What's the laplace inverse of this function? I'm completely stuck on how to do this one.  Any help is appreciated.
What is the inverse Laplace transform of:
$$\mathcal{L} ^ {-1} \left\{ \frac{e^{-2s}}{s-2} \right\} = f(t)$$
 A: Using the shift theorem:
$$\mathscr L\{f(t-a)\,\mathcal U(t-a)\}(s)=e^{-as}F(s).$$
Here $\mathcal U(t-a)$ is the Heaviside step function that is defined as $1$ for $t\geq a$, $0$ otherwise. What you've got is $e^{-as}F(s):a=3,F(s)=\dfrac{1}{s-2}.$ Therefore, applying the theorem yields
$$e^{2(t-3)}\,\mathcal U(t-3).$$
So the proof for the shift theorem (called second shift theorem (I guess)) is the following:
$$\mathscr L\{f(t-a)\,\mathcal U(t-a)\}(s)=\int_0^\infty e^{-st}f(t-a)\,\mathcal U(t-a)\, dt =\int_a^\infty e^{-st}f(t-a)\, dt. $$
Consider the substitution $u=t-a$, then:
$$\int_0^\infty e^{-s(u+a)} f(u)\,du=e^{-sa}F(s).$$
Recall that $F(s)$ is the laplace transform of $f(t)$.
A: We could also use the inverse Mellin transfer where the pole is at $s=2$ of order one. That is,
\begin{align}
\mathcal{L}^{-1}\Bigl\{\frac{e^{-2s}}{s-2}\Bigr\} &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{-2s}}{s-2}e^{st}ds\\
&=\sum\text{Res}\\
&=\lim_{s\to 2}(s-2)\frac{e^{s(t-2)}}{s-2}
\end{align}
From the limit, we have an exponential term of the form $e^{s(t-2)}$. Thus $\text{Re}(s) >2$ which means we need a unit step at $t = 2$.
$$
\mathcal{L}^{-1}\Bigl\{\frac{e^{-2s}}{s-2}\Bigr\} = e^{2(t-2)}\mathcal{U}(t-2)
$$
From the Wolfram link, you will see that the accepted answer is incorrect. Note that $\theta(t-2)$ is the Heaviside Theta which is the unit step. We even check our solution by taking the Laplace transform.
\begin{align}
\int_0^{\infty}e^{2(t-2)}\mathcal{U}(t-2)e^{-st}dt &= \int_2^{\infty}e^{2(t-2)}e^{-st}dt = e^{-4}\int_2^{\infty}e^{t(2-s)}dt\\ 
&= e^{-4}\frac{e^{t(2-s)}}{2-s}\Bigr|_2^{\infty}\\
&= \frac{e^{-2s}}{s-2}
\end{align}
