Polylogarithms against the arcsine weight #
This file proves the identity behind A011 and A016. Let s ≥ 2, and let w ≥ 0 on (0, 1]
have the moments ∫₀¹ yⁿ w(y) dy = 1/(n+1)^s. Then
∫₀¹ Li_s(x) / (x √(1 - x²)) dx = ∫₀^{π/2} w(sin φ) (π/2 + φ) dφ.
The weight w = -log (with s = 2) gives A011, and w = log²/2 (with s = 3) gives A016.
Main results #
LeanPolyLog.Arcsin.hasDerivAt_arcsin_div: for|x|, |y| < 1,d/dx arcsin ((x - y)/(1 - xy)) = √(1 - y²) / ((1 - xy) √(1 - x²)).LeanPolyLog.Arcsin.integral_one_div_mul_sqrt: for|y| < 1,∫₀¹ dx / ((1 - xy) √(1 - x²)) = (π/2 + arcsin y) / √(1 - y²).LeanPolyLog.Arcsin.integral_div_one_sub_mul: for0 < x < 1,∫₀¹ w(y) / (1 - xy) dy = Li_s(x)/x.LeanPolyLog.Arcsin.integral_Lir_div_mul_sqrt_eq_integral_arcsin:∫₀¹ Li_s(x) / (x √(1 - x²)) dx = ∫₀¹ w(y) (π/2 + arcsin y) / √(1 - y²) dy.LeanPolyLog.Arcsin.integral_mul_arcsin_div_sqrt: the substitutiony = sin φ, for anyw.LeanPolyLog.Arcsin.integral_Lir_div_mul_sqrt:∫₀¹ Li_s(x) / (x √(1 - x²)) dx = ∫₀^{π/2} w(sin φ) (π/2 + φ) dφ.
Proof sketch #
Li_s(x)/x = ∑ xⁿ/(n+1)^s = ∑ xⁿ ∫₀¹ yⁿ w(y) dy = ∫₀¹ w(y)/(1 - xy) dy. The termwise integration is justified because the terms are nonnegative and their integrals are summable.- So the integral is
∫₀¹ ∫₀¹ w(y) / ((1 - xy) √(1 - x²)) dy dx. The integrand is nonnegative and integrable on the square: for fixedxit is at most a multiple ofw(y), and its integral inyisLi_s(x) / (x √(1 - x²)) ≤ Li_s(1) / √(1 - x²). Fubini swaps the two integrals. - The integral in
xis elementary:arcsin ((x - y)/(1 - xy))is an antiderivative of√(1 - y²) / ((1 - xy) √(1 - x²)), and it runs from-arcsin yatx = 0toπ/2atx = 1. - Finally
y = sin φturnsdy / √(1 - y²)intodφandarcsin yintoφ.
In Lean √0 = 0 and a / 0 = 0, so the integrands are 0 at x = 1 or y = 1. Single points
do not affect the integrals, and the proofs work on the open square (0, 1)².
The kernel 1/((1 - xy) √(1 - x²)) #
x ↦ arcsin ((x - y)/(1 - xy)) is continuous on [0, 1] for |y| < 1.
1/√(1 - x²) is integrable on (0, 1].
Generating functions of moments #
Let w ≥ 0 on (0, 1] have the moments ∫₀¹ yⁿ w(y) dy = 1/(n+1)^s. Then for 0 < x < 1,
∫₀¹ w(y)/(1 - xy) dy = Li_s(x)/x: expand 1/(1 - xy) as a geometric series and integrate
term by term.
Swapping the order of integration #
Let s ≥ 2, and let w ≥ 0 on (0, 1] be measurable with the moments
∫₀¹ yⁿ w(y) dy = 1/(n+1)^s. Then
∫₀¹ Li_s(x) / (x √(1 - x²)) dx = ∫₀¹ w(y) (π/2 + arcsin y) / √(1 - y²) dy. Both sides equal
∫₀¹ ∫₀¹ w(y) / ((1 - xy) √(1 - x²)) dx dy, by Fubini.
The substitution y = sin φ:
∫₀¹ w(y) (π/2 + arcsin y) / √(1 - y²) dy = ∫₀^{π/2} w(sin φ) (π/2 + φ) dφ for any w.
Let s ≥ 2, and let w ≥ 0 on (0, 1] be measurable with the moments
∫₀¹ yⁿ w(y) dy = 1/(n+1)^s. Then
∫₀¹ Li_s(x) / (x √(1 - x²)) dx = ∫₀^{π/2} w(sin φ) (π/2 + φ) dφ.