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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?

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  • $\begingroup$ note that the original function is not defined at $1$, so you should break it apart as an improper integral. Once you do so the fact the integral doesn't converge should become clear! $\endgroup$ – Brevan Ellefsen Jul 1 '16 at 21:06
  • $\begingroup$ Maybe the question is the following: $$\int_{0}^{1}\frac{{\rm d}x}{\left \lfloor 1-\log_2(1-x) \right \rfloor}$$ where $\left \lfloor \cdot \right \rfloor$ denotes the floor function? $\endgroup$ – Tolaso Jul 1 '16 at 21:08
  • $\begingroup$ Yes, Sorry, you are correct, What does the floor function mean then, in this instance $\endgroup$ – Lewis Kelsey Jul 1 '16 at 21:12
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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*}

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Notice that $$\frac{1}{2^xx}\geq \frac{1}{2x}$$ on $[0,1]$ and

$$\int_{0}^{1}\frac{1}{x}dx = 0 -\lim _{\epsilon \rightarrow 0 }\ln \epsilon=\infty$$

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  • $\begingroup$ The original integral is convergent. The intermediate result is not correct. $\endgroup$ – Mark Viola Jul 2 '16 at 0:39
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Note the we can write

$$\begin{align} \int_0^1 \frac{1}{\lfloor1-\log_2(1-x) \rfloor }\,dx&=\int_0^1 \frac{1}{\lfloor \log_2\left(\frac{2}{1-x}\right) \rfloor }\,dx\\\\ &=\sum_{n=1}^\infty \int_{1-2^{1-n}}^{1-2^{-n}} \frac{1}{\lfloor \log_2\left(\frac{2}{1-x}\right) \rfloor }\,dx\\\\ &=\sum_{n=1}^\infty \frac{2^{-n}}{n}\\\\ &=\log_e(2) \end{align}$$

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