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 $$S_3 = \sum_{n=2}^{\infty} \sum_{m=2}^{\infty} \frac{1}{(m-1)n^m} = \sum_{m=2}^{\infty} \frac{\zeta(m) – 1}{m-1}$$
Best Answer
This number is Alladi-Grinstead constant (have a look here) $$c=0.78853056591150896106027632345455466647274966822328164975515640230178\cdots$$ If you want a "fuuny" approximation of it $$c \sim\frac{654 \pi ^2-449 \pi -1509}{341 \pi ^2+485 \pi-406}$$ which is in error of $1.41 \times 10^{-18} \text{ %}$