We have two positive and increasing integers $a$ and $b$. We assume that $a$ is a function of $b$ and that we have:
$$a \log a \sim b$$
where $\sim$ is the asymptotic equivalence.
I would like to show that $a = \Theta(\frac{b}{\log b})$.
Thank you.
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Note that the inverse of $x \log(x)$ is $\frac{x}{W(x)}$ where $W$ is the Lambert W function i.e. the inverse of $x e^x$. So you need asymptotics of $W(x)$ for large argument, specifically you want $W(x)=\Theta(\log(x))$. – Ian Jul 20 '17 at 15:43
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That is exactly the relation between the $n$-prime $p_n$ and the number $\pi(n)$ of primes less than or equal to $n$. See WIkipedia. – lhf Jul 20 '17 at 16:10
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$$\frac{b}{\log b}\sim\frac{a\log a}{\log a+\log \log a}\sim a.$$
This is even more precise than $\Theta$.
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Thank you very much for your answer. @lhf $a$ and $b$ are increasing and $a$ is a function of $b$. – Dingo13 Jul 20 '17 at 18:24