Mellin transform and Riemann zeta

The Endeavour 2024-10-26

The Mellin transform of a function f is defined as

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

For example, it follows directly from the definition that the gamma function Γ(s) is the Mellin transform of the function e^−x.

I ran across an exercise that states an impressive-looking theorem about the Mellin transform, namely that

${\cal M}\left( \sum_{k=1}^\infty f(kx) \right) = \zeta(s) F(s)$

where F(s) denotes the Mellin transform of f(x).

I suppose the theorem looks impressive, at least in part, because it involves the Riemann zeta function. But although the zeta function holds deep mysteries—a conjecture about the location of its zeros may be the most famous open question in mathematics—not everything involving the zeta function is deep. The zeta function has, for example, come up here many times in the context of fairly simple probability problems.

I don’t know whether the theorem above is important. It may be. It resembles the sampling theorem, an important theorem in signal processing, and maybe it has analogous applications. But in any case it is not difficult to prove. A simple change of variables applied to the definition shows that

${\cal M}\left(f(ax)\right) = a^{-s} F(s)$

The proof using the equation above is simply this:

${\cal M}\left( \sum_{k=1}^\infty f(kx) \right) = \sum_{k=1}^\infty {\cal M}\left(f(kx)\right) = \sum_{k=1}^\infty k^{-s} F(s) = \zeta(s) F(s)$

The post Mellin transform and Riemann zeta first appeared on John D. Cook.

Mellin transform and Riemann zeta

The Endeavour 2024-10-26

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