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](https://en.wikipedia.org/wiki/Logarithmic_integral_function#Asymptotic_expansion), 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.



$\mathrm{Li}(x)$ vs $x/\log x$
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 $x \geq x\_0$.


