6
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Why, you may find the ratio at infinity using the L'Hopital's rule. $\endgroup$ – Ivan Neretin Sep 13 '15 at 8:31
6
$\begingroup$

I suggest you try to find

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

using L'Hospital's rule.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
5
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • 7
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.