5
$\begingroup$

On the wikipedia page http://en.wikipedia.org/wiki/Mellin_transform

The second formula is an integral transformation for the inverse Mellin transform.

Being new to integral transforms, I wonder how that formula was reached. In fact, how do we prove that transform is indeed the inverse of the Mellin transform?

I know a bit about contour integration and Fourier series but I'm still confused.

Do we need to work with residues or can we just 'plug things in' ?

$\endgroup$
7
$\begingroup$

Mellin inversion is Fourier inversion in different coordinates. Although the assertion is not completely trivial, I have not seen any way to reduce this to complex-variable ideas, e.g., Cauchy's theorems and immediate corollaries. Rather, to my mind, the sane proof of Fourier inversion is proof with a Gaussian inserted to tweak things, for Schwartz functions, then extend by continuity upon observing Plancherel's identity.

In several regards one might perceive Fourier inversion (or questions about Fourier series, similarly, recovering the original function) as facts at a level of profundity "higher" than Leibniz-Newton (note the alphabetical order of authors) calculus, and "higher" than Cauchy's complex function-theory.

For reasons that I have yet to understand, people in the 19th century did believe the inversion formula for Fourier transform... with disclaimers. Dirichlet proved pointwise convergence of Fourier series (under hypotheses) early on... Similarly, people seemed to believe "Mellin inversion", despite not having a general argument, and perhaps despite not feeling that this was equivalent to Fourier inversion, since "change-of-coordinates" issues were (by far) not explicit, and (therefore) not easily understandable in the parlance of the times.

In summary: "Mellin inversion" is a changed-coordinates version of Fourier (transform) inversion, perhaps adding some analytic-continuation stuff... which in delicate circumstances can be highly meaningful... but, on most days, maybe everything is "just what it seems".

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.