A surprising integral result

$\begin{aligned} I & = \displaystyle\int_{0}^{\infty} \dfrac{dx}{ \lceil x \rceil 2^x} = \displaystyle\int_{0}^{1} \dfrac{dx}{1\cdot 2^x}+\displaystyle\int_{1}^{2} \dfrac{dx}{2\cdot 2^x}+\displaystyle\int_{2}^{3} \dfrac{dx}{3\cdot 2^x}+\cdots \\ & = \displaystyle\sum_{r=1}^{\infty}\int_{r-1}^{r} \dfrac{dx}{r\cdot 2^x} \\ & = \dfrac{1}{\ln{2}}\displaystyle\sum_{r=1}^{\infty} \dfrac{1}{r\cdot 2^r} \end{aligned}$

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$\begin{aligned} \ln{(1+x)} & = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}- \cdots \\ -\ln{(1-x)} & = x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+ \cdots \end{aligned}$

$\ \ln{2} = \dfrac{1}{2}+\dfrac{1}{2\cdot 2^2}+\dfrac{1}{3\cdot 2^3}+ \cdots = \displaystyle\sum_{r=1}^{\infty} \dfrac{1}{r\cdot 2^r}$

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$\boxed{I = \dfrac{1}{\ln{2}} \times \ln{2} =1}$