Answer by Michael Rubinstein for What does Mellin inversion "really mean"?

2013-02-19T17:11:48Z

Have a look at Zagier's appendix: http://people.mpim-bonn.mpg.de/zagier/files/tex/MellinTransform/fulltext.pdf

It provides a nice description of the Mellin transform when $f(x)$ is sufficiently smooth at $x=0$, and of rapid decay at infinity.

For example, assume $f(x) = \sum_0^\infty a_n x^n$, in some neighbourhood of the origin, and decays rapidly as $x \to \infty$, then its Mellin transform has meromorphic continuation to all of $\mathbb{C}$ with simple poles of residue $a_n$ at $s=-n$, $n=0,1,2,3,\ldots$. This is nicely explained in Zagier's appendix.

So, rate of decay issues aside, shifting the inverse Mellin transform to the left, i.e. letting $\sigma \to -\infty$, picks up the residues of the integrand at s=-n, i.e. $a_n x^n$, i.e. recovers the Taylor expansion about $x=0$ of $f(x)$.

Of course, it only applies to a limited class of functions $f$, but, in many practical examples, this reasoning gives one explanation of why the Mellin inversion formula is true, without resorting to Fourier inversion.

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Answer by Michael Rubinstein for What does Mellin inversion "really mean"?