Why $dt/t$ in Mellin transform I've noticed that often when people write the Gamma function $\Gamma(s) = \int_0^\infty t^{s-1}e^{-t}\,dt$, that they write it like 
$$
  \Gamma(s) = \int_0^\infty t^s e^{-t}\,\frac{dt}{t} ,
$$
where they group the $\frac 1t$ with the $dt$.
Other Mellin transforms are also often written similarly, with the $\frac{dt}{t}$ grouped together. 
My question is: 
In what sense is writing these integrals in this way "better" or "more meaningful" or "natural"?
 A: $\dfrac{dt}{t}$ is the (well, unique up to a constant factor, so "a", strictly) Haar measure on the topological group $(0,+\infty)$ (with multiplication). That makes some transformations particularly nice when written in that style,
$$\int_0^\infty f(at)\,\frac{dt}{t} = \int_0^\infty f(t)\,\frac{dt}{t}$$
for all $a > 0$, so
$$\int_0^\infty t^s g(at)\,\frac{dt}{t} = a^{-s}\int_0^\infty \left(at\right)^sg(at)\,\frac{dt}{t} = a^{-s}\int_0^\infty t^sg(t)\,\frac{dt}{t},$$
which may be a bit nicer on the eye than it would be for $\int_0^\infty t^{s-1}g(at)\,dt$.
When we use the variant of the Mellin transform$^1$ defined as
$$\mathscr{M}[f](s) = s\int_1^\infty x^{-s}f(x)\frac{dx}{x}$$
for locally integrable $f\colon [1,+\infty) \to \mathbb{C}$ with at most polynomial growth, we have
$$\mathscr{M}[f](s) = s\mathcal{L}[F](s),$$
where $F(t) = f(e^t)$ and $\mathcal{L}$ is the Laplace-transform. Thus it is rather easy to translate theorems about the Laplace transform into theorems about the Mellin transform.
These are not compelling reasons why writing $\dfrac{dt}{t}$ is the only right thing to do, but in this way, some things look a little nicer and perhaps more natural.

$^1$ The only one I have more than seen in passing, it's well suited to working with the Riemann $\zeta$-function.
