2019年6月23日 · $\begingroup$ @KevinNivek - I agree that would likely work. The main issue though is that a taylor expansion, to the best of my knowledge, utilizes an infinite series, which …

Integration by parts of the Logarithmic Integral, Li(x), can be calculated using the Exponential Integral, Ei(x), formula: li(x) = Ei(lnx) = (γ + ln(lnx)) + [the cumulative sum from n=1 to infinity …

2015年9月13日 · It immediately follows that x! = p≤x p[x/p]+[x/p2]+··· and log(x!) = p≤x x p + x p2 + x p3 + ··· log(p). Now log(x!) is asymptotic to x log(x) by Stirling’s asymptotic formula, and, …

2019年9月14日 · $\begingroup$ there is a standard definition (see Wikipedia) and then changing lower bounds makes it differ by a constant which is utterly irrelevant when one does Mellin …

$\begingroup$ Roughly, Chebyshev's theorem says that $\pi(x)$ and $\operatorname{Li}(x)$ come "arbitrarily close" infinitely often (for a certain value of "arbitrarily close"). The sign …

2022年12月24日 · $$\pi(x) > \operatorname{li}(x)$$ where $\pi(x)$ is the prime counting function and $\operatorname{li}(x)$ is the logarithmic integral function. The same Wikipedia article …

Minor correction: the $0$ at the lower limit should be $2$ so that the integral converges. The approximation for any trunca

2022年3月10日 · [This answer was posted by LurchiDerLurch to the original question, and was automatically deleted when that question was deleted.

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$\begingroup$ And $\log(s-1)/s$ is the contribution from the main singularity of $\log \zeta(s)/s$ which is almost the Mellin transform of $\pi(x)$.