Inverse Mellin transform

Remarks

The Lebesgue integral converges absolutely for every $x > 0$, because $\abs{x^{-c - iy}} = x^{-c}$. The result usually depends on $c$. When $\varphi$ is holomorphic on a vertical strip and decays fast enough, Cauchy's theorem shows that it does not change as $c$ moves inside the strip. Example: $\varphi = \Gamma$ and $c > 0$. Stirling's formula gives $\abs{\Gamma(c + iy)} = O(\abs{y}^{c - 1/2} e^{-\pi \abs{y}/2})$, so $\varphi$ is integrable on the line, and $\mathcal{M}^{-1}_c \Gamma(x) = e^{-x}$. Non-example: $\varphi(s) = 1/s$ and $c > 0$. Then $\abs{\varphi(c + iy)} = (c^2 + y^2)^{-1/2}$ is not integrable on $\R$, so the definition does not apply; the integral exists only as a limit of integrals over $[c - iT, c + iT]$. It is common to drop the index $c$ and write $\mathcal{M}^{-1} \varphi$.