332 lines
9.9 KiB
Plaintext
332 lines
9.9 KiB
Plaintext
import Mathlib.Analysis.SpecialFunctions.Integrals
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import Mathlib.Analysis.SpecialFunctions.Log.NegMulLog
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import Mathlib.MeasureTheory.Integral.CircleIntegral
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import Mathlib.MeasureTheory.Measure.Restrict
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open scoped Interval Topology
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open Real Filter MeasureTheory intervalIntegral
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lemma logsinBound : ∀ x ∈ (Set.Icc 0 1), ‖(log ∘ sin) x‖ ≤ ‖log ((π / 2)⁻¹ * x)‖ := by
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intro x hx
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by_cases h'x : x = 0
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· rw [h'x]; simp
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-- Now handle the case where x ≠ 0
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have l₀ : log ((π / 2)⁻¹ * x) ≤ 0 := by
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apply log_nonpos
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apply mul_nonneg
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apply le_of_lt
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apply inv_pos.2
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apply div_pos
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exact pi_pos
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exact zero_lt_two
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apply (Set.mem_Icc.1 hx).1
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simp
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apply mul_le_one
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rw [div_le_one pi_pos]
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exact two_le_pi
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exact (Set.mem_Icc.1 hx).1
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exact (Set.mem_Icc.1 hx).2
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have l₁ : 0 ≤ sin x := by
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apply sin_nonneg_of_nonneg_of_le_pi (Set.mem_Icc.1 hx).1
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trans (1 : ℝ)
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exact (Set.mem_Icc.1 hx).2
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trans π / 2
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exact one_le_pi_div_two
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norm_num [pi_nonneg]
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have l₂ : log (sin x) ≤ 0 := log_nonpos l₁ (sin_le_one x)
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simp only [norm_eq_abs, Function.comp_apply]
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rw [abs_eq_neg_self.2 l₀]
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rw [abs_eq_neg_self.2 l₂]
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simp only [neg_le_neg_iff, ge_iff_le]
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have l₃ : x ∈ (Set.Ioi 0) := by
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simp
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exact lt_of_le_of_ne (Set.mem_Icc.1 hx).1 ( fun a => h'x (id (Eq.symm a)) )
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have l₅ : 0 < (π / 2)⁻¹ * x := by
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apply mul_pos
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apply inv_pos.2
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apply div_pos pi_pos zero_lt_two
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exact l₃
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have : ∀ x ∈ (Set.Icc 0 (π / 2)), (π / 2)⁻¹ * x ≤ sin x := by
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intro x hx
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have i₀ : 0 ∈ Set.Icc 0 π :=
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Set.left_mem_Icc.mpr pi_nonneg
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have i₁ : π / 2 ∈ Set.Icc 0 π :=
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Set.mem_Icc.mpr ⟨div_nonneg pi_nonneg zero_le_two, half_le_self pi_nonneg⟩
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have i₂ : 0 ≤ 1 - (π / 2)⁻¹ * x := by
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rw [sub_nonneg]
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calc (π / 2)⁻¹ * x
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_ ≤ (π / 2)⁻¹ * (π / 2) := by
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apply mul_le_mul_of_nonneg_left
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exact (Set.mem_Icc.1 hx).2
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apply inv_nonneg.mpr (div_nonneg pi_nonneg zero_le_two)
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_ = 1 := by
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apply inv_mul_cancel
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apply div_ne_zero_iff.mpr
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constructor
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· exact pi_ne_zero
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· exact Ne.symm (NeZero.ne' 2)
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have i₃ : 0 ≤ (π / 2)⁻¹ * x := by
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apply mul_nonneg
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apply inv_nonneg.2
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apply div_nonneg
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exact pi_nonneg
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exact zero_le_two
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exact (Set.mem_Icc.1 hx).1
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have i₄ : 1 - (π / 2)⁻¹ * x + (π / 2)⁻¹ * x = 1 := by ring
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let B := strictConcaveOn_sin_Icc.concaveOn.2 i₀ i₁ i₂ i₃ i₄
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simp [Real.sin_pi_div_two] at B
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rw [(by ring_nf; rw [mul_inv_cancel pi_ne_zero, one_mul] : 2 / π * x * (π / 2) = x)] at B
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simpa
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apply log_le_log l₅
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apply this
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apply Set.mem_Icc.mpr
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constructor
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· exact le_of_lt l₃
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· trans 1
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exact (Set.mem_Icc.1 hx).2
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exact one_le_pi_div_two
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lemma intervalIntegrable_log_sin₁ : IntervalIntegrable (log ∘ sin) volume 0 1 := by
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have int_log : IntervalIntegrable (fun x ↦ ‖log x‖) volume 0 1 := by
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apply IntervalIntegrable.norm
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rw [← neg_neg log]
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apply IntervalIntegrable.neg
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apply intervalIntegrable_deriv_of_nonneg (g := fun x ↦ -(x * log x - x))
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· exact (continuous_mul_log.continuousOn.sub continuous_id.continuousOn).neg
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· intro x hx
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norm_num at hx
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convert ((hasDerivAt_mul_log hx.left.ne.symm).sub (hasDerivAt_id x)).neg using 1
