Mellin transform as a Fourier transform in logarithmic variables

Remarks

The proof is the substitution $x = e^{-u}$, with $dx = -e^{-u} \, du$, which maps $(0, \infty)$ onto $\R$ with reversed orientation. It gives $\int_0^\infty x^{\sigma - 1} \abs{f(x)} \, dx = \int_{\R} \abs{h_\sigma(u)} \, du$, and $$\int_0^\infty x^{s-1} f(x) \, dx = \int_{\R} e^{-su} f(e^{-u}) \, du = \int_{\R} h_\sigma(u) e^{-i\tau u} \, du.$$ Example: $f(x) = e^{-x}$ and $\sigma = 1$ give $h_1(u) = e^{-u} e^{-e^{-u}}$, and the statement says $\widehat{h_1}(\tau/2\pi) = \Gamma(1 + i\tau)$. Other sign conventions use $x = e^{u}$, which replaces $\tau / 2\pi$ by $-\tau / 2\pi$. For fixed $\sigma$, the map $f \mapsto h_\sigma$ is a bijection, so every theorem on the Fourier transform on $\R$ has a Mellin version on each vertical line.

Going deeper

For a student

This dictionary transports results from Fourier analysis. The Riemann–Lebesgue lemma gives the decay of $\mathcal{M}f(\sigma + i\tau)$ as $\abs{\tau} \to \infty$. The Fourier inversion theorem and the Plancherel theorem give the Mellin inversion theorem and the Parseval formula for Mellin transforms. Varying $\sigma$ as well, the same substitution identifies the Mellin transform with a two-sided Laplace transform of $u \mapsto f(e^{-u})$. It also identifies the strip of Mellin convergence with the strip of convergence of that Laplace transform.