All proofs
Project-declaredLean 4.32.0 · mathlib@81a5d257c8e4

Rectangle Integral HSplit

RectangleIntegralHSplit

Plain-language statement

A vertical splitting identity for rectangular contour integrals. Under integrability of ff along the two horizontal sides, splitting the rectangle with real coordinates from x0x_0 to x1x_1 at aa gives R(x0,x1;y0,y1)f=R(x0,a;y0,y1)f+R(a,x1;y0,y1)f.\int_{\partial R(x_0,x_1;y_0,y_1)}f=\int_{\partial R(x_0,a;y_0,y_1)}f+\int_{\partial R(a,x_1;y_0,y_1)}f. The two integrals over the shared vertical side cancel.

Exact Lean statement

lemma RectangleIntegralHSplit {a x₀ x₁ y₀ y₁ : ℝ}
    (f_int_x₀_a_bot : IntervalIntegrable (fun x => f (↑x + ↑y₀ * I)) volume x₀ a)
    (f_int_a_x₁_bot : IntervalIntegrable (fun x => f (↑x + ↑y₀ * I)) volume a x₁)
    (f_int_x₀_a_top : IntervalIntegrable (fun x => f (↑x + ↑y₁ * I)) volume x₀ a)
    (f_int_a_x₁_top : IntervalIntegrable (fun x => f (↑x + ↑y₁ * I)) volume a x₁) :
    RectangleIntegral f (x₀ + y₀ * I) (x₁ + y₁ * I) =
      RectangleIntegral f (x₀ + y₀ * I) (a + y₁ * I) +
      RectangleIntegral f (a + y₀ * I) (x₁ + y₁ * I)

Formal artifact

Lean source

Canonical source
Full Lean sourceLean 4
lemma RectangleIntegralHSplit {a x₀ x₁ y₀ y₁ : }    (f_int_x₀_a_bot : IntervalIntegrable (fun x => f (↑x + ↑y₀ * I)) volume x₀ a)    (f_int_a_x₁_bot : IntervalIntegrable (fun x => f (↑x + ↑y₀ * I)) volume a x₁)    (f_int_x₀_a_top : IntervalIntegrable (fun x => f (↑x + ↑y₁ * I)) volume x₀ a)    (f_int_a_x₁_top : IntervalIntegrable (fun x => f (↑x + ↑y₁ * I)) volume a x₁) :    RectangleIntegral f (x₀ + y₀ * I) (x₁ + y₁ * I) =      RectangleIntegral f (x₀ + y₀ * I) (a + y₁ * I) +      RectangleIntegral f (a + y₀ * I) (x₁ + y₁ * I) := by  dsimp [RectangleIntegral, HIntegral, VIntegral]  simp only [Complex.mul_re, Complex.mul_im, Complex.ofReal_re, Complex.ofReal_im, Complex.I_re,    Complex.I_im, mul_one, mul_zero, add_zero, zero_add, sub_self]  have h₁ := integral_add_adjacent_intervals f_int_x₀_a_bot f_int_a_x₁_bot  have h₂ := integral_add_adjacent_intervals f_int_x₀_a_top f_int_a_x₁_top  have h₁' :      (∫ (x : ) in x₀..a, f (↑x + ↑y₀ * I)) +          ∫ (x : ) in a..x₁, f (↑y₀ * I + ↑x) =        ∫ (x : ) in x₀..x₁, f (↑x + ↑y₀ * I) := by    simpa [add_comm] using h₁  have h₂' :      (∫ (x : ) in x₀..a, f (↑x + ↑y₁ * I)) +          ∫ (x : ) in a..x₁, f (↑y₁ * I + ↑x) =        ∫ (x : ) in x₀..x₁, f (↑x + ↑y₁ * I) := by    simpa [add_comm] using h₂  rw [ h₁',  h₂']  have hcomm₁ :      ∫ (x : ) in a..x₁, f (↑y₀ * I + ↑x) =        ∫ (x : ) in a..x₁, f (↑x + ↑y₀ * I) := by    apply intervalIntegral.integral_congr    intro x _    exact congrArg f (by ring)  have hcomm₂ :      ∫ (x : ) in a..x₁, f (↑y₁ * I + ↑x) =        ∫ (x : ) in a..x₁, f (↑x + ↑y₁ * I) := by    apply intervalIntegral.integral_congr    intro x _    exact congrArg f (by ring)  rw [hcomm₁, hcomm₂]  abel
Project
Prime Number Theorem and More
License
Apache-2.0
Commit
a93551347dce
Source
PrimeNumberTheoremAnd/ResidueCalcOnRectangles.lean:256-293

Reuse this declaration

Bring the exact result into your workflow

The import identifies the source module. Your project still needs the pinned package dependency shown on this page.

What this badge means

This completion status comes from the project or community source. It has not yet been represented here as an independent rebuild and axiom audit.

Continue in this project

Related declarations

Project-declaredLean 4.32.0

Admissible bound mono

admissible_bound.mono

Plain-language statement

For positive parameters A,B,C,RA,B,C,R, the classical error-bound function A(logxR)Bexp ⁣(ClogxR)A\left(\frac{\log x}{R}\right)^B\exp\!\left(-C\sqrt{\frac{\log x}{R}}\right) is nonincreasing once xexp ⁣(R(2B/C)2)x\ge \exp\!\left(R(2B/C)^2\right).

analytic number theoryprime numbersasymptotics

Source project: Prime Number Theorem and More

Person-level attribution pending.

View proof record
Project-declaredLean 4.32.0

Analytic On div Removable zero

AnalyticOn_divRemovable_zero

Plain-language statement

Let ff be analytic on an open set ss containing 00, and suppose f(0)=0f(0)=0. Define g(z)=f(z)/zg(z)=f(z)/z for z0z\ne0 and g(0)=f(0)g(0)=f'(0). Then the apparent singularity at 00 is removable and gg is analytic throughout ss.

analytic number theoryprime numbersasymptotics

Source project: Prime Number Theorem and More

Person-level attribution pending.

View proof record
Project-declaredLean 4.32.0

Analytic On div Removable zero closed Ball

AnalyticOn_divRemovable_zero_closedBall

Plain-language statement

Suppose R>0R>0 and ff is analytic on the closed disc zR|z|\le R with f(0)=0f(0)=0. Define g(z)=f(z)/zg(z)=f(z)/z for z0z\ne0 and g(0)=f(0)g(0)=f'(0). Then gg is analytic on the entire closed disc, including at the removed singularity.

analytic number theoryprime numbersasymptotics

Source project: Prime Number Theorem and More

Person-level attribution pending.

View proof record