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norm_num
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· intro x hx
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norm_num at hx
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rw [Pi.neg_apply, Left.nonneg_neg_iff]
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exact (log_nonpos_iff hx.left).mpr hx.right.le
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have int_log : IntervalIntegrable (fun x ↦ ‖log ((π / 2)⁻¹ * x)‖) volume 0 1 := by
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have A := IntervalIntegrable.comp_mul_right int_log (π / 2)⁻¹
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simp only [norm_eq_abs] at A
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conv =>
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arg 1
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intro x
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rw [mul_comm]
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simp only [norm_eq_abs]
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apply IntervalIntegrable.mono A
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simp
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trans Set.Icc 0 (π / 2)
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exact Set.Icc_subset_Icc (Preorder.le_refl 0) one_le_pi_div_two
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exact Set.Icc_subset_uIcc
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exact Preorder.le_refl volume
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apply IntervalIntegrable.mono_fun' (g := fun x ↦ ‖log ((π / 2)⁻¹ * x)‖)
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exact int_log
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-- AEStronglyMeasurable (log ∘ sin) (volume.restrict (Ι 0 1))
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apply ContinuousOn.aestronglyMeasurable
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apply ContinuousOn.comp (t := Ι 0 1)
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apply ContinuousOn.mono (s := {0}ᶜ)
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exact continuousOn_log
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intro x hx
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by_contra contra
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simp at contra
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rw [contra, Set.left_mem_uIoc] at hx
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linarith
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exact continuousOn_sin
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-- Set.MapsTo sin (Ι 0 1) (Ι 0 1)
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rw [Set.uIoc_of_le (zero_le_one' ℝ)]
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exact fun x hx ↦ ⟨sin_pos_of_pos_of_le_one hx.1 hx.2, sin_le_one x⟩
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-- MeasurableSet (Ι 0 1)
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exact measurableSet_uIoc
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-- (fun x => ‖(log ∘ sin) x‖) ≤ᶠ[ae (volume.restrict (Ι 0 1))] ‖log‖
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dsimp [EventuallyLE]
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rw [MeasureTheory.ae_restrict_iff]
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apply MeasureTheory.ae_of_all
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intro x hx
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have : x ∈ Set.Icc 0 1 := by
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simp
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simp at hx
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constructor
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· exact le_of_lt hx.1
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· exact hx.2
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let A := logsinBound x this
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simp only [Function.comp_apply, norm_eq_abs] at A
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exact A
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apply measurableSet_le
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apply Measurable.comp'
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exact continuous_abs.measurable
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exact Measurable.comp' measurable_log continuous_sin.measurable
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-- Measurable fun a => |log ((π / 2)⁻¹ * a)|
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apply Measurable.comp'
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exact continuous_abs.measurable
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apply Measurable.comp'
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exact measurable_log
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exact measurable_const_mul (π / 2)⁻¹
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lemma intervalIntegrable_log_sin₂ : IntervalIntegrable (log ∘ sin) volume 0 (π / 2) := by
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apply IntervalIntegrable.trans (b := 1)
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exact intervalIntegrable_log_sin₁
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-- IntervalIntegrable (log ∘ sin) volume 1 (π / 2)
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apply ContinuousOn.intervalIntegrable
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apply ContinuousOn.comp continuousOn_log continuousOn_sin
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intro x hx
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rw [Set.uIcc_of_le, Set.mem_Icc] at hx
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have : 0 < sin x := by
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apply Real.sin_pos_of_pos_of_lt_pi
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· calc 0
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_ < 1 := Real.zero_lt_one
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_ ≤ x := hx.1
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· calc x
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_ ≤ π / 2 := hx.2
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_ < π := div_two_lt_of_pos pi_pos
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by_contra h₁x
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simp at h₁x
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rw [h₁x] at this
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simp at this
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exact one_le_pi_div_two
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lemma intervalIntegrable_log_sin : IntervalIntegrable (log ∘ sin) volume 0 π := by
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apply IntervalIntegrable.trans (b := π / 2)
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exact intervalIntegrable_log_sin₂
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let A := IntervalIntegrable.comp_sub_left intervalIntegrable_log_sin₂ π
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simp at A
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let B := IntervalIntegrable.symm A
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have : π - π / 2 = π / 2 := by linarith
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rwa [this] at B
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lemma intervalIntegrable_log_cos : IntervalIntegrable (log ∘ cos) volume 0 (π / 2) := by
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let A := IntervalIntegrable.comp_sub_left intervalIntegrable_log_sin₂ (π / 2)
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simp only [Function.comp_apply, sub_zero, sub_self] at A
