Defining Upper and Lower Bounding Functions of $li(x)$ with $\displaystyle O\left(\sqrt{\frac{x}{\log(x)}}\right)$ Error Using Truncated Asymptotic Series (2408.10447v1)

Abstract: We introduce approximation functions of $li(x)$ for all $x\ge e$: (1) $\displaystyle li_{\underline{\omega},\alpha}(x) = \frac{x}{\log(x)}\left( \alpha\frac{\underline{m}!}{\log^{\underline{m}}(x)} + \sum_{k=0}^{\underline{m}-1}\frac{k!}{\log^{k}(x)} \right)$, and (2) $\displaystyle li_{\overline{\omega},\beta}=\frac{x}{\log(x)}\left( \beta\frac{\overline{m}!}{\log^{\overline{m}}(x)} + \sum_{k=0}^{\overline{m}-1}\frac{k!}{\log^{k}(x)} \right)$ with $0 < \omega < 1$ a real number, $\alpha \in { 0, \underline{\kappa}\log(x) }$, $\underline{m} = \lfloor \underline{\kappa}\log(x) \rfloor$, $\beta \in { \overline{\kappa}\log(x), 1 }$, $\overline{m} = \lfloor \overline{\kappa}\log(x) \rfloor$, and $\underline{\kappa} < \overline{\kappa}$ the solutions of $\kappa(1-\log(\kappa)) = \omega$. Since the error of approximating $li(x)$ using Stieltjes asymptotic series $\displaystyle li_{}(x) = \frac{x}{\log(x)}\sum_{k=0}^{n-1}\frac{k!}{\log^{k}(x)} + (\log(x)-n)\frac{xn!}{\log^{n+1}(x)}$, with $\displaystyle n = \lfloor \log(x) \rfloor$ for all $x\ge e$, satisfies $\displaystyle |\varepsilon(x)| = |li(x)-li_{}(x)| \le 1.265692883422\ldots$, by using Stirling's approximation and some facts about $\log(x)$ and floor functions, we show that $\displaystyle \varepsilon_{0}(x) = li(x) - li_{\underline{1/2},0}(x)$, $\displaystyle \underline{\varepsilon}(x) = li(x) - li_{\underline{1/2},\underline{\kappa}\log(x)}(x)$, $\displaystyle \overline{\varepsilon}(x) = li(x) - li_{\overline{1/2},\overline{\kappa}\log(x)}(x)$, and $\varepsilon_{1}(x) = li_{\overline{1/2},1}(x) - li(x)$ belong to $O\left(\sqrt{\frac{x}{\log(x)}}\right)$. Moreover, we conjecture that $li_{0}(x) \le \pi(x) \le li_{1}(x)$ and $\underline{li}(x) \le \pi(x) \le \overline{li}(x)$ for all $x \ge e$, here $\pi(x)$ is the prime counting function and we show that if one of those conjectures is true then the Riemann Hypothesis is true.