---
tag: 2TM14070HZ
kind: definition
name: Mellin convergence
msc: [44A15]
defines: SGZZ4Y0D36
uses: [X7GTF3D98F, 0RNGDMHT7M]
sources:
  - url: https://en.wikipedia.org/wiki/Mellin_transform
---

## Definition

Let $f \colon (0, \infty) \to \C$ be measurable and let $s \in \C$.
For $x > 0$, let $x^{s-1} = e^{(s-1)\log x}$ be the [[complex power|0RNGDMHT7M]].
The function $f$ is *Mellin convergent at $s$* if the function $x \mapsto x^{s-1} f(x)$ is [[integrable|X7GTF3D98F]] on $(0, \infty)$ with respect to Lebesgue measure, that is,
$$\int_0^\infty x^{\operatorname{Re} s - 1} \abs{f(x)} \, dx < \infty.$$

## Remarks

The two forms agree because $\abs{x^{s-1}} = x^{\operatorname{Re} s - 1}$.
So Mellin convergence at $s$ depends only on $\operatorname{Re} s$.
If $f$ is Mellin convergent at $\sigma_1$ and $\sigma_2$ with $\sigma_1 < \sigma_2$, then it is Mellin convergent at every $s$ with $\sigma_1 \le \operatorname{Re} s \le \sigma_2$, because $x^{\sigma - 1} \le x^{\sigma_1 - 1} + x^{\sigma_2 - 1}$ for $\sigma_1 \le \sigma \le \sigma_2$.
Example: $f(x) = 1/(1 + x)$ at $s = 1/2$.
Here $\int_0^\infty x^{-1/2}/(1 + x) \, dx = \pi$, by the substitution $x = t^2$, so $f$ is Mellin convergent at $1/2$; the same estimates show it is Mellin convergent exactly when $0 < \operatorname{Re} s < 1$.
Non-example: the same $f$ at $s = 1$.
The integral $\int_0^\infty dx/(1 + x)$ diverges at $\infty$, so $f$ is not Mellin convergent at $1$.