖log w − log w₀‖ ≤ (5/2)·(2/t)·... -/\n```\n\nHmm — actually let me parametrize cleanly. Set:\n- w := s/2, w₀ := (2 + t·I)/2 = 1 + (t/2)·I.\n- δ := w + 2 − w₀ = (s/2 + 2) − (1 + (t/2)I) = (σ/2 + 1) (purely real!) — since im(s/2) = t/2 = im w₀, the difference is real: δ = ((σ/2 + 1 : ℝ) : ℂ). ‖δ‖ = |σ/2 + 1| ≤ 5/2 (σ/2 + 1 ∈ [−1/4, 5/2]).\n\nThe segment z(y) = w₀ + y·δ: re = 1 + y(σ/2+1) ∈ [3/4, 7/2], im = t/2.\n\nLet me write the theorem:\n\n```lean\n/-- Piece 6 helper: the log difference bound along the constant-im\nsegment. For σ ∈ [−5/2, 3], t ≥ 10: with w = s/2, w₀ = 1 + (t/2)I:\n‖log (w + 2) − log w₀‖ ≤ 5/t. -/\ntheorem log_diff_le {t : ℝ} (ht : 10 ≤ t) {s : ℂ}\n    (hσlo : -5 / 2 ≤ s.re) (hσhi : s.re ≤ 3) (him : s.im = t) :\n    ‖Complex.log (s / 2 + 2) - Complex.log (1 + (t / 2) * Complex.I)‖ ≤ 5 / t := by\n  ...\n```\n\nWait — should I phrase with w₀ = 1 + (t/2)·I or (2+tI)/2? The Rexpr_decomp has digamma (s₀/2) with s₀ = 2 + t·I. So s₀/2 = (2 + t·I)/2. The bridge (2+t·I)/2 = 1 + (t/2)·I: by ring? (2 + t·I)/2 = 1 + (t·I)/2 = 1 + (t/2)·I: `by ring` should handle (field, char 0): (2 + t·I)/2 = 1 + (t/2)·I ✓ ring. Hmm — t·I/2 vs (t/2)·I: ring normalizes ✓.\n\nLet me write the full piece 6 now. Components:\n\n```lean\ntheorem digamma_diff_le {t : ℝ} (ht : 10 ≤ t) {s : ℂ}\n    (hσlo : -5 / 2 ≤ s.re) (hσhi : s.re ≤ 3) (him : s.im = t) :\n    ‖(Complex.digamma (s / 2) - Complex.digamma ((2 + t * Complex.I) / 2)) / 2‖\n      ≤ 6 / t := by\n  have ht0 : (0 : ℝ) < t := by linarith\n  have ht2 : (5 : ℝ) ≤ t / 2 := by linarith\n  -- w := s/2, w₀ := (2+tI)/2 = 1 + (t/2)I\n  set w : ℂ := s / 2 with hw\n  set w₀ : ℂ := (2 + t * Complex.I) / 2 with hw₀\n  have hw₀eq : w₀ = 1 + (t / 2) * Complex.I := by rw [hw₀]; ring\n  have hwim : w.im = t / 2 := by rw [hw]; simp [him]\n  have hw₀im : w₀.im = t / 2 := by rw [hw₀eq]; simp\n  have hwre : w.re = s.re / 2 := by rw [hw]; simp\n  have hw₀re : w₀.re = 1 := by rw [hw₀eq]; simp\n  have himw : w.im ≠ 0 := by rw [hwim]; positivity? (t/2 ≥ 5 > 0: linarith)\n  -- (i) the recurrence: ψ(w) = ψ(w+2) − w⁻¹ − (w+1)⁻¹\n  have hrec : Complex.digamma (w + 2) = Complex.digamma w + (w⁻¹ + (w + 1)⁻¹) := by\n    have h := digamma_add_natCast himw 2\n    rw [Finset.sum_range_succ, Finset.sum_range_succ, Finset.sum_range_zero] at h\n    -- h : ψ(w+2) = ψ(w) + ((w+0)⁻¹ + (w+1)⁻¹) — cast forms\n    convert h using 1 · ring_nf ... \n```\nHmm — digamma_add_natCast {w : ℂ} (hw : w.im ≠ 0) (n : ℕ) : digamma (w + n) = digamma w + ∑ j ∈ range n, (w + (j:ℂ))⁻¹. With n = 2: digamma (w + (2:ℕ)) = digamma w + ((w+0)⁻¹ + (w+1)⁻¹): the sum over range 2: Finset.sum_range_two? or sum_range_succ twice from range 0: ∑ j ∈ range 2, f j = f 0 + f 1: `Finset.sum_range_two`? hmm. The sum: range 2: sum_range_succ (range 1): ∑ range 1 + f 1 = (range 0 + f 0) + f 1 = 0 + f 0 + f 1. With f j = (w + (j:ℂ))⁻¹: f 0 = (w + 0)⁻¹ = w⁻¹ (Nat.cast_zero, add_zero); f 1 = (w+1)⁻¹ (Nat.cast_one). So:\n```lean\n    have h := digamma_add_natCast himw 2\n    rw [show w + ((2:ℕ):ℂ) = w + 2 by push_cast; ring] at h  -- hmm the cast (↑2:ℂ) vs (2:ℂ)\n    rw [Finset.sum_range_succ, Finset.sum_range_succ, Finset.sum_range_zero] at h\n    simp only [Nat.cast_zero, add_zero, Nat.cast_one] at h\n    -- h : ψ(w+2) = ψ(w) + (0 + w⁻¹ + (w+1)⁻¹)\n    linarith? no — ring_nf at h; exact? \n```\nThe sum form: ((0 + w⁻¹) + (w+1)⁻¹) — with the leading 0 from sum_range_zero: h : ψ(w+2) = ψ(w) + (0 + w⁻¹ + (w+1)⁻¹)-ish — normalize: `simp only [Nat.cast_zero, add_zero, Nat.cast_one, zero_add] at h` gives h : ψ(w+2) = ψ(w) + (w⁻¹ + (w+1)⁻¹) ✓.\n\nThe cast: digamma_add_natCast's statement has w + ↑n with n : ℕ — (2:ℕ) cast to ℂ: w + (↑(2:ℕ)) — I want w + 2: push_cast or the show bridge: rw [show w + ((2:ℕ):ℂ) = w + 2 by push_cast; ring] at h ✓.\n\n- (ii) Stirling at v := w + 2: re v = σ/2 + 2 ≥ 3/4 > 0: \n```lean\n  have hvre : (0 : ℝ) < (w + 2).re := by rw [Complex.add_re]; simp [hwre]; linarith [hσlo]\n```\n(w+2).re = w.re + 2 = s.re/2 + 2 ≥ −5/4 + 2 = 3/4 > 0 ✓ — via simp [Complex.add_re, hwre] then linarith [hσlo]: (w+2).re = w.re + (2:ℂ).re = s.re/2 + 2 ✓.\n\nhvim : (w+2).im = t/2: simp [Complex.add_im, hwim] ✓.\n```lean\n  have hStir1 := Zeta23.GammaFacts.digamma_stirling (w := w + 2) hvre (by\n    rw [hvim... |im| = t/2 ≥ 5 ≥ 1/2] ...)\n```\ndigamma_stirling (hw : 0 < w.re) (ht : 1/2 ≤ |w.im|) : ‖digamma w − log w + (1/2)/w‖ ≤ 3/w.im². The namespace: Zeta23.GammaFacts.digamma_stirling? or it's open Zeta23... the file opens Zeta23 — the lemma is at Zeta23/GammaFacts/StirlingVert.lean:483 — inside `namespace`? Let me check how LiReflect references it: LiReflect used \"StirlingVert digamma_stirling\" per the journal. Let me grep for the usage.\n\n- (iii) Stirling at w₀: re = 1 > 0 ✓, |im| = t/2 ≥ 1/2 ✓.\n\n- (iv) the log difference bound (the segment FTC).\n\n- (v) assembly.\n\nLet me check the digamma_stirling usage form in LiReflect.","signature":""},{"type":"text","text":"The patterns are all confirmed. Checking how `digamma_stirling` is referenced in LiReflect, then writing piece 6."},{"type":"tool_use","id":"Bash_216","name":"Bash","input":{"command":"cd D:/Python/Quantum/research/lean/NguyenRH1/NguyenRH1 && grep -rn \"digamma_stirling\" *.lean | head -4; grep -n \"norm_integral_le_of_norm_le_const\\|norm_integral_le\" LiReflect.lean | head -4","description":"Find digamma_stirling usage and integral norm bound forms"},"extra_content":null}]