User michael rubinstein - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:40:37Z http://mathoverflow.net/feeds/user/31582 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79868/what-does-mellin-inversion-really-mean/122325#122325 Answer by Michael Rubinstein for What does Mellin inversion "really mean"? Michael Rubinstein 2013-02-19T17:11:48Z 2013-02-19T17:11:48Z <p>Have a look at Zagier's appendix: <a href="http://people.mpim-bonn.mpg.de/zagier/files/tex/MellinTransform/fulltext.pdf" rel="nofollow">http://people.mpim-bonn.mpg.de/zagier/files/tex/MellinTransform/fulltext.pdf</a></p> <p>It provides a nice description of the Mellin transform when $f(x)$ is sufficiently smooth at $x=0$, and of rapid decay at infinity.</p> <p>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.</p> <p>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 $f(x)$.</p> <p>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. </p>