Lecture 8: Integration by Substitution for Complex Line Integrals

This file formalizes Theorem 1 from Lecture 8: the substitution rule for complex line integrals.

Main results

• complexLineIntegral — the complex line integral of f along a path γ.

• complex_line_integral_substitution — If φ is holomorphic on an open set Ω and γ is a differentiable path in Ω, then ∫_{φ(γ)} f(w) dw = ∫_γ f(φ(z)) φ'(z) dz.

• circleMap_zero_add_pi: The circle map centered at the origin satisfies circleMap 0 R (θ + π) = -circleMap 0 R θ, i.e., shifting by π negates the point.

• circleIntegral_comp_sq_eq_zero: Theorem 2, Lecture 8. For any function f : ℂ → ℂ and any real R, the circle integral ∮ z in C(0, R), f(z²) dz = 0.

  The proof uses the substitution z ↦ -z. Since (-z)² = z² and d(-z) = -dz, the integral picks up a minus sign. But the circle centered at 0 is invariant under z ↦ -z (rotation by π), so the integral equals its own negative, hence it is zero.

  In the textbook, this is stated for rational functions R with R(z²) ≠ 0 on the circle, but the proof works for any f : ℂ → ℂ (when the integrand is not integrable, circleIntegral returns 0 by convention).

noncomputable def complexLineIntegral (f : ℂ → ℂ) (γ : ℝ → ℂ) (a b : ℝ) : ℂ

The complex line integral of f along a path γ : ℝ → ℂ over the interval [a, b], defined as ∫ t in a..b, f(γ(t)) · γ'(t) dt.

theorem complex_line_integral_substitution (f φ : ℂ → ℂ) (γ : ℝ → ℂ) (a b : ℝ) (Ω : Set ℂ) (hΩ : IsOpen Ω) (hφ : DifferentiableOn ℂ φ Ω) (hγ_diff : ∀ t ∈ Set.uIcc a b, DifferentiableAt ℝ γ t) (hγ_mem : ∀ t ∈ Set.uIcc a b, γ t ∈ Ω) : complexLineIntegral f (φ ∘ γ) a b = complexLineIntegral (fun (z : ℂ) => f (φ z) * deriv φ z) γ a b

Theorem 1, Lecture 8 (Integration by Substitution for Complex Line Integrals).

Let φ be a holomorphic function on an open set Ω ⊆ ℂ, and let γ be a differentiable path in Ω parametrized on [a, b]. Then $$\int_{\varphi(\gamma)} f(w)\,dw = \int_{\gamma} f(\varphi(z))\,\varphi'(z)\,dz,$$ i.e., the complex line integral of f along the composed path φ ∘ γ equals the complex line integral of z ↦ f(φ(z)) · φ'(z) along γ.

The proof follows from the chain rule: (φ ∘ γ)'(t) = φ'(γ(t)) · γ'(t), which shows the two integrands are equal pointwise.

Theorem 2: Integral of R(z²) over circles centered at the origin

theorem circleMap_zero_add_pi (R θ : ℝ) : circleMap 0 R (θ + Real.pi) = -circleMap 0 R θ

Shifting the angle by π on a circle centered at the origin negates the point: circleMap 0 R (θ + π) = -circleMap 0 R θ. This follows from exp(πi) = -1.

theorem circleIntegral_comp_sq_eq_zero (f : ℂ → ℂ) (R : ℝ) : ∮ (z : ℂ) in C(0, R), f (z ^ 2) = 0

Theorem 2, Lecture 8. Let f : ℂ → ℂ be any function. Then the circle integral of f(z²) around a circle centered at the origin is zero: ∮ z in C(0, R), f(z²) dz = 0.

In the textbook, this is stated for a rational function R with the condition that R(z²) has no poles on the circle |z| = r. The proof uses the substitution z ↦ -z: since (-z)² = z² and the derivative d(-z) = -dz, the integrand is antiperiodic with period π in the parametrization angle, so the integral over [0, 2π] vanishes.