How does one show that the Laplace transform of $e^t$ converges uniformly?

Question

I know by definition that it converges uniformly if for any $\varepsilon>0$, there exist some number $\tau_0$ such that if $ \tau\ge\tau_0 $ then $$\left| \int_{\tau}^{\infty}e^{-st}f(t)dt \right| < \epsilon $$ for all $s$ in some complex domain.

But how can I proceed to show that the Laplace transform of $e^t$ converges uniformly?

Answer 1

If you take $f(t) = e^t$ then your expression becomes $$\left|\int_{\tau}^{\infty} e^{(-t(s+1))} dt\right|\leq \int_{\tau}^{\infty} \left| e^{(-t\,Re(s+1))}\right| dt = \int_{\tau}^{\infty} e^{(-t\,Re(s+1))}dt.$$

Now you find convergence on the domain $\{s\in\mathbb{C}|Re(s)>-1\}$ by convergence of the exponential integral.

Answer 2

More generally, suppose $f(t)$ is piecewise continuous on $[0,\infty)$ and of exponential type of order $a\in \mathbb R$, i.e. $|f(t)|\leq Ke^{at}$ for $K>0$ and $t\geq t_0$, then the Laplace transform converges uniformly if $Re(s)\geq b>a$. Indeed, \begin{align} \sup_{Re(s)\geq b}\left |\int_0^\infty f(t)e^{-st}\mathrm{d}t-\int_0^{t_0} f(t)e^{-st}\mathrm{d}t\right|&\leq \sup_{Re(s)\geq b}\int_{t_0}^\infty |f(t)e^{-st}|\mathrm{d}t \\ &\leq K \sup_{Re(s)\geq b}\int_{t_0}^\infty e^{-t(Re(s)-a)} \mathrm{d}t \\ &=\frac{Ke^{-t_0(b-a)}}{b-a}\to0,\end{align}as $t_0\to\infty$.