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I need an explicit lower bound for $\mathrm{Li}(x)$ in terms of $x$ and $\log x$. Say, Wikipedia gives $$ \mathrm{li}(x) >\frac x{\log x}+\frac x{(\log x)^2} $$ for $x>e^{11}$, see the logarithmic integral entry, and so

$$ \mathrm{Li}(x)= \mathrm{li}(x)-\log 2 >\frac x{\log x}+\frac x{(\log x)^2}-\log 2 $$

but it does not quote any sources. I need a similar bound but with exact quotation.

Of course, I can easily do it myself (and I will if I find nothing to quote) but I prefer having a reference.

Daniele Tampieri
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Yuri Bilu
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    See Rosser and Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math 6 (1962), 64-94. The function ${\rm li}(x)$ is defined in the usual way on p. 66. On p 81, Lemma 4 says for $x \geq e^5$ that $x/(\log x - 1/2) < {\rm li}(x) - {\rm li}(\sqrt{x})$, so $x/(\log x - 1/2) < {\rm li}(x)$ for $x \geq e^5$. Is that good enough? – KConrad May 24 '21 at 23:59
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    The lower bound I gave is weaker than what you mentioned. If you want to know the best bounds of this kind that have been published, consider writing to Tim Trudgian. I suspect he'll know the most recent relevant references. – KConrad May 25 '21 at 00:13
  • Keith, no this bound is perfectly fine. Shame on me, I had to check RS myself: it is one of the files permanently open on my desktop. – Yuri Bilu May 25 '21 at 01:05
  • Handy link for Rosser and Schoenfeld: https://doi.org/10.1215/ijm/1255631807 – David Roberts May 25 '21 at 04:44

1 Answers1

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Using ${\rm Li}(x) := \int_2^x dt/\log t$, as usual, here is an elementary argument that ${\rm Li}(x) > x/\log x$ for $x \geq 7$, so no need to appeal to a lower bound valid only starting at $e^5 \approx 148.4$ as in the Rosser-Schoenfeld paper from the comments.

For $x > 1$, let $f(x) = {\rm Li}(x) - x/\log x$. Then $$ f'(x) = \frac{1}{\log x} - \left(\frac{\log x - 1}{(\log x)^2}\right) = \frac{1}{(\log x)^2} > 0, $$ so $f$ is increasing on $(1,\infty)$. Since $f(7) \approx .1145$, for $x \geq 7$ we have $f(x) \geq f(7) > 0$, so ${\rm Li}(x) > x/\log x$.

Since $f(6) \approx -.1716$, $f$ vanishes somewhere between 6 and 7. That happens slightly below 6.58: from PARI, I find $f(6.58) \approx .000076$.

Similarly, for $x > 1$ let $g(x) = {\rm Li}(x) - x/\log x - x/(\log x)^2$. Then $$ g'(x) = \frac{2}{(\log x)^3} $$ so $g$ is increasing on $(1,\infty)$. From PARI, $g(20) \approx -.044$ and $g(21) \approx .028$, so ${\rm Li}(x) > x/(\log x) + x/(\log x)^2$ for $x \geq 21$. Since $g(20.65) \approx .0030005$, $g$ vanishes slightly below 20.65.

More generally, from repeated integration by parts $$ {\rm Li}(x) = \sum_{n=1}^N \frac{(n-1)!x}{(\log x)^n} + \int_2^x \frac{N!\, dt}{(\log t)^{N+1}} + c_N $$ for $N \geq 0$ and $x > 1$, where $c_N$ is a constant. Therefore the difference $$ g_N(x) := {\rm Li}(x) - \sum_{n=1}^N \frac{(n-1)!x}{(\log x)^n} $$ has $g_N'(x) = N!/(\log x)^{N+1} > 0$, so $g_N$ is increasing on $(1,\infty)$. Find an $x_0$ where $g_N(x_0) > 0$ and then ${\rm Li}(x) > \sum_{n=1}^N (n-1)!x/(\log x)^n$ for $x \geq x_0$.

KConrad
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