I see a proof in https://arxiv.org/abs/1805.11965 (equation 3.36) that uses the following.
$\log x = \int_0^{\infty} ds \left(\frac{1}{1+s} - \frac{1}{s+x}\right)$.
This seems to hinge on $\int \frac{1}{x} = \log_2 x$ (the context is information theory), as opposed to $\log_e(x)$. Why is this true?
On
Note that for $a > 0$, $$ \int_0^M \frac{1}{a+s}ds = \int_{a}^{M+a} \frac{1}{s} ds = \ln (M+a) - \ln a. $$ Therefore, $$ \int_0^M \left(\frac{1}{1+s} - \frac{1}{x+s}\right) ds = \ln (M+1) - \ln 1 - (\ln (M+x) - \ln x). $$ Therefore, by letting $M \to \infty$, we have $$ \int_0^\infty \left(\frac{1}{1+s} - \frac{1}{x+s}\right) ds = \ln x. $$
The notation in the paper is a little confusing because in the classical part entropy is measured in bits and $\log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)