# Tetration Forum

Full Version: Does the Mellin transform have a half-iterate ?
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Does the Mellin transform have a half-iterate ?

After all these half-iterates of functions , integrals and derivatives, I wonder ( again ) about half-iterates of integral transforms.

regards

tommy1729
Well I posted a reply to this but it deleted it :/

I have looked at this for so long Tommy. It's very closely related to tetration. I'll tell you how I can do it for some functions.

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

define

$\vartheta(w) = \sum_{n=0}^\infty M^n(f)(s) \frac{w^n}{n!}$

Then if

$\phi(z) = [\frac{d^z}{dw^z} \vartheta(w)]_{w=0}$

then

$\phi(z) = M^z (f)(s)$

However we have to show lots of conditions on convergence and what not.
I notice that Im not so comfortable with many iteration of the Mellin transform.
Convergeance issues seem to pop up out of nowhere.

For instance if we start with f(x) = exp(x) or f(x) = exp(-x) we get in trouble before we reach the 3rd iteration of the mellin transform.

And if we pick an elementary function between exp(x) and exp(-x), I seem to have trouble finding a closed form for every n th mellin transform.

Maybe I need to consider using hypergeometrics ?

What are standard tables of n th mellin transforms (for s,x > 0) , if that has even been made yet ?

regards

tommy1729
Also of intrest to me are integral transforms that are cyclic :

Like for almost all f,s and some transform M :

M^[3](f)(s) = f(s) but M^[1](f)(s) =/= f(s) and M^[2](f)(s) =/= f(s).

I assume this is consistant with Schwartz kernel theorem.

As Kernel I assume abelian functions could be used. And maybe some others too ?

regards

tommy1729
Ok this has given me an idea for tetration ...

The James-tommy method is in progress

regards

tommy1729