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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 $x$.

added:

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.

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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

1 Answer 1

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 $1/N$.

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