Mellin transform

Remarks

Since $\abs{x^{s-1}} = x^{\operatorname{Re} s - 1}$, this set depends only on $\operatorname{Re} s$ and is a union of vertical lines; it is a vertical strip, possibly empty or degenerate. Example: $f(x) = e^{-x}$. Near $0$, $\abs{x^{s-1} e^{-x}} \le x^{\operatorname{Re} s - 1}$, which is integrable on $(0, 1)$ when $\operatorname{Re} s > 0$; near $\infty$, the factor $e^{-x}$ beats every power. So $\mathcal{M}f(s) = \Gamma(s)$ for $\operatorname{Re} s > 0$, the gamma function. Non-example: $f(x) = 1$. Integrability of $x^{\operatorname{Re} s - 1}$ on $(0, 1)$ needs $\operatorname{Re} s > 0$, and on $(1, \infty)$ it needs $\operatorname{Re} s < 0$, so $f$ is Mellin convergent at no $s$. Other notations are $\tilde f(s)$, $F(s)$ and $\set{\mathcal{M} f}(s)$. Some authors call the open strip of convergence the fundamental strip and write it $\langle \alpha, \beta \rangle$.