# Why do number theorists care so much about how well $\text{Li}(x)$ approximates $\pi(x)$ if it's not our best approximation?

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.

• "...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. Commented May 17, 2023 at 21:24
• 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. Commented May 17, 2023 at 21:48

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 $$1.24 \times 10^{24}$$?