Mellin inversion theorem

Remarks

Put $x = e^{u}$ and $g(u) = e^{cu} f(e^u)$. Mellin convergence at $c$ says that $g$ is integrable on $\R$, and $\mathcal{M}f(c + iy) = \int_{\R} g(u) e^{iyu} \, du$ is a Fourier transform of $g$. The Fourier inversion theorem then recovers $g$ almost everywhere, hence at every point where $g$ is continuous, and this is the formula above. Example: $f(x) = e^{-x}$ with $c > 0$ gives $e^{-x} = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \Gamma(s) x^{-s} \, ds$, the Cahen–Mellin integral. Continuity at $x$ is needed: if $f(x) = e^{-x}$ is changed at the single point $x = 1/2$, its Mellin transform is still $\Gamma$. So the formula gives $e^{-1/2}$ there and not the changed value. For $f = \mathbf{1}_{(0, 1]}$, the transform $\mathcal{M}f(s) = 1/s$ is not integrable on vertical lines, so the integrability hypothesis fails. The formula still holds at $x \ne 1$ as a limit of integrals over $[c - iT, c + iT]$, but that needs a separate argument.

Going deeper

For a student

The theorem is the Fourier inversion theorem written on the multiplicative group $(0, \infty)$. It is how one passes from a Mellin transform back to a function, as in Perron's formula and in Barnes integrals. There the transform is often a product of gamma functions or a Dirichlet series. In practice one then shifts the line of integration and collects residues, which turns the inversion integral into series or asymptotic expansions. Without integrability of $\mathcal{M}f$ on the line, as for jump functions, the integral must be taken as a symmetric limit. For functions of bounded variation near a jump, this limit is the average of the one-sided limits.