---
tag: Z61W394YFA
kind: definition
name: Asymptotic expansion from poles of a Mellin transform
msc: [44A15, "30E15", 41A60]
defines: B7EFVHCEQV
uses: [N5PVGRJE1T, 1JAA21FHHQ, XJWC13NY9E, 0ZZ7ZYDR8W, 066E7PJCYE, 0PHN07KVQT, BJQTEPEZ9G]
sources:
  - url: https://en.wikipedia.org/wiki/Mellin_transform
---

## Definition

Let $f \colon (0, \infty) \to \C$ be a function whose [[Mellin transform|N5PVGRJE1T]] $f^*(s) = \int_0^\infty f(x)\, x^{s-1}\, dx$ converges absolutely in a [[fundamental strip|1JAA21FHHQ]] $\alpha < \operatorname{Re} s < \beta$.
Let $\gamma < \alpha$, and assume that $f^*$ extends to a [[meromorphic function|0ZZ7ZYDR8W]] on the strip $\gamma < \operatorname{Re} s < \beta$ with finitely many [[poles|066E7PJCYE]] $\rho_1, \dots, \rho_m$ there.
For each pole $\rho$, write the principal part of $f^*$ at $\rho$ as
$$\sum_{k=0}^{K} \frac{c_{\rho, k}}{(s - \rho)^{k+1}}.$$
The *pole expansion of $f$ at $0$* is
$$\sum_{j=1}^{m} \sum_{k} c_{\rho_j, k}\, \frac{(-1)^k}{k!}\, x^{-\rho_j} (\log x)^k,$$
where the inner sum runs over $0 \le k \le K$ for each pole.
It is the sum of the [[residues|0PHN07KVQT]] of $f^*(s)\, x^{-s}$ at $\rho_1, \dots, \rho_m$.
The expansion at $\infty$ is defined in the same way from the poles to the right of the strip, with the opposite sign.

## 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|BJQTEPEZ9G]] of $f$ to that order.
The proof writes $f$ by the [[Mellin inversion theorem|XJWC13NY9E]] 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.