Mellin transform of rescaled delta distributions

Question

There's something about the Mellin transform I don't get, so hopefully someone can tell me what it is that I'm doing wrong.

Let's define the Mellin transform of $f(t)$ as $\mathcal{M}\{f(t)\}(s) = \int_0^\infty f(t) t^{s-1} dt$.

First, let $\delta(t-n)$ be the Dirac delta distribution that spikes at $t=n$. According to Mathworld, the Mellin transform of such a distribution is

• $\mathcal{M}\{\delta(t-n)\}(s) = n^{s-1}$

Second, multiple references, such as this book, list the "rescaling property" of the Mellin transform as

$\mathcal{M}\{f(at)\}(s) = a^{-s} \mathcal{M}\{f(t)\}(s)$

But these two things don't seem to play nicely together.

For instance, let $g(t) = \delta(t-n)$ for some n. Then $g(at) = \delta(at-n) = \delta(t-\tfrac{n}{a})$. But:

• $\mathcal{M}\{\delta(t-\tfrac{n}{a})\} = (\tfrac{n}{a})^{s-1}$
• $\mathcal{M}\{g(at)\} = a^{-s} \mathcal{M}\{g(t)\} = a^{-s} n^{s-1}$

and it's clear that $(\tfrac{n}{a})^{s-1} \neq a^{-s} n^{s-1}$.

What am I doing wrong?

Answer 1

It's not true that $\delta(at-n) = \delta(t-\frac{n}{a})$.

With the usual interpretation of what rescaling of a distribution should mean, $$ \delta(ct) = \frac{1}{|c|}\delta(t). $$ See for example wikipedia for more on the properties of $\delta$.