Mellin transform under inversion of the variable

Remarks

The proof is the substitution $u = 1/x$, with $dx = -u^{-2} \, du$: $$\int_0^\infty x^{s-1} f(1/x) \, dx = \int_0^\infty u^{1-s} f(u) \, u^{-2} \, du = \int_0^\infty u^{-s-1} f(u) \, du.$$ The same computation with $\abs{f}$ gives the equivalence of convergence. So the strip of convergence is reflected in the imaginary axis: if $f$ converges for $\alpha < \operatorname{Re} s < \beta$, then $g$ converges for $-\beta < \operatorname{Re} s < -\alpha$. Example: $f(x) = e^{-x}$ gives $\int_0^\infty x^{s-1} e^{-1/x} \, dx = \Gamma(-s)$ for $\operatorname{Re} s < 0$.

Going deeper

For a student

This rule is used with functional equations of the form $\theta(1/x) = \sqrt{x}\, \theta(x)$, where the behaviour at $0$ is read from the behaviour at $\infty$. It also explains why the Mellin transform of a function and of its reflection have mirror strips of convergence. Together with the rules for dilations and powers, it shows how the Mellin transform reacts to the maps $x \mapsto a x^{b}$ with $a > 0$ and $b \ne 0$.