Lets consider the signal whose Laplace Transform is $\hat{X}(s) = \frac{1}{s + 3}$. If the signal is causal, then we have

\begin{align*}
x(t) &= e^{-3t}u(t) \\
RoC(x) &= \{ s \st \Real{s} \gt -3 \} \\
\end{align*}

This signal decays as $t$ approaches $\infty$. This is a stable signal, which can be confirmed by the fact that the signal's region of convergence includes the $i\omega$-axis. If this signal signal was anti-causal, then we have

\begin{align*}
x(t) &= -e^{-3t}u(-t) \\
RoC(x) &= \{ s \st \Real{s} \lt -3 \} \\
\end{align*}

Notice that the region of convergence is flipped, and doesn't contain the $i\omega$-axis. This signal is unstable. This brings us to the statement:

\begin{claim}
$x$ is a stable signal if and only if the region of convergence contains the $i\omega$-axis.
\end{claim}

\begin{proof}

Assume that $x$ is stable. This means that $x$ is absolutely integrable.
\begin{align*}
&\infint \abs{x(t)} dt \lt \infty \\
&\infint \abs{x(t)}\abs{e^{-i\omega t}} dt \lt \infty \quad \mbox{multiply by 1}\\
&\infint \abs{x(t) e^{-(\sigma + i\omega) t}} dt \lt \infty \quad \mbox{for } \sigma = 0\\
&\infint \abs{x(t) e^{-st}} dt \lt \infty \\
&\Rightarrow \mbox{for the chosen values of $s = i\omega$, the Laplace Transform converges}
\end{align*}

Therefore the stable signal $x$'s absolute integrability implies that $i\omega \in RoC(x)$ since $\infint \abs{x(t) e^{-st}} dt \lt \infty$ for $\sigma = 0$. The reverse direction is practically the same. Assume that $i\omega \in RoC(x)$. This implies that $\infint \abs{x(t) e^{-st}} dt \lt \infty$ for $s = 0 + i\omega$.

\begin{align*}
&\infint \abs{x(t) e^{-st}} dt \lt \infty \\
&\infint \abs{x(t) e^{-i\omega t}} dt \lt \infty \quad s = i\omega \\
&\infint \abs{x(t)}\abs{e^{-i\omega t}} dt \lt \infty  \\
&\infint \abs{x(t)}dt \lt \infty \\
&\Rightarrow \mbox{ $x$ is absolutely integrable } \\
\end{align*}

Lets compare the CTFT and the Laplace Transform:

\begin{align*}
X(\omega) &= \infint x(t) e^{-i\omega t} dt \\
\hat{X}(s) &= \infint x(t) e^{-s t} dt \\
\end{align*}

If $i\omega \in RoC(x)$, then 

\begin{align*}
X(\omega) &= \left. \hat{X}(s) \right|_{s = i\omega} \\ 
\end{align*}

\end{proof}

\begin{example}

Given the following signal, 
\begin{align*}
x(t) &= e^{-3t}u(t) \ltp \frac{1}{s + 3} \quad \Real{s} \gt -3 \\
\end{align*}

we can evaluate the Laplace Transform at $s = i\omega$ to find the Fourier Transform:

\begin{align*}
\ftrans{x(t)} &= \left. \hat{X}(s) \right|_{s=i\omega} = \frac{1}{3 + i\omega} \\
\end{align*}

\end{example}

Stability