I am currently trying to understand Watson-Nevanlinna (WN) theorem, which gives sufficent conditions for a function $f(z)$ to be equal to the Borel sum of its asymptotic expansion as $z\to0$.
The theorem is stated as follows.
Let $f(z)$ be analytic in the circle $C_R=\{z:\mathrm{Re}(z^{-1})>R^{-1}\}$, $R>0$, and let $f(z)\sim\sum_{n=0}^\infty f_n z^n$ be its asymptotic expansion as $z\to 0$. If the remainder after summing $N-1$ terms satisfies
$$ \vert R_N(z)\vert\leq A\sigma^N N! |z|^N,\quad A>0,\,{}\sigma>0, $$
uniformly in $N$ and $z\in C_R$, then:
- $B(t)=\sum_{n=0}^\infty f_n z^n/n!$ converges in $S_\sigma=\{t:\mathrm{dist}(t,\mathbb{R}^+)<1/\sigma\}$, and
- $f(z)$ equals the convergent integral $f(z)=(1/z)\int_0^\infty e^{-tz}B(t)\mathrm{d}t$ for any $z\in C_R$.
However, from reading some reviews on asymptotic expansions, I was under the impression that if $\sum_{n=0}^\infty f_n z^n$ is asymptotic to $f(z)$ as $z\to0$ along paths in the $\mathrm{Re}(z)>0$ part of the complex plane of $z$ (where the circle $C_R$ is contained), then it is also asymptotic to $f(z)+e^{-1/z}$, as $e^{-1/z}$ has an expansion in powers of $z$ that vanishes at all orders. Due to this, I was taking $e^{-1/z}$ as some sort of lower boundary in the accuracy in obtaining a function $f(z)$ from its asymptotic expansion as $z\to0$ in powers of $z$.
WN theorem seems to suggest that, provided the conditions on $f(z)$ are satisfied, one can recover $f(z)$ exactly from the asymptotic expansion, and I am having trouble understanding that. Indeed, the function $e^{-1/z}$ itself seems to satisfy the conditions of the theorem, but its Borel transform $B(t)$ is identically zero and so is the integral. The theorem then would ensure $e^{-1/z}=0$ in $C_R$, which is not true.
Thanks in advance!
Edit: for further clarification. The reviews and books on asymptotic expansions that I have been reading emphasize the fact that the function $e^{-1/z}$ has, for $\mathrm{Re}(z)>0$, the following asymptotic expansion as $z$ tends to $0$ and with respect to the sequence $\{z^n\}^\infty_{n=0}$: $$e^{-1/z}\sim 0+0\cdot z^2+0\cdot z^3+...$$ Due to this, they remark that more than one function can have the same asymptotic expansion, and the example for asymptotic expansions as $z$ tends to $0$ in the $\mathrm{Re}(z)>0$ part of the complex plane and with respect to the sequence $\{z^n\}^\infty_{n=0}$ are the functions $f(z)$ and $f(z)+e^{-1/z}$. My suprise comes when Watsons theorem or Watson-Nevanlinna theorem state conditions under which an asymptotic power series at $z=0$ equals a function, as I fail to see why it wouldn't be possible to add $e^{-1/z}$ terms to obtain a different function with the same asymptotic expansion.
Taking $w:=1/z$, you have to verify $\left| {w^N {\rm e}^{ - w} } \right| \le A\sigma ^N N!$ for all $N\geq 1$ and $\operatorname{Re} (w) > r: = R^{ - 1}$, with some positive $R$, $A$ and $\sigma$. But taking $w = 2r + \mathrm{i}T$ (note that $\operatorname{Re} (w) =2r > r$) $$ \left| {w^N {\rm e}^{ - w} } \right| = \left| {(2r + \mathrm{i}T)^N } \right|{\rm e}^{ - 2r} \ge T^N {\rm e}^{ - 2r}, $$ which can be arbitrarily large if $T$ is large enough.