Fundamental strip of a Mellin transform

Remarks

Let $\sigma = \operatorname{Re} s$. Choose $0 < \delta < 1 < R$ with $\abs{f(x)} \le C x^{-b}$ on $(0, \delta]$ and $\abs{f(x)} \le C x^{-a}$ on $[R, \infty)$. Then $x^{\sigma - 1} \abs{f(x)} \le C x^{\sigma - b - 1}$ on $(0, \delta]$, which is integrable since $\sigma - b > 0$. Also $x^{\sigma - 1} \abs{f(x)} \le C x^{\sigma - a - 1}$ on $[R, \infty)$, which is integrable since $\sigma - a < 0$. On $[\delta, R]$, $f$ is integrable and $x^{\sigma - 1}$ is bounded.

The largest open strip $\alpha < \operatorname{Re} s < \beta$ on which $f$ is Mellin convergent is called the fundamental strip and is written $\langle \alpha, \beta \rangle$. Example: $f(x) = e^{-x}$ is $O(x^0)$ at $0$ and $O(x^{-a})$ at $\infty$ for every $a$, so its fundamental strip is $\langle 0, \infty \rangle$, where $\mathcal{M}f = \Gamma$. Example: $f(x) = 1/(1 + x)$ is $O(1)$ at $0$ and $O(x^{-1})$ at $\infty$, giving the strip $\langle 0, 1 \rangle$. The strict inequalities are needed: for this $f$ at $s = 1$, the bound $x^{\sigma - a - 1} = x^{-1}$ is not integrable at $\infty$, and indeed $\int_0^\infty dx/(1+x)$ diverges.

Going deeper

For a student

This is the first step in every computation with Mellin transforms: one finds the strip before anything else. The edges of the strip are tied to the asymptotics of $f$. Conversely, under suitable conditions, poles of the continued transform beyond the strip give terms of the asymptotic expansion of $f$ at $0$ or at $\infty$. On the open strip the transform is also holomorphic, which is the companion result on differentiability of the Mellin transform. Under the substitution $x = e^{-u}$, the statement becomes the familiar fact that a function with exponential decay at both ends of $\R$ has a two-sided Laplace transform on a strip.