An alleged primary motivator for the RH is so that we can bound the error term $|\text{Li}(x) - \pi(x)|$ by a factor of $O(\sqrt{x}\log x)$. However, I also learned about Riemann's explicit formula $R(x)$ that converges, upon addition of the harmonics $C_i(x, \theta_i)$, where $\theta_i$ is the Riemann spectrum. Thus, we basically know that the Zeta function encodes enough information to construct an exact prime counting function and to know the distribution of the primes, $\pi(x)$, exactly. And even if we put the harmonics aside, $R_0(x)$ itself (no harmonics) is a much better approximation to $\pi(x)$ than just $\text{Li}(x)$, so why do we care so much about bounding the error between $\text{Li}(x)$ and $\pi(x)$? I get that $R(x)$ of course is a function of $\text{li}(x)$, but my point is shouldn't $R(x)$ be the fundamental object of study in the study of the distribution of primes, then?

Another, very related, question I have is why a square-root bound is good at all -- the Prime Number Theorem tells us that $\left|\frac{\pi(x)}{\text{Li}(x)}\right| \to 1$, so $\text{Li}(x)$ gets arbitrarily close to $\pi(x)$ for large enough $x$, so what does this (in my mind, very weak) square-root bound tell us that PNT doesn't already? These questions are probably very naïve -- I have taken a course in complex analysis, but not much number theory.

  • $\begingroup$ "...what does this square-root bound tell us that PNT doesn't already?" the way I see it, the estimate $\pi(x)\sim\text{Li}(x)$ tells us that $\text{Li}(x)$ approximates $\pi(x)$ better and better for larger and larger $x$, but it doesn't tell us how large $x$ needs to be for this approximation to hold within a given degree of accuracy. Bounds like $\pi(x)-\text{Li}(x)=O(\sqrt{x}\ln x)$ help in this regard because they refine $\pi(x)\sim\text{Li}(x)$, going beyond saying this estimate holds for large $x$ and giving information on how big $x$ needs to be for this to occur. $\endgroup$ Commented May 17, 2023 at 21:24
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    $\begingroup$ You mix the absolute and the relative error. The relative error is extremely small and tends to $0$ , if $n$ tends to $\infty$ , but the absolute error gets ever larger. $\endgroup$
    – Peter
    Commented May 17, 2023 at 21:48

1 Answer 1


It’s a decent approximation that can be calculated quickly. So it doesn’t help with limits, it helps with actually finding individual values. How many prime numbers approximately between $1.23 \times 10^{24}$ and $1.24 \times 10^{24}$?


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