Talk:Mellin transform

The definition of the two-sided Laplace transform is

:$\backslash left\backslash \{\backslash mathcal\{B\}\; f\backslash right\backslash \}(s)\; =\; \backslash varphi(s)\; =$

\int_{-\infty}^{\infty} e^{-sx} f(x) dx

Under the substitution $x=-\backslash ln\backslash ,\; t$, $t\; \backslash in\; [0,\backslash infty]$ this transforms into

:$\backslash left\backslash \{\backslash mathcal\{B\}\; f\backslash right\backslash \}(s)\; =\; \backslash varphi(s)\; =$

\int_{0}^{\infty} t^{s} f(-\ln\,t) {dt \over t} \ .

So the relationship between Mellin transform and two-sided Laplace transform should be

:$\backslash left\backslash \{\backslash mathcal\{B\}\; f\backslash right\backslash \}(s)=\backslash left\backslash \{\backslash mathcal\{M\}\; f(-\backslash ln\backslash ,x)\backslash right\backslash \}(s)$

rather than the stated

:$\backslash left\backslash \{\backslash mathcal\{B\}\; f\backslash right\backslash \}(s)=\backslash left\backslash \{\backslash mathcal\{M\}\; f(e^\{-x\})\backslash right\backslash \}(s)\; \backslash \; ?$

--212.18.24.11 14:58, 22 August 2005 (UTC)

Hello my question is ..let's suppose we know the Mellin inverse transform of F(s) then..how could we calculate the Mellin inverse transform of F(as) when a is a real number, i believe that if g(x) has F(s) as its Mellin transform then $g(x^\{1/a\})x^\{1/a-1\}$ is the inverse of F(as) but i'm not pretty sure --Karl-H 14:20, 1 October 2006 (UTC)

Hello, I have a question -- the example transform of the function

:$f(x)\; =\; \backslash frac\{1\}\{1+x\}$

is apparently not convergent on the positive real line since the integral of $\backslash lim\_\{R\; \backslash rightarrow\; \backslash infty\}\; \backslash int\_\{0\}^\{R\}\; \backslash frac\{1\}\{1+x\}\; dx\; =\; \backslash lim\_\{R\; \backslash rightarrow\; \backslash infty\}\; \backslash ln\; (R)$ diverges - but you state that the function must be integrable on the positive real line to have a Mellin transform? Could you correct this apparent inconsistency or explain the condition for having a Mellin transform a bit more clearly? --81.146.66.114 14:38, 24 March 2007 (UTC)

:: It's supposed to be locally integrable. Corrected. -Zahlentheorie 23:26, 25 March 2007 (UTC)

I think this article could use some restructuring.

:-I don't think that the fundamental strip is nearly as important as the relationship between the Mellin Transform and Laplace Transform.

:-Some talk about applications to signal/image processing of the Mellin Transform would be nice.

:-Discussion of the intuitive meaning of the transform, as a hyperbolic expansion compared to the Fourier Transform's sinusoidal expansion...

Your thoughts?

--Joe056 02:19, 15 March 2007 (UTC)

:: The importance of the fundamental strip is most evident in the use of the Mellin-Perron formula and general Mellin inversion, which is used a lot in number theory and the analysis of algorithms (harmonic sums). Unfortunately the article on Mellin-Perron is a "stub" as of today. It would need additional detailed material to back up this claim (the importance of the fundamental strip). If you don't compute the fundamental strip, you won't know which residues will contribute to the inversion integral. -Zahlentheorie 10:51, 15 March 2007 (UTC)

:: Fair enough. I'll incorporate some of the suggestions I have into an "Applications" section. Joe056 14:00, 22 March 2007 (UTC)

:::Hello there, I do not have the time at the moment for a more detailed comment, but in harmonic sums and hence in CPSC, the key property is not only scale invariance, but the fact that when you scale the argument of the source function by, say, $\backslash mu$ (the frequency), you get a factor of $1/\backslash mu^s$ in the transform, which produces Dirichlet series when you sum over integer frequencies. -Zahlentheorie 17:34, 22 March 2007 (UTC)

I think there's a mistake here. The article says

:The Mellin Transform is widely used in computer science because of its scale invariance property. The magnitude of the Mellin Transform of a scaled function is identical to the magnitude of the original function. This scale invariance property is analogous to the Fourier Transform's shift invariance property. The magnitude of a Fourier transform of a time-shifted function is identical to the original function.

First of all, I presume that what is intended is The magnitude of the Mellin Transform of a scaled function is identical to the magnitude of the Mellin Transform of the original function. But I don't think even this is correct. Just as the shift invariance of the magnitude of the Fourier transform does not hold for the Laplace transform, the scale invariance property as stated here does not hold for the Mellin transform (though I suspect it holds for purely imaginary z). Consider for examle, the Mellin transform of exp(-at), as given here [http://mathworld.wolfram.com/MellinTransform.html]. 72.75.103.224 03:24, 2 July 2007 (UTC)

The article indicates the following:

The two-sided Laplace transform may be defined in terms of the Mellin

transform by

:$\backslash left\backslash \{\backslash mathcal\{B\}\; f\backslash right\backslash \}(s)\; =\; \backslash left\backslash \{\backslash mathcal\{M\}\; f(-\backslash ln\; x)\; \backslash right\backslash \}(s)$

and conversely we can get the Mellin transform from the two-sided Laplace transform by

:$\backslash left\backslash \{\backslash mathcal\{M\}\; f\backslash right\backslash \}(s)\; =\; \backslash left\backslash \{\backslash mathcal\{B\}\; f(e^\{-x\})\backslash right\backslash \}(s).$

This seems inconsistent with the following link which indicates the subsequent text:

Laplace Transform

The Mellin transform and its inverse are related to the two-sided Laplace transform by a simple change of variables.

If in the Mellin transform

: $G(s)\; =\; \backslash mathcal\{M\}\backslash \{g(\backslash theta)\backslash \}\; =\; \backslash int\_0^\backslash infty\; \backslash theta^s\; g(\backslash theta)\; \backslash ,\; \backslash frac\{d\backslash theta\}\; \backslash theta$

we set {{math|1=θ = e^−t}} we get a two-sided Laplace transform.

StvC (talk) 21:12, 27 November 2017 (UTC)

It looks like there is a problem in the 7th row, 2nd column of the "properties" table of this article:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \nu^{-s} \tilde{f}(s) }

Not sure how to fix this. Looks like Wikipedia has a locally hosted server for converting the MathML into images, and it failed for this particular bit of math. It might be enough to just trigger a re-render by making an edit to this bit of math (I'll give it a try). Frpzzd0 (talk) 18:08, 17 June 2025 (UTC)

:This appears to have worked. Frpzzd0 (talk) 18:10, 17 June 2025 (UTC)