Abscissa of absolute convergence

Remarks

For $\operatorname{Re} s > \sigma_a$ the Laplace integral converges as a Lebesgue integral, and for $\operatorname{Re} s < \sigma_a$ it does not converge absolutely. Always $\sigma_c \le \sigma_a$, where $\sigma_c$ is the abscissa of convergence. Example: for $f(t) = e^{at}$ with $a \in \R$, $\abs{f} = f$, so $\sigma_a = \sigma_c = a$. Example with $\sigma_c < \sigma_a$: let $f(t) = e^t \sin(e^t)$. The substitution $u = e^t$ turns $\int_0^\infty f(t) e^{-\sigma t}\, dt$ into $\int_1^\infty \sin(u)\, u^{-\sigma}\, du$. This converges for $\sigma > 0$ by the Dirichlet test and diverges for $\sigma \le 0$, so $\sigma_c = 0$. It converges absolutely exactly for $\sigma > 1$, since $\int_1^\infty \abs{\sin u}\, u^{-\sigma}\, du < \infty$ if and only if $\sigma > 1$; so $\sigma_a = 1$. For a function of exponential order $a$ that is integrable on bounded intervals, $\sigma_a \le a$.