inverse of $y=\frac{x}{\log{x}}$?

Question

By Prime Number theorem $\pi(x)=\frac{x}{\log{x}}$ for large x

Putting $x=p_n$ where $p_n$ denotes $n^{th}$ prime number,
We have, $\pi(p_n)=\frac{p_n}{\log{p_n}}$,
$\because \pi(p_n)=n$,
$\therefore \frac{p_n}{\log{p_n}}=n$,
$\therefore p_n=\pi^{-1}(n)$,
Thus, finding inverse of $y=\frac{x}{\log{x}}$ would help us to find $n^{th}$ prime number for large $n$
Please provide clues to find it?

Answer 1

I don't know about prime numbers, but the easiest way to find the inverse is usually to use substitution:

$$\pi(x)=\frac{x}{\log{x}}\land u=\pi^{-1}(x)\implies x=\frac{u}{\log{u}}$$

$$\implies x\log{u}=u\implies u^x=e^u\implies u=-xW_{-1}\left(-\frac{1}{x}\right)$$

(there's a step between $u^x=e^u$ and $u=-xW(-1/x)$ that I skipped, but I would never do it by hand anyway.)

Where $W$ is the product-logarithm or Lambert W function, as it is also called. $W$ and $W_{-1}$ are both built into Wolfram Alpha, so for particular values, I would just enter it there.

You can input it as -x ProductLog(-1,-1/x).

Answer 2

An asymptotically correct inverse of $\frac{x}{\log x}$ is $x \log x$. It has been shown by Rosser that $n\log n$ is always an underestimate for the $n$th prime. Better estimates, due to Rosser and Dusart, show that the $n$th prime lies between $$ n(\log n + \log\log n - 1) $$ and $$ n(\log n + \log\log n) $$ for all $n\ge 6$.

For example, taking $n = 10^6$, this shows that the millionth prime is between $15441302$ and $16441302$. In fact, it is nearer the lower value: $p_{1000000} = 15485863$.

Answer 3

Apart from the solution of $y=x/\log(x)$ in terms of the Lambert $W$ function, you can also solve it numerically by the simple iterative procedure

$$ x_{n}= y \log(x_{n-1})$$

with $x_0 = y \log(y)$. It converges fast, especially for large $n$.