Logarithmic integral, $π(x)$ and $x/(\ln x)$

Question

The function $\text{Li}$ (logarithmic integral) is defined for $x>0$ by $$ \text{Li}(x)=\int_2^{x}\frac{dt}{\ln t}. $$ The prime number theorem, proven by Hadamard and de la Vallée-Poussin in 1896 asserts that $π(x) \sim \text{Li}(x)$ when $x$ goes to $+\infty$ where $ π(x)=\text{Card}\{2\le p\in \mathbb N, \text{ $p$ is prime}\}. $ It was proven by J. Littlewood in 1914 that the difference $d(x)=\text{Li}(x)-π(x)$ changes sign an infinite number of times, although for $x\le 10^{25}$, $d(x)>0$. (I guess that the $10^{25}$ could be enlarged.) I think that it is also known that for $x$ quite large, say in the interval $[10^{300}, 10^{500}]$, there exists $x$ such that $d(x)<0$. Thanks to the answers below, I have understood that there is no explicit value of $x$ for which we know that $d(x)<0$.

A related question: is it true that for $x$ large enough $$ π(x)\ge \frac{x}{\ln x}? $$

Answer 1

No explicit values are known.

For a recent reference, see "A still sharper region where $π(x) − \text{li}(x)$ is positive", by Y. Saouter, T.S. Trudgian and P. Demichel (2014). An extract from the last section of the paper:

The last point we shall address is the existence of smaller values of $x$ for which $π(x)>\text{li}(x)$. The is no practical way to test an isolated value of $x$. Analytical methods, such as the one described here, can only give intervals of positivity.

PS. Yes, $ π(x)\ge \frac{x}{\ln x}$ for all $x\geq17$. See "Approximate formulas for some functions of prime numbers" by Rosser and Schoenfeld, page 6, for example.

Answer 2

To your added question: yes, $\pi(x)\ge \frac{x}{\ln x}$ certainly holds for $x\geq 59$ by Theorem 1 in Rosser-Schoenfeld: Approximate formulas for some functions of prime numbers.

Answer 3

Just to address the question $\pi(X) \geq \frac{X}{\log X}$? The above answers are sufficient -- let's just see why that is elementarily.

Let's assume RH just to illustrate the following idea/approximation for $\pi(X)$. Thus, we have $\pi(X) = Li(X) + O(X^{\frac{1}{2} + \varepsilon})$. If we rewrite $Li(X)$ via integration by parts, we see $Li(X) = \frac{X}{\log X} + \int_2^{X} \frac{1}{\log^2 t} dt + O(1) = \frac{X}{\log X} + \frac{X}{\log^2 X} + O\left(\frac{X}{\log^3 X}\right)$. Thus, $\pi(X) - \frac{X}{\log X} \sim \frac{X}{\log^2 X}$; now just take $X$ sufficiently large!

Note: one can iteratively expand via integration by parts, and the RH assumption emphasizes no term $\frac{X}{\log^N X}$ is outraced by the error term. However, it is overkill as the classical PNT estimate is also small enough.