Question

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.

added:

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.

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 $1/N$.