I'm looking for the asymptotic expansion as $\ x \rightarrow \infty$ for $\ f(x)$ for small $\alpha$. Ideally, I'd like to get the asymptotic expansion for all orders. How would I go about doing this? Of course, if $ \alpha$ is somewhat large, $\ f(x) \rightarrow 2\alpha$ quickly, but the question is what happens when $\alpha$ is small.

$\ f(x)=\frac{\log(x)}{\frac{\log(x)}{2\alpha}-\log(\log(x))}$ where $\ 0<\alpha<1/2$


1 Answer 1


Multiply numerator and denominator by $\frac{2a}{\log x}$: $$ \frac{2a}{1-2a\frac{\log\log x}{\log x}}. $$ $a<1/2$ and $\log(x) < x$; so $2a\log\log x/\log x < 1$, and $$ f(x)=2a\sum_{n=0}^\infty \left(2a\frac{\log\log x}{\log x}\right)^n. $$ So, $f(x)\sim 2a$.

  • 1
    $\begingroup$ Just a typo: there shouldn't be a $(-1)^n$ in the sum. $\endgroup$ Jul 24, 2015 at 17:53
  • $\begingroup$ @AntonioVargas, yes, indeed. Thanks $\endgroup$ Jul 25, 2015 at 1:00

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