Spence function

[Albert Chan]

Very beautiful ! We can proof I the same way, just different integral limits.

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)

When we fold the second term, it disappeared !

∫(f(x), x=a..b) = ∫(f(x) + f(a+b-x), x=a..b) / 2

int(ln(cos(θ-pi/4)) - ln(cos(θ)) + ln(cos(-θ)) - ln(cos(pi/4-θ)), θ=-pi/4 .. pi/2) / 2 = 0

All is left is the first term, I = 3/8*ln(2)*pi