$\lim_{s\to 0^+}\int_0^\infty a(t) e^{-st} dt $

Question

$$\int_0^\infty a(t) e^{-st} dt = f(s)$$ What is the meaning of the limit of this integral as $s\to 0^+.$

Answer 1

I am guessing that you would like to know $\lim_{s \to 0+} f(s) = \lim_{s \to 0+} \int_0^\infty a(t) e^{-st}dt$. Let $(s_n)$ be any sequence of positive numbers converging to $0$ and set $f_n(t) = a(t)e^{-s_nt} \to a(t)$ as $n \to \infty$. Moreover, $|f_n(t)| \leq |a(t)|$ so if $a(t)$ is Lebesgue integrable, then the limit equals $\int_0^\infty a(t) dt$ by Lebesgue's Dominated convergence theorem. On the other hand, if $a(t)$ is not Lebesgue integrable, then the limit need not exist. Consider $a(t) = \frac{|\sin(t)|}{t}$, for example.