<s>[INST] Evaluate $\int_0^1 \frac 1 {\lfloor 1-\log_2(x) \rfloor}\,\mathrm{d}x$ Evaluate $$\int_{0}^{1} \frac{1}{\lfloor 1-\log_2(x) \rfloor}\,\mathrm{d}x$$ where $\lfloor \cdot \rfloor$ is the Greatest Integer Function (GIF)
I learnt that the solution is probably $\displaystyle\sum_{r=1}^\infty \frac 1 {r \cdot 2^r}.$ 
But how does that happen? 
 [/INST] Let $$f(x)=\frac {1}{\lfloor 1-\log_2 x\rfloor}$$
For $x\in \left(1/2,1\right)$, $f(x)=1$
For $x\in \left(1/4,1/2\right)$, $f(x)=1/2$
For $x\in \left(1/8,1/4\right)$, $f(x)=1/3$
So the integral turns out to be $$\frac 11\cdot\frac 12+\frac 12\cdot\frac 14+\frac 13\cdot\frac 18\cdots= \frac {1}{1\cdot 2^1}+\frac {1}{2\cdot 2^2}+\frac {1}{3\cdot 2^3}\cdots
=    \sum_{r=1}^{\infty} \frac {1}{r\cdot 2^r}$$
On further computation you might notice that the answer is simply $\ln 2$
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