Region of Covergence of $x(t) = e^{-2(t-3)} u(t-3)$ I was answering a question bank just to see what I have become and how my math skills fare after understanding some topics when a wild question
 appeared. It goes like this:

Determine the region of convergence of the Laplace transform of $$x(t) = e^{-2(t-3)} u(t-3)$$

My work
I got to find first the Laplace transform of $x(t) = e^{-2(t-3)}
   u(t-3)$. But there's a problem: I don't know how to find the Laplace
   transform of  $$x(t) = e^{-2(t-3)} u(t-3)$$ because of the pesky
   expression $t-3$, which I think, is a time delay. I barely had
   experiences with those kinds of  getting Laplace transforms
   before...so I turned here for help.
How to get the Laplace transform of $$x(t) = e^{-2(t-3)} u(t-3)$$,
   and ultimately, getting its region of convergence?
 A: The unit step function determines from which value of $t$ your function will be different of $0$.
So, if we have:
$$
u(t) =
\begin{cases}
1,  & \text{$t \geq 0$} \\
0, & \text{$t<0$}
\end{cases}
$$
With the time shift we have:
$$
u(t-3) =
\begin{cases}
1,  & \text{$t \geq 3$} \\
0, & \text{$t<3$}
\end{cases}
$$
Applying the Laplace Transform integral, we have:
$$
\begin{align}
X(s)&=\int_{3}^{\infty}e^{-2(t-3)}e^{-st}dt\\
&=\int_{3}^{\infty}e^{-t(2+s)}e^{6}dt\\
&=e^{6}\int_{3}^{\infty}e^{-t(2+s)}dt\\
&=e^{6}\left[-\frac{e^{-t(2+s)}}{(2+s)}\right]_{3}^{\infty}\\
&=\frac{e^{6}}{s+2}\left[e^{-3(s+2)}-\color{blue}{\lim \limits_{t \to \infty}e^{-t(s+2)}}\right]\\
&=\frac{e^{-3s}}{s+2}
\end{align}$$
Now, we use the part in $\color{blue}{\text{blue}}$ to find the $ROC$ of this Laplace Transform. In essence, we can say that:
$$ROC=\left\{s:\left|\lim \limits_{t \to \infty}e^{-t(s+2)}\right|<\infty\right\}$$
So
$$
\begin{gather}
\left|\lim \limits_{t \to \infty}e^{-t(s+2)}\right|<\infty\\
\left|\lim \limits_{t \to \infty}e^{-2t}e^{-st}\right|<\infty\\
\left|\lim \limits_{t \to \infty}e^{-2t}e^{-t(\sigma+j\omega)}\right|<\infty\\
\left|\lim \limits_{t \to \infty}e^{-t(2+\sigma)}e^{-j\omega t}\right|<\infty\\
\lim \limits_{t \to \infty}\left|e^{-t(2+\sigma)}\right|<\infty\\
\end{gather}
$$
This will only be true if:
$$
\begin{align}
2+\sigma&>0\\
\sigma&>-2
\end{align}
$$
So the region of convergence comprises all values of $s=\sigma+j\omega=\Re\{s\}+j\Im\{s\}$ so that $\Re\{s\}>-2$ and $\Im\{s\}\in \mathbb{R}$
Which is correct for a causal signal.
