TITLE: Proving $\int_0^1 \frac{\mathrm{d}x}{1-\lfloor \log_2(1-x)\rfloor} = 2 \log 2 - 1$.
QUESTION [7 upvotes]: By testing in maple I found that 
$$
\int_0^1 \frac{\mathrm{d}x}{1-\lfloor \log_2(1-x)\rfloor} = 2 \log 2 - 1
$$
Does there exists a proof for this? I tried rewriting it as an series but no luck there. 
$$
\frac{1}{1-\lfloor \log_2(1-x)\rfloor}
= \sum_{k=0}^\infty \lfloor \log_2(1-x) \rfloor^k
= \sum_{k=0}^\infty \Big \lfloor -\frac{1}{\log 2} \sum_{n=0}^\infty \frac{x^n}{n} \Big\rfloor^k
$$
However this expression seems to diverge

REPLY [2 votes]: A related technique. Here is an approach. Let's make the change of variables $\log_2(1-x)=-u$ which results in the integral
$$ I  = \int_0^1 \frac{{d}x}{1-\lfloor \log_2(1-x)\rfloor} = \int_0^{\infty} \frac{2^{-u}{d}u}{1-\lfloor -u\rfloor} =\ln(2)\sum_{k=0}^{\infty}\int_k^{k+1}\frac{2^{-u}{d}u}{1-(-k-1)} $$
$$ = \ln(2)\sum_{k=0}^{\infty}\frac{1}{k+2}\int_k^{k+1}{2^{-u}{d}u}= \sum_{k=0}^{\infty}\frac{2^{-k-1}}{k+2}=2\ln(2)-1. $$
Note: 

$$ \lfloor -u \rfloor = -k-1 \quad \mathrm{if} \quad -k-1 \leq   -u  <-k.  $$