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Mathematicians back in 19th century tried to find a function that satisfies $$\lim_{x\to\infty}\frac{\pi(x)}{f(x)}=1$$ and $f(x)$ turns out to be $\frac{x}{\ln x}$, or any function asymptotic to it(like $\text{Li}(x)$). They proved it rigorously and now it is known as the Prime Number Theorem.

However, I don’t see much work on finding a function that satisfies $$\lim_{x\to\infty}(\pi(x)-g(x))=0$$ As far as I know, $$\lim_{x\to\infty}(\pi(x)-\frac{x}{\ln x})=\infty$$ so $\frac{x}{\ln x}$ cannot be a candidate of $g(x)$.

Moreover, if such function is discovered, it will be very useful in the sense that estimation of number of primes below some large $N$ can become more and more precise as $N$ grows. This would surely be more powerful than PNT.

Why only little work has done by mathematicians to figure out $g(x)$?

Szeto
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    Well, even the Riemann hypothesis implies only $\pi(x) - \text{li}(x) = O(\sqrt{x}\log x).$ Since $\text{li}(x)$ is continuous and $\pi(x)$ is highly discontinuous, studying the difference isn't very profitable. It seems to be to be almost equivalent to asking, "what numbers are prime?" – Dzoooks Jul 24 '18 at 00:52
  • But, we can contrast the study of $\pi(x)$ with the partition function: https://en.wikipedia.org/wiki/Partition_(number_theory)#Partition_function , which has (an exact!) $g(x)$ in the form of an infinite series. – Dzoooks Jul 24 '18 at 00:57
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    Such a function would essentially encode all information about sufficiently large prime numbers. For example, you could tell whether $x$ was prime by looking at $g(x)$ and $g(x-1)$. – Jair Taylor Jul 24 '18 at 01:19
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    @Jair: So basically, it'd have to be $\pi(x)$ up to some lower order error term that vanishes in the limit. That's actually pretty boring, unless you've got a closed form of some kind. – Kevin Jul 24 '18 at 05:11

2 Answers2

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In brief, because what you're suggesting is overwhelmingly more powerful than the multiplicative difference, to the point where none of the known techniques can even come close. It's not that this isn't studied; indeed, 'additive differences' on the PNT and related functions - but as noted in a comment, they're usually only as good as being able to say $\pi(x)=f(x)+O(x^\alpha)$ for some $\alpha$ (typically with $\alpha\gt 1/2$). Note that these sorts of asymptotics imply the 'multiplicative' equalities you mention (since then $\pi(x)/f(x)=1+O^*(x^{\alpha-1})$), but the additive results you're requesting are even stronger than reducing $\alpha$ to zero.

Steven Stadnicki
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The question you are suggesting, in its exact form, would basically be a lost fight. Should you be able to find such function $g(x)$, this would mean $$\forall \epsilon > 0\ \exists x_0 \in \mathbb{N}\ \forall x \in \mathbb{N}, x > x_0: |\pi(x) - g(x)| < \epsilon.$$ If you plug in $\epsilon = 1/2$, this tells you that from some $x_0$ on the function value $g(x)$ differs from the $x$-th prime by less than a half, in other words, that you could just round it to the nearest integer to get the $x$-th prime. I doubt anyone nowadays hopes that a closed form like that could be found.

The Vee
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