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Let the function $f(x,\nu)$, with $x>0$ and $\nu\in[0,+\infty)$, have the asymptotic expansion \begin{equation} f(x,\nu)\sim\sum_{n=0}^{\infty}a_{n}(\nu)x^{n}\;, \end{equation} as $x\to 0$. Assume also that $|a_{n}(\nu)|\leq M_{n}$ for some constant $M_{n}$ for $\nu\in[0,+\infty)$. Can one conclude that the above asymptotic expansion is uniform in the variable $\nu$? namely can one conclude that \begin{equation} f(x,\nu)=\sum_{n=0}^{N}a_{n}(\nu)x^{n}+O(x^{N+1})\;, \end{equation} with $O(x^{N+1})$ independent of $\nu$?

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  • $\begingroup$ @Antonio Vargas. I can see that there might be problems with uniformity when the function depends on the ratio $\nu/x$. Let us say, then, that for the functions $a_{n}(\nu)$ we have $a_{n}(\nu)=e^{-\nu}P_{n}(\nu)$ with $\nu\in[0,+\infty)$. $P_{n}(\nu)$ denotes a polynomial of degree less or equal than $n$. Can, in this case, one conclude that the expansion is uniform in $\nu$? Basically what I would like to do is justify the term-by-term integration w.r.t $\nu$ of the asymptotic series (provided that the integrals exist) to get an asymptotic series of $\int_{0}^{\infty}f(x,\nu)d\nu$. $\endgroup$
    – user154259
    May 30, 2014 at 18:25
  • $\begingroup$ I think my example is still a counterexample for that case. If I'm not mistaken, just multiply it by $e^{-\nu}$. I'm skeptical that any conclusion about uniformity can be drawn from the coefficients of the asymptotic series. $\endgroup$ May 30, 2014 at 20:42

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In general no. For example, $e^{-\nu^2/x^2} = O(x^n)$ as $x \to 0$ for all $n \geq 0$ and $\nu \neq 0$ fixed, so we have

$$ \frac{1}{1-x} + e^{-\nu^2/x^2} \sim \sum_{n=0}^{\infty} x^n$$ as $x \to 0$ for all fixed $\nu \neq 0$. This asymptotic is not uniform with respect to $\nu$ when, say, $\nu = x$.

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