Domain of Laplace transform

Question

According to Wikipedia, if the Laplace transform $$\mathcal{L}\{f\}(s)=\lim_{b\to\infty}\int_0^b e^{-st}f(t)\,dt$$ converges for some $s_0\in\mathbb{R}$, then it automatically converges for all $s\in\mathbb{R}$ with $s>s_0$. (I'm considering the Laplace transform on the real line.) How can I prove this? If $$\lim_{b\to\infty}\int_0^be^{-st}|f(t)|\,dt$$ converges, then I can prove it using comparison test, but what if it converges conditionally?

Answer 1

This answer is essentially @RRL's answer given here, re-written for more novice readers: I worked within real numbers, and also I did not use Riemann-Stieltjes integral.

Let $\alpha:[0,\infty)\to\mathbb{R}$ be a function defined by $$\alpha(t)\mathrel{\mathop:}=\int_0^te^{-s_0t}f(t)\,dt.$$ Let $s>s_0$. Since $\alpha$ is continuous, integration by parts gives \begin{align}\int_0^be^{-st}f(t)\,dt&=\int_0^be^{-(s-s_0)t}e^{-s_0t}f(t)\,dt\\ &=[e^{-(s-s_0)t}\alpha(t)]_0^b+(s-s_0)\int_0^be^{-(s-s_0)t}\alpha(t)\,dt.\tag{$\ast$}\label{lap} \end{align} By assumption, $\lim_{t\to\infty}\alpha(t)$ exists. In particular, $|\alpha(t)|$ is bounded, say $|\alpha|\leq M$. Then $$|e^{-(s-s_0)t}\alpha(t)|\leq e^{-(s-s_0)t}M\quad\xrightarrow{t\to\infty}\quad0,$$ so the first term in \eqref{lap} goes to zero as $b\to\infty$. On the other hand, $$\int_0^be^{-(s-s_0)t}|\alpha(t)|\,dt\leq M\int_0^be^{-(s-s_0)t}\,dt\leq M\int_0^\infty e^{-(s-s_0)t}\,dt,$$ so $\int_0^\infty e^{-(s-s_0)t}|\alpha(t)|\,dt$ converges, hence $\int_0^\infty e^{-(s-s_0)t}\alpha(t)\,dt$ converges. It follows that $\int_0^\infty e^{-st}f(t)\,dt$ converges.