# How to find $\int_0^1 \frac {\mathrm dx}{\left \lfloor{1-\log_2(1-x)}\right \rfloor}$

We want to evaluate;

$$\int_0^1 \frac {\mathrm dx}{\left \lfloor{1-\log_2(1-x)}\right \rfloor}$$

The $\left \lfloor{x}\right \rfloor$ is the floor function. I have made no progress so far.

• Is the floor on the entire denominator? Aug 21, 2014 at 21:37
• Yes. Fixing latex, just wait. Aug 21, 2014 at 21:39
• I'll fix it for you 1 sec. Aug 21, 2014 at 21:40
• Definitely not the right stackexchange, but why do you use $$instead of  and how do you make it bigger and centered? Aug 21, 2014 at 21:45 • The two dollar signs make it centered and bigger, but it's a block DOM object whereas one dollar sign is an inline-block and thus smaller. Aug 21, 2014 at 21:46 ## 1 Answer The integrand function is constant on a sequence of intervals, in particular:$$I=\int_{0}^{1}\frac{dx}{\left\lfloor 1-\log_2 x\right\rfloor}=\sum_{n=1}^{+\infty}\frac{2^{1-n}-2^{-n}}{n}=\sum_{n=1}^{\infty}\frac{2^{-n}}{n}=\color{red}{\log 2.}

• I can see by the graph why it would be a sum and I was trying to find similarities and paterns, how did you end up with that particular sum so fast? Aug 21, 2014 at 22:04
• By just looking on which intervals the integrand function equals $\frac{1}{1},\frac{1}{2},\ldots$. Aug 21, 2014 at 22:15
• Didn't you consider $n = 0$ ?. Because $\large\int_{0}^{1}{{\rm d}x \over \left\lfloor\,1 - \log_{2}\left(x\right)\,\right\rfloor} = \ln\left(2\right)\int_{0}^{\infty}{2^{1 - x} \over \left\lfloor\,x\,\right\rfloor}\,\dd x$ Aug 23, 2014 at 22:20