A radial argument on the upper half disk #
Mathlib has no Cauchy theorem for a half disk. This file replaces it, for the one function A015 needs, by differentiating the half-circle mean with respect to the radius.
Let F be holomorphic on the open unit disk, and put H(ρ) = ∫₀^π F(ρ e^{iθ}) dθ. Differentiating
under the integral sign and then integrating in θ,
H'(ρ) = ∫₀^π F'(ρ e^{iθ}) e^{iθ} dθ = (F(ρ e^{iπ}) - F(ρ)) / (iρ) = i (F(ρ) - F(-ρ))/ρ.
For F(z) = log³((1+z)/2) the values F(±ρ) are real, so h(ρ) = Im H(ρ) has derivative
(F(ρ) - F(-ρ))/ρ and h(0) = 0. At ρ = 1 the function h is continuous by dominated
convergence: for ρ > 1/2, |1 + ρ e^{iθ}| ≥ ρ sin θ bounds |Re log((1 + ρ e^{iθ})/2)| by
log 4 + |log sin θ|, and Im log³ = 3 Re² Im - Im³ only involves the square of the real part,
which is integrable.
Main results #
LeanPolyLog.Radial.hasDerivAt_integral: forFwith a continuous derivativeF'on the unit disk and|ρ| < 1,d/dρ ∫₀^π F(ρ e^{iθ}) dθ = ∫₀^π F'(ρ e^{iθ}) e^{iθ} dθ.LeanPolyLog.Radial.integral_deriv_mul_exp: for0 < |ρ| < 1,∫₀^π F'(ρ e^{iθ}) e^{iθ} dθ = i (F(ρ) - F(-ρ))/ρ.LeanPolyLog.Radial.integral_im_logCube:∫₀^π Im log³((1 + e^{iθ})/2) dθ = ∫₀¹ (log³((1+r)/2) - log³((1-r)/2))/r dr, provided the integrand on the right is integrable on[0, 1].LeanPolyLog.Radial.im_logCube_exp_two_mul: for|x| < π/2,Im log³((1 + e^{2ix})/2) = 3x log² cos x - x³, becauselog((1 + e^{2ix})/2) = log cos x + ix.
Holomorphic functions on the unit disk #
Differentiation under the integral sign: if F has the continuous derivative F' on the unit
disk, then for |ρ₀| < 1, d/dρ ∫₀^π F(ρ e^{iθ}) dθ = ∫₀^π F'(ρ₀ e^{iθ}) e^{iθ} dθ at ρ₀.
The fundamental theorem of calculus in θ: for 0 < |ρ| < 1,
∫₀^π F'(ρ e^{iθ}) e^{iθ} dθ = (F(-ρ) - F(ρ))/(iρ) = i (F(ρ) - F(-ρ))/ρ.
The function log³((1+z)/2) #
For ‖z‖ < 1, (1 + z)/2 lies in the slit plane: its real part is positive.
d/dz log³((1+z)/2) = 3 log²((1+z)/2)/(1+z) on the unit disk.
3 log²((1+z)/2)/(1+z) is continuous on the unit disk.
The half-circle mean of Im log³((1+z)/2) #
For |ρ| < 1, θ ↦ log³((1 + ρ e^{iθ})/2) is continuous.
Near any |ρ| < 1, halfCircleIm is the imaginary part of ∫₀^π log³((1 + ρ e^{iθ})/2) dθ.
The complex half-circle mean has a derivative at every |ρ| < 1.
halfCircleIm 0 = 0: the integrand is Im log³(1/2) = 0.
halfCircleIm is continuous at 1 within [0, 1], by dominated convergence.
halfCircleIm is continuous on [0, 1].
The radial identity: if (log³((1+r)/2) - log³((1-r)/2))/r is integrable on [0, 1], then
∫₀^π Im log³((1 + e^{iθ})/2) dθ = ∫₀¹ (log³((1+r)/2) - log³((1-r)/2))/r dr.