The Mellin transform of a function *f* is defined as

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

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

The proof using the equation above is simply this:

Should the identity following “namely that” have an $s$ instead of $x$ on the left?

I don’t think so, unless I’m missing something.

fis a functionxandFis a function ofs.The notation is standard, but a bit informal. To be more precise, we’d say the Mellin transform of the function that maps

xto … is the function that mapssto ….I think Aidan’s suggestion is to add (s) right after the big parenthesis, and before the equal sign. So as to have M(…)(s), just like in the very first equation.

@Authur: OK, I see. That makes sense.