The Mellin transform of a function is given by:

$$\mathcal{M}[f](s) = \int_0^{\infty}x^{s-1}f(x)dx$$

Supposedly, the magnitude of the Mellin transform is invariant to scaling, analogous to how the magnitude of the Fourier transform is invariant to translation. Yet, when I try to derive this I get:

$g(x) = f(kx)$

$|\mathcal{M}[g](s)| = |\int_0^{\infty}x^{s-1}f(kx)dx|=|\int_0^{\infty}(\frac{x'}{k})^{s-1}f(x')\frac{dx'}{k}|=|k^{-s}||\mathcal{M}[f](s)|$

which is not invariant, though it does have a simple relation between them. Where did I go wrong?

  • $\begingroup$ only for scale invariant functions $ f(kx)=f(x) $ for the rest this is not true $\endgroup$ – Jose Garcia Sep 12 '13 at 17:45
  • $\begingroup$ But that's a trivial case. Any transform of a scale-invariant function will be scale-invariant, by definition. $\endgroup$ – David Pfau Sep 12 '13 at 17:54
  • 3
    $\begingroup$ If $s$ were purely imaginary, then $|k^{-s}|=1$, ... $\endgroup$ – paul garrett Sep 12 '13 at 19:09
  • $\begingroup$ That must be the case being referred to when talking about the "scale-invariant" Mellin transform. $\endgroup$ – David Pfau Sep 12 '13 at 19:11

$k^{-s}$ is proportional to $s$ coefficient so you can extract and compare $k$ by $\log \dfrac{g(s)}{f(s)}$ for each integral of $s$ like you do with shift on fourier.


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