Proof for x/log(x)~ Li(x) Could anyone please show the steps for proving $\frac{x}{log(x)}$~ Li(x) or at least point to a source. Where ~ is the asymptotic equivalence. Any help would be appreciated.
 A: $\lim_{x\to \infty } \, \dfrac{\frac{x}{\log x}}{\text{Li}(x)}=1$
Use L'Hopital rule
$$\lim_{x\to \infty } \, \frac{\frac{1}{\log (x)}-\frac{1}{\log ^2(x)}}{\frac{1}{\log (x)}}=\lim_{x\to \infty } \, \left(1-\frac{1}{\log (x)}\right)=1$$
Hope this helps
A: Repeated integration by parts gives an asymptotic expansion:
$$\newcommand{\Li}{\operatorname{Li}}
\begin{align}
\Li(x)
&=\int_2^x\frac{\mathrm{d}t}{\log(t)}\\
&=\frac{x}{\log(x)}\left(1+\frac{1}{\log(x)}+\frac{2}{\log(x)^2}+\dots+\frac{k!}{\log(x)^k}+O\left(\frac{1}{\log(x)^{k+1}}\right)\right)
\end{align}
$$
Note that the constants introduced by the integration by parts are included in the $O\!\left(\frac{x}{\log(x)^{k+2}}\right)$ term, which grows faster than the sum of those constants.
A: I am not entirely convinced that all conditions for using L'Hôpital's rule are met in the accepted answer from @Raffaele. Therefore I would like to share a proof from Peter Bundschuh's book "Einführung in die Zahlentheorie", page 295. It uses integration by parts:
$$
\newcommand{\Li}{\operatorname{Li}}
\Li x - \Li 2 = \int_2^x 1 \cdot \frac{1}{\log t} \, \mathrm{d}t = \frac{x}{\log x} - \frac{2}{\log 2} + \int_2^x \frac{\mathrm{d}t}{\log^2 t}
$$
$$
\Rightarrow | \Li x - \frac{x}{\log x} | \leq \int_\sqrt{x}^{x} \frac{\mathrm{d}t}{\log^2 t} + \mathcal{O}(\sqrt{x}) = \mathcal{O} \left( \frac{x}{\log^2 x} \right)
$$
