Spence function

[Albert Chan]

(04-11-2021 03:22 AM)Albert Chan Wrote:  I = int(ln(1+x)/(1+x²), x=-1 .. ∞)       // x=tan(θ), dx = sec(θ)^2 dθ
  = int(ln(1+tan(θ)), θ=-pi/4 .. pi/2)
  = int(ln(√2), θ=-pi/4 .. pi/2) + int(ln(cos(θ-pi/4)) - ln(cos(θ)), θ=-pi/4 .. pi/2)

Simplify steps, tan(θ) = sin(θ)/cos(θ), then cos(θ) + sin(θ) = √(2)*cos(θ-pi/4) can all be skipped.
Folding tan(θ) integrand directly is easier.

I = ∫(ln(1+tan(θ)), θ=-pi/4 .. pi/2)
2*I = ∫(ln(1+tan(α)) + ln(1+tan(β)), α=-pi/4 .. pi/2)                   // where α = θ, β = pi/4 - α
      = ∫(ln(1 + tan(α) + tan(β) + tan(α)*tan(β)), α=-pi/4 .. pi/2)

tan(α+β) = (tan(α) + tan(β)) / (1 - tan(α)*tan(β)) = tan(pi/4) = 1

1 + tan(α) + tan(β) + tan(α)*tan(β) = 1 + (1 - tan(α)*tan(β)) + tan(α)*tan(β) = 2

2*I = ∫(ln(2), α=-pi/4 .. pi/2)

I = ln(2) * (pi/2 + pi/4) / 2 = 3/8*ln(2)*pi