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.
$\endgroup$ 13 Answers
$\begingroup$I suggest you try to find
$$\lim_{x\to\infty}\frac{\mathrm{li}(x)}{\frac{x}{\log x}}$$
using L'Hospital's rule.
$\endgroup$ 6 $\begingroup$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 $$\int_{2}^{x}\frac{dt}{\log^{2}\left(t\right)}\leq\frac{1}{\log\left(2\right)}\int_{2}^{x}\frac{dt}{\log\left(t\right)}. $$
$\endgroup$ $\begingroup$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
