## integrate of functions involving floor

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

$$\int_2^x\frac{dt}{[\frac{\log x}{\log t}]\log t}$$

where [x] is the greatest integer less than or equal to x.

When I use

$$x-1<[x]\le x$$ I get $$\frac{x-2}{\log x}=\int_2^x\frac{dt}{\log x}\leq \int_2^x\frac{dt}{[\frac{\log x}{\log t}]\log t}\le \int_2^x\frac{dt}{\log x-\log t}$$ 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 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 at 16:03 Why do you integrate from $2$ and not for $e$? – Davide Giraudo Oct 25 at 16:14 @Davide Giraudo, @Gerhard Paseman, I added some phrases. – asd Oct 25 at 19:19

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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