Inverse Mellin transform as inverse Fourier transform

Remarks

Substitute $y = 2\pi t$ in the definition: $$\mathcal{M}^{-1}_c \varphi(x) = \frac{1}{2\pi} \int_{\R} x^{-c - iy} \varphi(c + iy) \, dy = x^{-c} \int_{\R} e^{-2\pi i t \log x} \, \psi(t) \, dt,$$ and the last integral is $\check{\psi}(-\log x)$. The function $\psi$ is integrable because $\int \abs{\psi(t)} \, dt = \frac{1}{2\pi} \int \abs{\varphi(c + iy)} \, dy$. Example: $\varphi = \Gamma$ and $c = 1$ give $x^{-1} \check{\psi}(-\log x) = e^{-x}$, that is, $\check{\psi}(u) = e^{-u} e^{-e^{-u}}$ for $\psi(t) = \Gamma(1 + 2\pi i t)$. Combined with the statement that the Mellin transform is a Fourier transform in logarithmic variables, this reduces the Mellin inversion theorem to the Fourier inversion theorem.

Going deeper

For a student

This identity is the inverse half of the dictionary between Mellin and Fourier analysis on $\R$. It turns questions about the inverse Mellin integral, such as continuity in $x$, decay or behaviour under smoothing, into the corresponding questions for the inverse Fourier transform. For instance, it shows at once that $x \mapsto x^{c} \mathcal{M}^{-1}_c \varphi(x)$ is bounded and continuous on $(0, \infty)$.