Is there any exact formula or at least exact inequalities for the following intehral
$$ \int_{2}^{x}{{\rm d}t \over \left\lfloor\vphantom{\Large A}% \log\left(x\right)/\log\left(t\right)\right\rfloor \log\left(t\right)} $$$$ \int_{2}^{x}{{\rm d}t \over \left\lfloor\vphantom{\large h}% \log(x)/\log(t)\right\rfloor \log\left(t\right)} $$
where [x]$\lfloor x\rfloor$ is the greatest integer less than or equal to x$x$.
added:
When I use
$$ x-1<[x]\le x $$$$ x-1<\lfloor x\rfloor\le x $$ I get $$ \frac{x-2}{\log x}=\int_2^x\frac{dt}{\log x}\leq \int_2^x\frac{dt}{[\frac{\log x}{\log t}]\log t}\le \int_2^x\frac{dt}{\log x-\log t} $$$$ \frac{x-2}{\log (x)}=\int_2^x\frac{{\mathrm d}t}{\log (x)}\leq \int_2^x\frac{{\mathrm d}t}{\left\lfloor\vphantom{\large h}\log (x)/\log (t)\right\rfloor\log (t)}\le \int_2^x\frac{\mathrm dt}{\log (x)-\log (t)} $$ but they are not exact enough. I need more closer bounds.
