Spence function

[Albert Chan]

Here is the proof for non-zero integer k, I = pi^2/4

Let A(w) = re(-Li2(1/(w-1)) + Li2(1/(1-w)) - Li2(w-1) + Li2(1+w))

Code:

  -Li2(1/(w-1)) - Li2(w-1)  +            Li2(1/(1-w))             + Li2(1+w)
= (pi*pi/6 + ln(1-w)^2/2)   + (-pi*pi/6 - ln(w-1)^2/2 - Li2(1-w)) + Li2(1+w)
= (ln(1-w)^2 - ln(w-1)^2)/2 + (Li2(1+w) - Li2(1-w))
= ln(-(1-w)^2) * (ln(1-w)-ln(w-1))/2 + (Li2(1+w) - Li2(1-w))

We like to show A(w) + A(-w) = pi^2/2, which implied I = average(A(W)) = pi^2/4
Using symmetry, we avoided calculations of Li2(x), since all will cancelled out. (*)

First factor, let w = c+s*i, where c=cos(x), s=sin(x), so that |w|=1

Code:

  -(1-w)^2 
= -(1-c-s*i)^2 
= -(1-c)^2     +   s^2   + 2*(1-c)*s*i
= -(1-2*c+c^2) + (1-c^2) + 2*(1-c)*s*i
= 2*(1-c)*c              + 2*(1-c)*s*i
= 2*(1-c)*w

Similar way, -(1+w)^2 = -2*(1+c)*w

Second factor, by definition, ln(x) = ln(|x|) + arg(x) * i

ln(x) - ln(-x) = (arg(x) - arg(-x)) * i = ±pi * i

Sign depends on phase angle. If we assume arg(w) > 0, we have

ln(1-w) - ln(-(1-w)) = -pi*i
ln(1+w) - ln(-(1+w)) = pi*i

re(ln(-(1-w)^2) * (ln(1-w) - ln(-(1-w)))) = arg(w) * pi
re(ln(-(1+w)^2) * (ln(1+w) - ln(-(1+w)))) = arg(-w) * -pi = (pi - arg(w)) * pi

A(w)  = arg(w)*pi/2          + re(Li2(1+w) - Li2(1-w))
A(-w) = (pi - arg(w))*pi/2 + re(Li2(1-w) - Li2(1+w))

→ A(w) + A(-w) = pi^2/2       QED

Note that k does not appear anywhere in the proof.
So, as long as we pair w with -w, even for non-integer k, above should hold.

---
(*) For odd k, w=-1 does not have anyone to pair with, but evaluation is simple.

A(w) = re(ln(-(1-w)^2) * (ln(1-w)-ln(w-1))/2 + (Li2(1+w) - Li2(1-w)))
A(-1) = pi^2/2 - re(Li(2))

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

Li2(-1) = -pi^2/12
Li2(2) = (pi^2/6 - Li2(-1)) - ln(2)*pi*i = pi^2/4 - ln(2)*pi*i

A(-1) = p^2/2 - pi^2/4 = pi^2/4

To keep the symmetry, we can assign A(1) = pi^2/4