Spence function

[Albert Chan]

(01-07-2021 09:37 PM)Valentin Albillo Wrote:  

The funny thing about this integral is that its value remains the same if you replace 2.021 by any positive real value !!

I think Spence's function (Dilogarithms) is the key to proof I = pi^2/4, for any k

After transformation, we have I = ∫(g(t) dt, t = 0 .. 1)

\(\exp(g(t)·k\,t) = \Large
{(1+(1+t)^k)·(1+({1\over t}+1)^k) \over (1+(1-t)^k)·(1+({1\over t}-1)^k)} =
{(1+(1+t)^k)·(t^k+(1+t)^k) \over (1+(1-t)^k)·(t^k+(1-t)^k)}\)

For now, assume k is positive integer → numerator is polynomial of t, with degree 2k.

Let W = exp((2n+1)/k*pi*i), n = 0 .. k-1. In other words, W = roots of x^k = -1

roots of numerator   , t = W-1, 1/(W-1)
roots of denominator, t = 1-W, 1/(W+1)

Consider just 1 of W (we have k of them), factor it out, scaled away k, and integrate:

\(\displaystyle \int _0^1 \ln\left( {(1-{t \over w-1})·(1-t(w-1)) \over
(1-{t \over 1-w})·(1-t(w+1)) }\right) {dt \over t} \normalsize
= -Li_2({1\over w-1}) + Li_2({1\over 1-w}) - Li_2(w-1) + Li_2(w+1) \)

I = average of all the k pieces, dropped imaginary parts (since I is real)

For example, for k = 7, these are the integral pieces.

>>> from mpmath import *
>>> Li2 = lambda x: polylog(2, x)
>>> k = 7
>>> W = [exp((2*n+1)/k*pi*1j) for n in range(k)]
>>> for w in W: print -Li2(1/(w-1)) + Li2(1/(1-w)) - Li2(w-1) + Li2(w+1)
...
(2.46740110027234 + 5.0951007958138j)
(2.46740110027234 + 2.21291211873152j)
(2.46740110027234 + 0.940187304494618j)
(2.46740110027234 + 1.22410719457408e-16j)
(2.46740110027234 - 0.940187304494618j)
(2.46740110027234 - 2.21291211873152j)
(2.46740110027234 - 5.0951007958138j)

All the real parts have the same size = pi^2/4

What happened if k is not integer ? Lets try k = 2.021
Amazingly, for any non-zero real k, real part of the integral piece still = pi^2/4

>>> w = exp(pi*1j/2.021)
>>> print -Li2(1/(w-1)) + Li2(1/(1-w)) - Li2(w-1) + Li2(w+1)
(2.46740110027234 + 1.85770404643526j)

Note: this only show how integral evaluated to the same value, for different k's
It is still not a proof, but much closer than what I had before ...