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simp_rw [sin_pi_div_two_sub] at A
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have : (fun x => log (cos x)) = log ∘ cos := rfl
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apply IntervalIntegrable.symm
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rwa [← this]
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lemma integral_log_sin : ∫ (x : ℝ) in (0)..(π / 2), log (sin x) = -log 2 * π/2 := by
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have t₀ {x : ℝ} : sin (2 * x) = 2 * sin x * cos x := sin_two_mul x
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have t₁ {x : ℝ} : log (sin (2 * x)) = log 2 + log (sin x) + log (cos x) := by
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rw [sin_two_mul x, log_mul, log_mul]
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exact Ne.symm (NeZero.ne' 2)
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sorry
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sorry
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sorry
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have t₂ {x : ℝ} : log (sin x) = log (sin (2 * x)) - log 2 - log (cos x) := by
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rw [t₁]
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ring
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conv =>
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left
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arg 1
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intro x
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rw [t₂]
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rw [intervalIntegral.integral_sub, intervalIntegral.integral_sub]
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rw [intervalIntegral.integral_const]
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rw [intervalIntegral.integral_comp_mul_left (c := 2) (f := fun x ↦ log (sin x))]
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simp
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have : 2 * (π / 2) = π := by linarith
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rw [this]
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have : ∫ (x : ℝ) in (0)..π, log (sin x) = 2 * ∫ (x : ℝ) in (0)..(π / 2), log (sin x) := by
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sorry
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rw [this]
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have : ∫ (x : ℝ) in (0)..(π / 2), log (sin x) = ∫ (x : ℝ) in (0)..(π / 2), log (cos x) := by
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sorry
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rw [← this]
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simp
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linarith
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exact Ne.symm (NeZero.ne' 2)
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-- IntervalIntegrable (fun x => log (sin (2 * x))) volume 0 (π / 2)
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sorry
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-- IntervalIntegrable (fun x => log 2) volume 0 (π / 2)
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simp
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-- IntervalIntegrable (fun x => log (sin (2 * x)) - log 2) volume 0 (π / 2)
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apply IntervalIntegrable.sub
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-- -- IntervalIntegrable (fun x => log (sin (2 * x))) volume 0 (π / 2)
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sorry
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-- -- IntervalIntegrable (fun x => log 2) volume 0 (π / 2)
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simp
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-- -- IntervalIntegrable (fun x => log (cos x)) volume 0 (π / 2)
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exact intervalIntegrable_log_cos
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lemma int₁₁ : ∫ (x : ℝ) in (0)..π, log (4 * sin x ^ 2) = 0 := by
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sorry
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lemma int₁ :
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∫ x in (0)..(2 * π), log ‖circleMap 0 1 x - 1‖ = 0 := by
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have {x : ℝ} : log ‖circleMap 0 1 x - 1‖ = log (4 * sin (x / 2) ^ 2) / 2 := by
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dsimp [Complex.abs]
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rw [log_sqrt (Complex.normSq_nonneg (circleMap 0 1 x - 1))]
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congr
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calc Complex.normSq (circleMap 0 1 x - 1)
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_ = (cos x - 1) * (cos x - 1) + sin x * sin x := by
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dsimp [circleMap]
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rw [Complex.normSq_apply]
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simp
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_ = sin x ^ 2 + cos x ^ 2 + 1 - 2 * cos x := by
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ring
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_ = 2 - 2 * cos x := by
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rw [sin_sq_add_cos_sq]
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norm_num
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_ = 2 - 2 * cos (2 * (x / 2)) := by
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rw [← mul_div_assoc]
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congr; norm_num
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_ = 4 - 4 * Real.cos (x / 2) ^ 2 := by
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rw [cos_two_mul]
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ring
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_ = 4 * sin (x / 2) ^ 2 := by
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nth_rw 1 [← mul_one 4, ← sin_sq_add_cos_sq (x / 2)]
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ring
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simp_rw [this]
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simp
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have : ∫ (x : ℝ) in (0)..2 * π, log (4 * sin (x / 2) ^ 2) = 2 * ∫ (x : ℝ) in (0)..π, log (4 * sin x ^ 2) := by
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have : 1 = 2 * (2 : ℝ)⁻¹ := by exact Eq.symm (mul_inv_cancel_of_invertible 2)
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nth_rw 1 [← one_mul (∫ (x : ℝ) in (0)..2 * π, log (4 * sin (x / 2) ^ 2))]
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rw [← mul_inv_cancel_of_invertible 2, mul_assoc]
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let f := fun y ↦ log (4 * sin y ^ 2)
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have {x : ℝ} : log (4 * sin (x / 2) ^ 2) = f (x / 2) := by simp
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conv =>
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left
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right
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right
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arg 1
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intro x
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rw [this]
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rw [intervalIntegral.inv_mul_integral_comp_div 2]
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simp
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rw [this]
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simp
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exact int₁₁
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