Mellin convergence

Remarks

The two forms agree because $\abs{x^{s-1}} = x^{\operatorname{Re} s - 1}$. So Mellin convergence at $s$ depends only on $\operatorname{Re} s$. If $f$ is Mellin convergent at $\sigma_1$ and $\sigma_2$ with $\sigma_1 < \sigma_2$, then it is Mellin convergent at every $s$ with $\sigma_1 \le \operatorname{Re} s \le \sigma_2$, because $x^{\sigma - 1} \le x^{\sigma_1 - 1} + x^{\sigma_2 - 1}$ for $\sigma_1 \le \sigma \le \sigma_2$. Example: $f(x) = 1/(1 + x)$ at $s = 1/2$. Here $\int_0^\infty x^{-1/2}/(1 + x) \, dx = \pi$, by the substitution $x = t^2$, so $f$ is Mellin convergent at $1/2$; the same estimates show it is Mellin convergent exactly when $0 < \operatorname{Re} s < 1$. Non-example: the same $f$ at $s = 1$. The integral $\int_0^\infty dx/(1 + x)$ diverges at $\infty$, so $f$ is not Mellin convergent at $1$.