MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there any exact formula or at least exact inequalities for the following intehral

$$ \int_{2}^{x}{{\rm d}t \over \left\lfloor\vphantom{\large h}% \log(x)/\log(t)\right\rfloor \log\left(t\right)} $$

where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.


When I use

$$ x-1<\lfloor x\rfloor\le x $$ I get $$ \frac{x-2}{\log (x)}=\int_2^x\frac{{\mathrm d}t}{\log (x)}\leq \int_2^x\frac{{\mathrm d}t}{\left\lfloor\vphantom{\large h}\log (x)/\log (t)\right\rfloor\log (t)}\le \int_2^x\frac{\mathrm dt}{\log (x)-\log (t)} $$ but they are not exact enough. I need more closer bounds.

share|cite|improve this question
Why not replace log t by a (temporary) constant c, and then sum rectangle areas? Gerhard "Ask Me About System Design" Paseman, 2012.10.25 – Gerhard Paseman Oct 25 '12 at 15:59
To answer my own question, because it is dt, not dx. Gerhard "So Sorry About That, Chief" Paseman, 2012.10.25 – Gerhard Paseman Oct 25 '12 at 16:03
Why do you integrate from $2$ and not for $e$? – Davide Giraudo Oct 25 '12 at 16:14
@Davide Giraudo, @Gerhard Paseman, I added some phrases. – asd Oct 25 '12 at 19:19
Can you use z-(1/2+atan(tan(pi*(z-1/2)))/pi) instead of the floor function, replacing z with log(x)/log(t)? Such a thing may not necessarily work and could introduce other problems, but perhaps it could help? – user78169 Aug 13 '15 at 22:17

Note that $\left\lfloor \dfrac{\log x}{\log t} \right\rfloor = n$ for $x^{1/(n+1)} < t < x^{1/n}$, so if $m = \left\lceil \dfrac{\log x}{\log 2} \right \rceil$

$$ \eqalign{\int_{2}^x &\frac{dt}{\left\lfloor \frac{\log x}{\log t}\right\rfloor \log t} = \int_2^{x^{1/m}} \dfrac{dt}{m \log t} + \sum_{j=1}^{m-1} \int_{x^{1/(j+1)}}^{x^{1/j}} \dfrac{dt}{j \log t} \cr &= \frac{1}{m} \text{Ei}(1,-\log 2) - \text{Ei}(1,-\log x)+ \sum_{j=2}^{m} \frac{1}{j(j-1)} \text{Ei}(1,-\frac{1}{j} \log x) \cr}$$

You might also note that $$ \frac{x^{1/j} - x^{1/(j+1)}}{\log x} \le \int_{x^{1/(j+1)}}^{x^{1/j}} \dfrac{dt}{j \log t} \le \frac{j+1}{j} \frac{x^{1/j} - x^{1/(j+1)}}{\log x} $$

so using these bounds for all $j > N$ and the exact values for $j \le N$ will give you approximations with relative error at most $1/N$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.