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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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    $\begingroup$ I'm getting an answer close to $0.788531$. Your work in finding alternate forms for this number is very good. But I don't see any reason to think that we have an exact expression for this number. $\endgroup$ Nov 28, 2019 at 6:22
  • $\begingroup$ It is true that there is no reason for an exact expression, but I'm hoping that with the many expressions for the series, at least one will be able to be easily simplified. $\endgroup$ Nov 28, 2019 at 6:31

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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 $1.41 \times 10^{-18} \text{ %}$

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  • $\begingroup$ Thanks for this! Unfortunate that it doesn't have a closed-form expression. $\endgroup$ Nov 28, 2019 at 7:49
  • $\begingroup$ @automaticallyGenerated. You are welcome ! If it had a closed-form expression, it would not be one of the so many constants we work with !! You did a good job. Cheers and $\to +1$.. $\endgroup$ Nov 28, 2019 at 7:53
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    $\begingroup$ @automaticallyGenerated. I added an approximation of it (just for the fun). $\endgroup$ Nov 28, 2019 at 8:14
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    $\begingroup$ @ClaudeLeibovici How did you obtain that approximation? $\endgroup$
    – J.G.
    Nov 28, 2019 at 8:15
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    $\begingroup$ @J.G. In my former group, during his spare time, a guy built a code to find the best rational approximation of a given number as the ratio of two quadratics of $\pi$ or $e$. $\endgroup$ Nov 28, 2019 at 8:19

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