An alternating harmonic series with trigonometry

The ceiling function is $1$ , if and only if $sin(1/x) > 0$

Therefore the integral becomes a sum of the intervals, where $\sin {1 \over x}$ is non negative, thus one gets:

$\left(1-\frac{1}{\pi}\right)+\left(\frac{1}{2\pi}-\frac{1}{3\pi}\right)+\left(\frac{1}{4\pi}-\frac{1}{5\pi}\right)+\ldots \\ = 1-\frac{1}{\pi}\left(1-\frac12+\frac13-\frac14+\ldots\right) \\ = 1-\frac{1}{\pi}\ln(1+1)=1-\frac{\ln(2)}{\pi} \approx 0.7993...$

Which leads directly to the desired answer: $799$