Simple approx. formula for Inverse of the Logarithmic integral function

Question

A pretty good simple approx. formula for inverse of the Logharitmic integral function is:
$li^{-1}(x)\approx -\frac{1}{li(\frac{1}{x})}$
It is also an acurate approx. for the n-th prime number
$p_n \approx -\frac{1}{li(\frac{1}{n})}$.
Table ( integer parts ) :

$n\qquad\qquad 10\qquad 10^2\qquad\quad 10^3\qquad\quad 10^4\quad\qquad 10^5\qquad\qquad 10^6$

$li^{-1}(n)\qquad 20\quad 488\qquad 7762\qquad 104270\qquad 1298170\qquad 15479100$

$-\frac{1}{li(\frac{1}{n})}\quad 30\qquad 546\qquad 7803\qquad 101264\qquad 1244281\qquad 14755260$

$p_n\qquad\quad 29\qquad 541\qquad 7919\qquad 104729\qquad 1299709\qquad 15485863$
Is this approx. usefull and how to estimate the error $p_{n}-(-\frac{1}{li(\frac{1}{n})})$ ? Is it allways positive for big enough $n$?

Answer 1

A more accurate large-$x$ approximation of $\text{Li}^{-1}(x)$, and containing only elementary functions, is $$g(x)=x \left(\frac{\ln (\ln x)-2}{\ln x}+\ln x+\ln (\ln x)-1\right),$$ see https://mathworld.wolfram.com/PrimeFormulas.html

I compare these approximations in the table below (all entries are rounded to the nearest integer).

function	p=4	p=5	p=6	p=7	p=8
$\text{Li}^{-1}(10^p)$	$104270$	$1298172$	$15479066$	$179415191$	$2038024425$
$g(10^p)$	$104546$	$1299492$	$15486599$	$179464275$	$2038374432$
$-1/\text{Li}(1/10^p)$	$101264$	$1244281$	$14755261$	$170652555$	$1937361573$