Spence function

[Albert Chan]

This post extended A(w) + A(-w) = pi^2/2, to a proof that A(w) = pi^2/4

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

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

Li2(1+w) - Li2(1-w) = ln(w)*ln(1-w) - ln(-w)*ln(1+w) + (Li2(w) - Li2(-w))

The reason for the conversion is because of this amazingly neat identity

\(\displaystyle\Re(Li_2(e^{iθ})) = \sum_{n=1}^∞ {\cos(nθ) \over n^2}
= {\pi^2\over6} - {\pi θ \over 2} + {θ^2 \over 4},\qquad 0 ≤ θ ≤ 2\pi\)

Applying formula for arg(w = e^(ix)) range of (-pi, pi], we have a sawtooth pattern:

Li2(w) - Li2(-w) = pi^2/4 - |arg(w)|*pi/2

From previous post, we have (1±w)^2 = ±2*(1-c)*w, where c = re(w).
Thus, arg(1±w) = arg(w)/2, with possible shifts.

Excluding w = 1, i.e. range(arg(w)) = (-pi,0) + (0,pi]
Let a = arg(w), s = sign(a) = ±1

arg(w) = a
arg(-w) = a - s*pi
arg(1-w) = a/2 - s*pi/2
arg(1+w) = a/2

ln(±w) is pure imaginary, this make pulling out real parts of terms easy.

re(ln(w)*ln(1-w) - ln(-w)*ln(1+w)) = (-a)*(a/2-s*pi/2) + (a-s*pi)*(a/2) = 0

A(w) = re(-Li2(1/(w-1)) + Li2(1/(1-w)) - Li2(w-1) + Li2(1+w))
        = |arg(w)|*pi/2 + re(Li2(1+w) - Li2(1-w))       // previous post, extended to all w
        = |arg(w)|*pi/2 + re(Li2(w) - Li2(-w))             // all log terms disappeared
        = |arg(w)|*pi/2 + pi^2/4 - |arg(w)|*pi/2
        = pi^2/4

Comment: This may be enough to show I = pi^2/4, for any non-zero real k.