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.

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    $\begingroup$ Why, you may find the ratio at infinity using the L'Hopital's rule. $\endgroup$ – Ivan Neretin Sep 13 '15 at 8:31

I suggest you try to find

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

using L'Hospital's rule.

  • $\begingroup$ One could also note that the two functions are asymptotically bounded to the prime counting function $\pi(x)$ , so they must be asymptotically bounded to each other. But how does limit this imply the OP? $\endgroup$ – Linus S. Sep 13 '15 at 8:38
  • $\begingroup$ I have proved this using L'Hospital's rule, as you suggested. Am I right in saying now that all that is needed to show that $\mathrm{li}(x) = O\left(\frac{x}{\log x}\right)$ , is the reverse triangle inequality? $\endgroup$ – Aneesh Sep 13 '15 at 8:42
  • $\begingroup$ @LinusS. With this limit being $L$, you know that, for a large enough $x$, you have $$\frac{\mathrm{li}(x)}{\frac{x}{\log x}}< L + \epsilon$$ and multiplying the inequality gives the OP what he needs (knowing that all the factors here are positive for a large enough $x$. $\endgroup$ – 5xum Sep 13 '15 at 8:48
  • $\begingroup$ @moorish I don't know what you mean by the reverse triangle inequality... $\endgroup$ – 5xum Sep 13 '15 at 8:48
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    $\begingroup$ @5xum Using the fact that $|x|-|y| < |x-y|$ we can prove that $\mathrm{li}(x) = O\left(\frac{x}{\log x}\right)$ is what I am saying. Of course we need to multiply the inequality like you suggested, before we use this fact. $\endgroup$ – Aneesh Sep 13 '15 at 8:52

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 $$\int_{2}^{x}\frac{dt}{\log^{2}\left(t\right)}\leq\frac{1}{\log\left(2\right)}\int_{2}^{x}\frac{dt}{\log\left(t\right)}. $$


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

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    $\begingroup$ Hi and welcome to Math.SE. Please use mathjax to write questions/answers, since it gets more readable for all of us... $\endgroup$ – mickep Sep 13 '15 at 9:53
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 13 '15 at 10:24

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