An integral is defined as converging if it yields a finite value upon application of limits of integration. It is divergent otherwise.
Now sticking to the mathematical notation of Laplace transform, we have for a causal function $x(t) = u(t)$: $$ X(s) = \int_0^\infty x(t)e^{-st}dt = \int_0^\infty e^{-st}dt $$
$$ X(s) = -\frac{1}{s}e^{-st}\Biggr|_{0}^{\infty} = \frac{1}{s} = \frac{1}{(\sigma + jw)} $$
By multiplying numerator and denominator with $(\sigma - jw)$, we get: $$ X(s) = \frac{(\sigma - jw)}{(\sigma^2 +\omega^2)} $$
For laplace transform not to exist, the denominator must become 0. Hence in this contrived example, both $\sigma$ and $w$ must be 0.
Conversely, if $\sigma = 0$ and $w \neq 0$, $X(s)$ exists as $\frac{-j}{\omega}$ with a magnitude of $\frac{1}{\omega}$.
Unless my basics are messed up, why in the literature is $jw$ disregarded as influencing $X(s)$? In other words, why is the ROC dependent only on $\sigma$?