The integral ∫₀^{π/4} x log (2 + 2 cos 2x) dx #
This file computes ∫₀^{π/4} x log (2 + 2 cos 2x) dx = (π/4) G - (21/64) ζ(3), the heart of A007
(and hence of A008 and A009), by an Abel-limit argument.
Main results #
LeanPolyLog.Fourier.hasSum_neg_pow_mul_cos_div: for0 ≤ r < 1,∑_{n ≥ 1} (-r)ⁿ cos (nθ) / n = -log (1 + 2r cos θ + r²) / 2.LeanPolyLog.Fourier.integral_mul_cos_two_mul:∫₀^{π/4} x cos (2nx) dx = (π/8) sin (nπ/2) / n + (cos (nπ/2) - 1) / (4n²)forn ≠ 0.LeanPolyLog.Fourier.hasSum_integral_mul_log: for0 ≤ r < 1,∫₀^{π/4} x log (1 + 2r cos 2x + r²) dx = ∑ₙ coeff n * rⁿ.LeanPolyLog.Fourier.hasSum_one_div_odd_cube:∑_{k ≥ 0} 1/(2k+1)³ = (7/8) ζ(3).LeanPolyLog.Fourier.hasSum_coeff:∑ₙ coeff n = (π/4) G - (21/64) ζ(3).LeanPolyLog.Fourier.integral_mul_log_two_add_two_cos:∫₀^{π/4} x log (2 + 2 cos 2x) dx = (π/4) G - (21/64) ζ(3).
Proof sketch #
- For
0 ≤ r < 1, the real part of∑ zⁿ/n = -log (1 - z)atz = -r e^{iθ}is the cosine series oflog (1 + 2r cos θ + r²)above. Its terms are bounded byrⁿ, so after puttingθ = 2xand multiplying byxit can be integrated term by term over[0, π/4]. This writesL(r) = ∫₀^{π/4} x log (1 + 2r cos 2x + r²) dxas a power series∑ coeff n * rⁿ. - The complex logarithm is never used at
r = 1. Instead,Lis continuous atr = 1by dominated convergence (forx ∈ [0, π/4]andrnear1the argument of the logarithm lies in[1, 9]), and the power series tends to∑ coeff nasr → 1⁻by Abel's limit theorem (Real.tendsto_tsum_powerSeries_nhdsWithin_lt). HenceL(1) = ∑ coeff n. - Splitting
nby its residue mod4, thesin (nπ/2)terms give(π/4) Gand thecos (nπ/2)terms give-(1/2 - 1/8) ∑ 1/(2k+1)³ = -(21/64) ζ(3).
∫₀^{π/4} x cos (2nx) dx = (π/8) sin (nπ/2) / n + (cos (nπ/2) - 1) / (4n²) for n ≠ 0.
Termwise integration for r < 1 #
Continuity at r = 1 and Abel's limit #
r ↦ ∫₀^{π/4} x log (1 + 2r cos 2x + r²) dx is continuous at r = 1, by dominated
convergence: for r near 1 the integrand is bounded by (π/4) log 9.