The order of li(x)

Question

So I've been investigating $\mathrm{li}(x)$, that is the logarithmic integral function.

I am unsure if this is true, but it seems as if $\mathrm{li}(x) = O\left(\frac{x}{\log x}\right)$ for $x$ sufficiently large.

Specifically, it seems as if for $x$ sufficiently large, there exists an $M>0$, such that

$$|\mathrm{li}(x)|< M\left|\left(\frac x{\log x}\right)\right|.$$

Wolfram Alpha seems to agree with me, although I cannot be sure if this holds for all $x$.

I would like to prove this behavior myself, and so would rather have hints regarding how I could prove that $\mathrm{li}(x) = O\left(\frac{x}{\log x}\right)$ rather than proofs themselves as answers.

Answer 1

I suggest you try to find

$$\lim_{x\to\infty}\frac{\mathrm{li}(x)}{\frac{x}{\log x}}$$

using L'Hospital's rule.

Answer 2

You can also use the “european” definition $$\textrm{Li}\left(x\right)=\int_{2}^{x}\frac{dt}{\log\left(t\right)}=\textrm{li}\left(x\right)-\textrm{li}\left(2\right) $$ and integrating by parts we have $$\int_{2}^{x}\frac{dt}{\log\left(t\right)}=\left.\frac{t}{\log\left(t\right)}\right|_{2}^{x}+\int_{2}^{x}\frac{dt}{\log^{2}\left(t\right)}=$$ $$=\frac{x}{\log\left(x\right)}-\frac{2}{\log\left(2\right)}+\int_{2}^{x}\frac{dt}{\log^{2}\left(t\right)}$$ and note that $$\int_{2}^{x}\frac{dt}{\log^{2}\left(t\right)}\leq\frac{1}{\log\left(2\right)}\int_{2}^{x}\frac{dt}{\log\left(t\right)}. $$

Answer 3

x p + x p2 + x p3 + ··· 1 where [t] denotes the greatest integer less than or equal to t. It immediately follows that x! = p≤x p[x/p]+[x/p2]+··· and log(x!) = p≤x x p + x p2 + x p3 + ···
log(p). Now log(x!) is asymptotic to x log(x) by Stirling’s asymptotic formula, and, since squares, cubes, ... of primes are comparatively rare, and [x/p] is almost the same as x/p, one may easily infer that x p≤x log(p) p = x log(x) + O(x) from which one can deduce that π(x) is of order x log(x) . T