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I am looking to evaluate the integral $$\int_0^1 \frac{dx}{\lfloor 1 - \log_2(1 - x) \rfloor}$$ Investigating the graph of the function, I found that the area under the curve equaled $$\sum_{n = 1}^\infty \frac{1}{n \cdot 2^n}$$ I checked the series on my calculator and found that it was equal to $\ln 2$. I also found it on Wikipedia under "Natural logarithm of 2" and "Bailey-Borwein-Plouffe formula", on Wolfram|Alpha and on Wolfram MathWorld. However, none of them gave a derivation.

How can I find the value of the infinite sum? Alternatively, how else can I evaluate the integral?

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$$\sum_{n=1}^{+\infty}\frac{1}{n\cdot 2^n}=\sum_{n=1}^{+\infty}\int_{0}^{1}\frac{x^{n-1}}{2^n}\,dx =\int_{0}^{1}\frac{dx}{2-x}=\log 2.$$

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