Spence function

[Albert Chan]

(01-08-2021 01:28 AM)Albert Chan Wrote:  For k=1, g(t) is simple enough that Wolfram Alpha can proof it (after a few seconds)

I = ∫(ln((2+t)/(2-t)*(2t+1))/t, t = 0 .. 1) = pi^2/4

I just learned how to proof this, using Spence's function (dilogarithm)



Break-up Integral to parts, and integrate each:

∫(ln(1+t/2)/t - ln(1-t/2)/t + ln(1+2t)/t) dt
= (-Li2(-t/2)) - (-Li2(t/2)) + (-Li2(-2t)) + C

I = -Li2(-1/2) + Li2(1/2) - Li2(-2)
  = LI2(1/2) - (Li2(-2) + Li2(-1/2))

Identity:    Li2(z) + Li2(1-z) = pi^2/6 - ln(z)*ln(1-z)
Let z=1/2: Li2(1/2) * 2 = pi^2/6 - ln(1/2)^2 = pi^2/6 - ln(2)^2

Identity:    Li2(z) + Li2(1/z) = -pi^2/6 - ln(-z)^2/2
Let z=-2:   Li2(-2) + Li2(-1/2) = -pi^2/6 - ln(2)^2/2

I = (pi^2/6 - ln(2)^2)/2 - (-pi^2/6 - ln(2)^2/2) = pi^2/4