The prime counting function $\pi(x)$ is defined as
\begin{equation}
\pi(x)=\sum_{p\leq x}1
\end{equation}
where $p$ runs over primes.
I have seen many bounds for $\pi(x)$ such as
\begin{equation}
\frac{x}{\log x}\left(1+\frac{1}{2\log x}\right)<\pi(x)<\frac{x}{\log x}\left(1+\frac{3}{2\log x}\right)
\end{equation}
\begin{equation}
\frac{x}{\log x  1/2}<\pi(x)<\frac{x}{\log x + 3/2}
\end{equation}
\begin{equation}
\frac{x}{\log x+2}<\pi(x)<\frac{x}{\log x  4}
\end{equation}
Till now, what are the best known upper and lower bounds for the primecounting function? Is there a better bound that $\mathrm{Li}(x)$?

1$\begingroup$ Some better bounds are here $\endgroup$ – Wojowu Oct 10 at 10:10

3$\begingroup$ ${\rm Li}(x)$ is neither an upper bound nor a lower bound for $\pi(x)$, so it's not clear to me how you can call it a bound. Maybe what you want is $f(x)$ such that $f(x)\pi(x)<{\rm Li}(x)\pi(x)$ for all $x$, or for all sufficiently large $x$? $\endgroup$ – Gerry Myerson Oct 10 at 11:32

2$\begingroup$ $f(x)=\pi(x)$ is the best approximation to $\pi(x)$. A close relative is $f(x)=\pi(x)+1$. $\endgroup$ – GH from MO Oct 10 at 15:03

3$\begingroup$ mathguy, do you know the prime number theorem with its error term? That will show that functions of the form $\mathop{\rm Li}(x) \pm C x\exp(c\sqrt{\log x})$ are upper and lower bounds for $\pi(x)$, and explicit values for the constants $C$ and $c$ can be found. This is nearly the best known bound (the power of $\log x$ inside the exponential has been improved). Certainly no bounds of the form $\mathop{\rm Li}(x) \pm Cx^{1\delta}$ are known. $\endgroup$ – Greg Martin Oct 10 at 19:51

1$\begingroup$ @mathguy: In my response below, I added some explicit bounds of the type Greg Martin mentioned. Check them out! $\endgroup$ – GH from MO Oct 10 at 21:18
The following explicit version of the Prime Number Theorem was proved by Trudgian: $$ \pi(x)\mathrm{li}(x)<x e^{0.39\sqrt{\ln x}},\qquad x\geq 229.\tag{$\ast$}$$ In fact Trudgian's Theorem 2 is somewhat stronger than $(\ast)$, and with Mathematica it is straightforward to extend the validity of $(\ast)$ to $x\geq 2$.