Is there any general method of finding the lower bound of $x$ that satisfy the inequality?

Question

Is there any general method to find a real valued function $f(x)$ such that,

$$\dfrac{x}{\ln x -(1+\epsilon)}>\pi(x)>\dfrac{x}{\ln x -(1-\epsilon)}$$

for all $x \geq f(\epsilon)$. The value of this $f(\epsilon)$ needn't be the smallest lower bound of $x$ but just a lower bound. For example this paper and this paper gives some estimates of this kind but don't indicate any general method.

Any idea as to how to progress in this problem? Has there already been any progress regarding this problem?

Answer 1

Per the discussion on MO meta, I’m going to ignore “general method”, and just exhibit an explicit function with this property.

Up to simple manipulations, it is in fact given in Dusart’s thesis linked in the question. Let $\epsilon_x$ be the number such that $$\pi(x)=\frac x{\log x-(1+\epsilon_x)}.$$ Théorème 1.10, item 5 tells you that $\epsilon_x\ge0$ for $x\ge5393$. On the other hand, for $x\ge355991$, we have $$\pi(x)\le\frac x{\log x}\left(1+\frac1{\log x}+\frac{2{.}51}{\log^2x}\right).$$ Since $$\pi(x)=\frac x{\log x\left(1-\frac{1+\epsilon_x}{\log x}\right)}>\frac x{\log x}\left(1+\frac{1+\epsilon_x}{\log x}+\frac{(1+\epsilon_x)^2}{\log^2x}\right),$$ this implies $$\epsilon_x<\frac{1{.}51}{\log x}.$$ One can easily verify with a computer that this also holds for all $x\le355991$. However, $\epsilon_x$ may be negative for very small $x$; nevertheless, $\epsilon_x>-1{.}51$ for $x\ge227$. Thus, $$\lvert\epsilon_x\rvert<\frac{1{.}51}{\log x}\qquad\text{for }x\ge227,$$ which means the property in the question holds with $$f(\epsilon)=\max\left\{227,\exp\left(\frac{1{.}51}\epsilon\right)\right\}.$$ Note that this is not far from optimal: using the asymptotic expansion of $\operatorname{Li}(x)$, we have $$\pi(x)=\frac x{\log x}\left(1+\frac1{\log x}+\frac2{\log^2x}+O\left(\frac1{\log^3x}\right)\right),$$ from which it follows by an easy computation that $$\log x=\frac1{\epsilon_x}+O(1)$$ as $x\to+\infty$, hence the best possible $f(\epsilon)$ is of the order $\exp(1/\epsilon)$ up to a multiplicative constant.

Answer 2

As noted by Emil Jeřábek in a comment at MO meta, it is easy to extract such a function $f$ from Dusart's thesis that you linked to above, Théorème 1.10, parts 5 and 7. From part 5 we have

$$\frac{x}{\log x - (1-\epsilon)} < \frac{x}{\log x - 1} < \pi(x)$$

as soon as $x \geq 5393$, and provided that we arrange

$$\frac{x}{\log x}\left(1 + \frac1{\log x} + \frac{2.51}{(\log x)^2}\right) < \frac{x}{\log x - (1+\epsilon)}$$

for sufficiently large $x \geq 355991$ depending on $\epsilon$, part 7 gives us

$$\pi(x) < \frac{x}{\log x - (1+\epsilon)}.$$

Writing

$$\frac{x}{\log x - (1+\epsilon)} = \frac{x}{\log x}\left(\frac1{1-\frac{1+\epsilon}{\log x}}\right)$$

we just need to arrange the first inequality in

$$1 + \frac1{\log x} + \frac{2.51}{(\log x)^2} < 1 + \frac{1+\epsilon}{\log x} < \frac1{1-\frac{1+\epsilon}{\log x}}$$

or in other words that $\frac{2.51}{\log x} < \epsilon$. Hence we may take

$$f(\epsilon) = \max\{355991, \exp(\frac{2.51}{\epsilon})\}$$

to achieve the desired inequality.