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

Question

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.

Definitely not the right stackexchange, but why do you use $$ instead of $ and how do you make it bigger and centered? — UserX, Aug 21 '14 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. — Shahar, Aug 21 '14 at 21:46
I don't know what you mean by graphing. If you mean a numerical solution, no I'm looking for an analytical one. — UserX, Aug 21 '14 at 21:51
No I meant that to solve where the denominator change, but I can do it without it (I think) — Shahar, Aug 21 '14 at 21:52

score 5 · Accepted Answer · edited Aug 21 '14 at 22:38

5

The integrand function is constant on a sequence of intervals, in particular: $$I=\int_{0}^{1}\frac{dx}{\left\lfloor 1-\log_2 x\right\rfloor}=\sum_{n=1}^{+\infty}\frac{2^{1-n}-2^{-n}}{n}=\sum_{n=1}^{\infty}\frac{2^{-n}}{n}=\color{red}{\log 2.}$$

edited Aug 21 '14 at 22:38

user84413

27,747

answered Aug 21 '14 at 21:58

Jack D'Aurizio

361,689

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? – UserX Aug 21 '14 at 22:04
2

By just looking on which intervals the integrand function equals $\frac{1}{1},\frac{1}{2},\ldots$. – Jack D'Aurizio Aug 21 '14 at 22:15
Didn't you consider $n = 0$ ?. Because $\large\int_{0}^{1}{{\rm d}x \over \left\lfloor,1 - \log_{2}\left(x\right),\right\rfloor} = \ln\left(2\right)\int_{0}^{\infty}{2^{1 - x} \over \left\lfloor,x,\right\rfloor},\dd x$ – Felix Marin Aug 23 '14 at 22:20

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

1 Answers1