AlexKontorovich/PrimeNumberTheoremAnd
Source indexedtheorem · leanprover/lean4:v4.32.0
Li2Bounds.setDiff_integral_eq_split
PrimeNumberTheoremAnd.IEANTN.Li2Bounds · PrimeNumberTheoremAnd/IEANTN/Li2Bounds.lean:199 to 224
Source documentation
The Set.diff integral equals the split interval integrals.
Exact Lean statement
theorem setDiff_integral_eq_split (ε x : ℝ) (hε : 0 < ε) (hε1 : ε < 1) (hx : 2 ≤ x) :
∫ t in Set.Ioc 0 x \ Set.Ioo (1 - ε) (1 + ε), 1 / log t =
(∫ t in (0 : ℝ)..(1 - ε), 1 / log t) + ∫ t in (1 + ε)..x, 1 / log tComplete declaration
Lean source
Full Lean sourceLean 4
theorem setDiff_integral_eq_split (ε x : ℝ) (hε : 0 < ε) (hε1 : ε < 1) (hx : 2 ≤ x) : ∫ t in Set.Ioc 0 x \ Set.Ioo (1 - ε) (1 + ε), 1 / log t = (∫ t in (0 : ℝ)..(1 - ε), 1 / log t) + ∫ t in (1 + ε)..x, 1 / log t := by rw [setDiff_decompose ε x hε hx, setIntegral_union (by grind) measurableSet_Icc, intervalIntegral.integral_of_le (by linarith), integral_Icc_eq_integral_Ioc, intervalIntegral.integral_of_le (by linarith)] · have hlog_neg : log (1 - ε) < 0 := Real.log_neg (by linarith) (by linarith) have hcont : ContinuousOn (fun t => 1 / log t) (Set.Ioc 0 (1 - ε)) := ContinuousOn.div continuousOn_const (Real.continuousOn_log.mono fun x hx => ne_of_gt hx.1) fun x hx => Real.log_ne_zero_of_pos_of_ne_one hx.1 (by linarith [hx.2]) refine IntegrableOn.of_bound measure_Ioc_lt_top (hcont.aestronglyMeasurable measurableSet_Ioc) (-1 / log (1 - ε)) ?_ filter_upwards [ae_restrict_mem measurableSet_Ioc] with t ht simp only [Set.mem_Ioc] at ht rw [norm_div, norm_one, Real.norm_eq_abs, abs_of_neg (Real.log_neg ht.1 (by linarith [ht.2])), show (-1 : ℝ) / log (1 - ε) = 1 / (-log (1 - ε)) from by ring] exact one_div_le_one_div_of_le (neg_pos.mpr hlog_neg) (by linarith [Real.log_le_log ht.1 ht.2]) · exact ContinuousOn.integrableOn_compact isCompact_Icc (ContinuousOn.div continuousOn_const (Real.continuousOn_log.mono fun x hx => ne_of_gt (by linarith [hx.1] : (0 : ℝ) < x)) fun x hx => Real.log_ne_zero_of_pos_of_ne_one (by linarith [hx.1] : 0 < x) (by linarith [hx.1]))