Evaluate $\int_0^1\frac{1}{\lfloor1-\log_{2}(1-x)\rfloor}dx$ Problem:
$$\int_0^1\frac{1}{1-\log_{2}(1-x)}dx$$
Which I simplified down to this:
$$2\log(2)\int_0^1\frac{1}{2^xx} dx$$
Now I am stuck as the answer is supposed to be computed without the use of a calculator, meaning non standard use of Ei(x) would not be applicable, methods such as Feynmann don't seem to work in this case either, please could someone assist me. Thanks
Edit : The denominator is floored , sorry
$$\int_0^1\frac{1}{\lfloor1-\log_{2}(1-x)\rfloor}dx$$
I am not sure what this now means, any help?
 A: Notice that 
$$\frac{1}{2^xx}\geq \frac{1}{2x}$$ on $[0,1]$ and 
$$\int_{0}^{1}\frac{1}{x}dx = 0 -\lim _{\epsilon \rightarrow 0 }\ln \epsilon=\infty$$
A: The answer given by @clark is correct for the originally given integral.  Things change quite a bit when we bring in the floor function.
In a comment on the question you asked what the floor function means in this instance.  It means the same thing it always does:  $\lfloor y \rfloor$ is the greatest integer less than or equal to $y$.  Therefore $\lfloor 1 - \log_2(1-x) \rfloor$ is the greatest integer less than or equal to $1 - \log_2(1-x)$.
Note that the integral is from $0$ to $1$.  Therefore we only care about the integrand on the interval $[0,1]$.  Also, $1-\log_2(1-x)$ is a strictly increasing function of $x$ on $(0,1)$, because its derivative is $\frac{1}{1-x}$, which is positive on $(0,1)$.  And since we have $1 - \log_2(1-0) = 1$, then the integral (if it exists, which it does) must be positive.
So now we need to better understand $\lfloor 1 - \log_2(1-x) \rfloor$ before we can do the integral.  Of course we know that $\lfloor 1 - \log_2(1-0) \rfloor = \lfloor 1 \rfloor = 1$.  But for which value of $x$ does this expression equal 2?  3?  4?  etc.?
$\lfloor 1-\log_2(1-x) \rfloor = 2$ when $1-\log_2(1-x) \in [2,3)$.  This basically gives us two inequalities:
$$
  1-\log_2(1-x) \ge 2 \qquad \text{and} \qquad 1-\log_2(1-x) < 3
$$
Solving these inequalities gives us $\dfrac{1}{2} \le x < \dfrac{3}{4}$.
More generally:
$\lfloor 1-\log_2(1-x) \rfloor = n$ when $1-\log_2(1-x) \in [n,n+1)$.  This gives us two inequalities:
$$
  1-\log_2(1-x) \ge n \qquad \text{and} \qquad 1-\log_2(1-x) < n+1
$$
Solving gives us $1-\dfrac{1}{2^{n-1}} \le x < 1 - \dfrac{1}{2^n}$.
So what we see then is
$$
  \lfloor 1 - \log_2(1-x) \rfloor = n \quad \text{if} \quad 1-\dfrac{1}{2^{n-1}} \le x < 1 - \dfrac{1}{2^n}.
$$
Note also that
$$
[0,1] = \bigcup_{n=1}^{+\infty} \left[1-\frac{1}{2^{n-1}}, 1-\frac{1}{2^n}\right].
$$
Therefore we have:
\begin{align*}
  \int_0^1 \frac{dx}{\lfloor 1 - \log_2(1-x)\rfloor}
    &= \sum_{n=1}^{+\infty}\int_{1-1/2^{n-1}}^{1-1/2^n} \frac{1}{n} \, dx\\[0.3cm]
    &= \sum_{n=1}^{+\infty} \left(\frac{1}{n} \cdot x \bigg|_{x=1-1/2^{n-1}}^{x=1-1/2^n}\right)\\[0.3cm]
    &= \sum_{n=1}^{+\infty} \frac{1}{n2^n}\\[0.3cm]
    &= \ln 2
\end{align*}
A: Note the we can write
$$\begin{align}
\int_0^1 \frac{1}{\lfloor1-\log_2(1-x) \rfloor }\,dx&=\int_0^1 \frac{1}{\lfloor \log_2\left(\frac{2}{1-x}\right) \rfloor }\,dx\\\\
&=\sum_{n=1}^\infty \int_{1-2^{1-n}}^{1-2^{-n}}  \frac{1}{\lfloor \log_2\left(\frac{2}{1-x}\right) \rfloor }\,dx\\\\
&=\sum_{n=1}^\infty \frac{2^{-n}}{n}\\\\
&=\log_e(2)
\end{align}$$
