Asymptotic expansion from poles of a Mellin transform

Remarks

The justification is the converse mapping theorem. Assume moreover that $f^*(s) = O(\abs{s}^{-r})$ for some $r > 1$ as $\abs{\operatorname{Im} s} \to \infty$, uniformly in the closed strip $\gamma \le \operatorname{Re} s \le c$ for some $c$ with $\alpha < c < \beta$. Then, as $x \to 0^+$, $f(x)$ equals the sum above plus $O(x^{-\gamma})$, so this sum is an asymptotic expansion of $f$ to that order. The proof writes $f$ by the Mellin inversion theorem as an integral over the line $\operatorname{Re} s = c$, moves the line to $\operatorname{Re} s = \gamma$, and collects residues. The residue of $x^{-s} (s - \rho)^{-k-1}$ at $\rho$ is $x^{-\rho} (-\log x)^k / k!$, which gives the coefficients above.

Example: $f(x) = e^{-x}$ has $f^*(s) = \Gamma(s)$ on $\operatorname{Re} s > 0$. The poles at $s = -n$ are simple with residue $(-1)^n / n!$, and $\Gamma$ decays exponentially on vertical lines. So for every $N$ one gets $e^{-x} = \sum_{n=0}^{N} (-1)^n x^n / n! + O(x^{N+1/2})$ as $x \to 0^+$, as Taylor's formula confirms.

Non-example: $f(x) = 1$ for $0 < x \le 1$ and $f(x) = 0$ for $x > 1$ has $f^*(s) = 1/s$ on $\operatorname{Re} s > 0$. The pole at $0$ gives the term $1$, which is correct for $x < 1$, but $1/s$ decays only like $\abs{s}^{-1}$, so the decay hypothesis fails. The jump of $f$ at $x = 1$ is what spoils the decay.