Kenna Stanton

2022-11-20

$\underset{s\to {0}^{+}}{lim}{\int }_{0}^{\mathrm{\infty }}a\left(t\right){e}^{-st}dt$
${\int }_{0}^{\mathrm{\infty }}a\left(t\right){e}^{-st}dt=f\left(s\right)$
What is the meaning of the limit of this integral as $s\to {0}^{+}.$

Samsonitew7b

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