Mellin transform of a dilation

Remarks

The proof is the substitution $u = ax$: $$\int_0^\infty x^{s-1} f(ax) \, dx = \int_0^\infty \left(\frac{u}{a}\right)^{s-1} f(u) \, \frac{du}{a} = a^{-s} \int_0^\infty u^{s-1} f(u) \, du.$$ The same computation with $\abs{f}$ and $\operatorname{Re} s$ gives the equivalence of convergence. Example: $f(x) = e^{-x}$ gives $\int_0^\infty x^{s-1} e^{-ax} \, dx = a^{-s} \Gamma(s)$ for $\operatorname{Re} s > 0$. Applied to $a = n$ and summed over $n$, this is the step that links the gamma function with Dirichlet series. The hypothesis $a > 0$ is needed so that $x \mapsto ax$ maps $(0, \infty)$ onto itself.

Going deeper

For a student

This is the reason the Mellin transform suits problems with scale invariance, such as homogeneous kernels and harmonic sums $\sum_k \lambda_k f(\mu_k x)$. For a harmonic sum the rule gives $\mathcal{M}\left[\sum_k \lambda_k f(\mu_k x)\right](s) = \left(\sum_k \lambda_k \mu_k^{-s}\right) \mathcal{M}f(s)$, when the sum and integral can be exchanged. The factor in brackets is a Dirichlet series, and its poles drive the asymptotic behaviour of the sum.