Spence function

[Albert Chan]

Chaos in Numberland: The secret life of continued fractions, by John D. Barrow

In this great article, there was an identity that we can prove, using Spence function.
(section: What is chaos ?, equation 36, where h = Kolmogorov, or metric, entropy of the mapping)



Do integration by parts, where \(u = \ln(x), v = \ln(1+x) \)

\(\displaystyle \int {\ln(x) \over 1+x}\,dx
= \int \ln(x)\,d(\ln(1+x))
= \ln(x)\ln(1+x)\;+ \int {-\ln(1+x) \over x}\,dx
\)

Adding limit of integration, x = 0 to 1, last term = Li2(-1)

At x=1, uv = ln(1) ln(2) = 0
At x=0, uv = ln(0) ln(1) = -∞ * 0 = ?

Note: At x=0, v = ln(1+x) = 0 , v' = 1/(1+x) = 1

\(\displaystyle \lim_{x \to 0} {u \over v^{-1}}
= \lim_{x \to 0} {u' \over -v^{-2}\;v'}
= \lim_{x \to 0} {-v^2 \over x}
= \lim_{x \to 0} {-2v\,v' \over 1}
= 0 \)

Or, we can avoid repeated applications of L'Hospital rule, by letting x = 1/n

\(\displaystyle \lim_{ x \to 0}\,(uv)
= \lim_{n \to ∞} \left[ \left({-\ln(n) \over n}\right)
\ln\left(\left(1+{1\over n}\right)^n\right) \right]
= (-0) (1) = 0 \)

For n → ∞, n ≫ ln(n) ; e = limit((1+1/n)^n, n=inf), by definition

Either way, uv term disappered, and we are left with last term, Li2(-1) = -pi^2/12

\(\displaystyle h
= {-2 \over \ln(2)} \int_0^1 {\ln(x) \over 1+x}\,dx
= {-2 \over \ln(2)} {-\pi^2 \over 12}
= {\pi^2 \over 6\ln(2)} \)