Evaluating $\int_{0}^1 \ln \left\lfloor \frac{1}{x} \right\rfloor dx$

Inspired by this question, I decided to ask about $$I = \int_{0}^1 \ln \left\lfloor \frac{1}{x} \right\rfloor dx$$

This can be easily converted to an infinite summation by considering each segment: the $n$th segment, starting at $n = 1$ from the right and going to the left, has length $\frac{1}{n}-\frac{1}{n+1}$ and height equal to $\ln(n)$. This then means that the integral is equal to $$S_1 = \sum_{n=1}^{\infty} \left( \frac{1}{n}-\frac{1}{n+1} \right) \ln(n)$$

Through telescoping, this can be rewritten as $$S_2 = \sum_{n=2}^{\infty} \frac{\ln(n)-\ln(n-1)}{n}$$

I don't know how $S_2$ or either of the other representations can be solved, but I am fairly certain that it converges since numerical approximations have given me an answer around $0.788$.

Any help in solving the integral would be appreciated.

Edit: By using $$\frac{\ln(n)-\ln(n-1)}{n} = \sum_{m=2}^{\infty} \frac{1}{(m-1)n^m}$$ the series can be rewritten as $$S_3 = \sum_{m=2}^{\infty} \sum_{m=2}^{\infty} \frac{1}{(m-1)n^m} = \sum_{m=2}^{\infty} \frac{\zeta(m) - 1}{m-1}$$

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