$\mathrm{Li}(x)$ vs $x/\log x$

Question

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.

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
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

KConrad · Accepted Answer · 2021-05-25T02:24:43.950

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

$x \geq x_0$ .

Li(x)\mathrm{Li}(x) vs x/logxx/\log x

1 Answers1

$\mathrm{Li}(x)$ vs $x/\log x$