Let $\pi(x)$ denote the prime counting function. It is well known that $\log \zeta(s) = \int_{2}^{\infty} \dfrac{s\pi(x)}{x(x^s - x)} \mathrm d{x}$ where $\Re(s)\geq 2$.

Inserting $s=4$, we have

$\log \zeta(4) = \log \dfrac{\pi^4}{945} = \int_{2}^{\infty} \dfrac{4\pi(x)}{x^4 - x} \mathrm d{x}$

From such a formula, why can't we determine the exact value (or at least sharper bounds), for $\pi (x)$ ?

EDIT: Infact it seems from this, we can verify the inequality $\mid \pi(x) - Li(x) \mid \leq \sqrt x \log x$ for all $x$ such that $\pi (x) > Li(x)$. We sketch the argument as follows:

Inserting $s=2$, we have $\log \zeta(2) = \log \dfrac{\pi^2}{6} = \int_{2}^{\infty} \dfrac{2\pi(x)}{x^2- x} \mathrm d{x}$. Suppose on the contradiction that $x$ is the least counterexample such that $\pi (x) > Li(x) + \sqrt x \log x$. Observe that this would imply that

$\log \zeta(2) > \int_{2}^{\infty} \dfrac{2(Li(x) + \sqrt x) \log x}{x^3- x} \mathrm d{x}$.

By the ineqaulity $Li(x)>\dfrac{x}{\log x}$ for all $x\geq 2$, the right hand side is $>$ $\int_{2}^{\infty} \dfrac{2(x + \sqrt x (\log x)^2)}{(x^2- x)\log x} \mathrm d{x}$, and by Wolfram one quickly finds that this is equal to $2.4296\cdots$.

But we now have $\log \zeta (2) = \log \dfrac{\pi^2}{6} = 0.498 > 2.4296$, an absurdity ?

  • $\begingroup$ You cannot extract any local information about how $\pi(x)$ behave just from knowing the total integral. Said differently: there are infinitely many different function '$\pi(x)$' that give rise to the same value for the integral. $\endgroup$
    – Kibble
    Jan 22, 2016 at 1:09
  • $\begingroup$ You say "suppose on the contradiction that $x$ is the least counterexample" and then you continue to integrate over all $x$ assuming this holds for all $x$. Does not make sense. All this argument shows (when done correctly) is that we cannot have $\pi(x) > \text{li}(x) + \sqrt{x}\log(x)$ for all $x$. This is true and it is for example known that there are infinitely many values $x$ for which $\pi(x) = \text{li}(x)$. $\endgroup$
    – Kibble
    Jan 22, 2016 at 1:57
  • $\begingroup$ @Kibble, thanks, i think your explanation is clearer. So where exactly does the argument need to be corrected such that it proves what you stated, that is: the inequality $\pi(x) > li(x) + \sqrt x \log x$ cannot hold for all $x$ ? $\endgroup$
    – User1
    Jan 22, 2016 at 2:03
  • $\begingroup$ Change "suppose on the contradiction that $x$ is the least counterexample" to "Assume $\pi(x) > ...$ for all $x$ then ..." $\endgroup$
    – Kibble
    Jan 22, 2016 at 2:04
  • $\begingroup$ Thank you very much, really interesting. I'm a prospective undergraduate student sir. Can i write a paper on this as an ''elementary proof that...'' and perhaps submit it to an undergrad/less ''prestigious'' journal which i think will surely strengthen my undergrad freshman application? $\endgroup$
    – User1
    Jan 22, 2016 at 2:08

1 Answer 1


Your mistake starts with "suppose by contradiction that..."

The opposite of "something is true for all $x$" is NOT "something is false for all $x$".

Your statement should be $\exists x$ such that $\pi (x) > Li(x) + \sqrt x \log x$. But then, since this is not true for all $x$ you don't get the integral inequality anymore.

  • $\begingroup$ I think the supposition is: ''the inequality is false for that particular $x$'', not all $x$. $\endgroup$
    – User1
    Jan 22, 2016 at 1:47
  • $\begingroup$ @User1 If the inequality is false for that particular $x$, then the integral inequality doesn't hold anymore. You are abusing notation here, which leads to the confusion: you denote by $x$ the particular fixed number for which the inequality fails, and the variable of integration.... $\endgroup$
    – N. S.
    Jan 22, 2016 at 2:05
  • $\begingroup$ okay thank you. I now see where my mistake is. Thank you very much. $\endgroup$
    – User1
    Jan 22, 2016 at 2:10

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