Integral transforms, operational calculus Mellin transform Proposition

Mellin transform of a power times a function

Builds on Mellin convergence Mellin transform Complex power

Statement

Let $f \colon (0, \infty) \to \C$ be measurable, let $a \in \C$ , and let $g(x) = x^{a} f(x)$ for $x > 0$ , where $x^{a} = e^{a \log x}$ is the complex power. Let $s \in \C$ . Then $g$ is Mellin convergent at $s$ if and only if $f$ is Mellin convergent at $s + a$ . In that case $\mathcal{M}g(s) = \mathcal{M}f(s + a),$ where $\mathcal{M}$ is the Mellin transform.

Remarks

The proof is the identity $x^{s-1} \cdot x^{a} = x^{(s + a) - 1}$ for $x > 0$ , which holds because $\log x$ is real. Taking absolute values gives $\abs{x^{s-1} g(x)} = x^{\operatorname{Re}(s + a) - 1} \abs{f(x)}$ , hence the equivalence of convergence. So the strip of convergence is shifted by $-\operatorname{Re} a$ . Example: $f(x) = e^{-x}$ and $a = 1$ give $\mathcal{M}[x e^{-x}](s) = \Gamma(s + 1)$ for $\operatorname{Re} s > -1$ .

Going deeper

For a student

The rule shows that the choice of the exponent $s - 1$ rather than $s$ in the Mellin kernel is a convention: changing it only shifts $s$ . With the dilation rule and the power substitution rule, it gives the Mellin transforms of all functions $x^{b} f(c x^{a})$ from that of $f$ . It is also how a weight $x^{\sigma}$ is moved in and out of Plancherel-type identities for the Mellin transform.

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Complex power concept
Mellin convergence concept
Mellin transform concept
Mellin transform of a power times a function statement

Tag	`RPQNSM5HKC`
URI	`m2://RPQNSM5HKC`
Kind	statement, proposition
Writing	W0 stub
MSC	44A15
Status	This copy of the site has no build data. Run `m2 status RPQNSM5HKC`.
About	Mellin transform
Uses	Mellin convergence, Mellin transform, Complex power