Integral transforms, operational calculus Definition

Function having a given Mellin transform

Builds on Mellin convergence Mellin transform

Definition

Let $f \colon (0, \infty) \to \C$ be measurable, let $s \in \C$ and let $m \in \C$ . The function $f$ has Mellin transform $m$ at $s$ if both conditions hold:

$f$ is Mellin convergent at $s$ , that is, $x \mapsto x^{s-1} f(x)$ is integrable on $(0, \infty)$ .
The Mellin transform of $f$ at $s$ equals $m$ : $\int_0^\infty x^{s-1} f(x) \, dx = m$ .

Remarks

Example: $f = \mathbf{1}_{(0, 1]}$ , the indicator function of $(0, 1]$ , has Mellin transform $1/s$ at every $s$ with $\operatorname{Re} s > 0$ . Indeed $\int_0^1 x^{\operatorname{Re} s - 1} \, dx = 1/\operatorname{Re} s < \infty$ , and $\int_0^1 x^{s-1} \, dx = 1/s$ . Non-example: the same $f$ at $s = 0$ fails clause 1, because $\int_0^1 x^{-1} \, dx$ diverges. So $f$ does not have Mellin transform $m$ at $0$ for any $m$ , although the function $1/s$ has a pole there and no value. In Mathlib this predicate is HasMellin f s m.

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Mellin convergence concept
Mellin transform concept
Function having a given Mellin transform definition

Tag	`4HMBNA4RMN`
URI	`m2://4HMBNA4RMN`
Kind	definition
Writing	W0 stub
MSC	44A15
Defines	Function having a given Mellin transform
Uses	Mellin convergence, Mellin transform

Function having a given Mellin transform

Definition

Remarks

Reading path

Further reading

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