Inverse Laplace Step Function I've been asked to find the inverse Laplace Transform of: $$\mathcal L^{-1}\left\lbrace e^{-2s}\over {s^3}\right\rbrace$$
I lost my notes, so I'm going off of examples I have found online. I got stuck on the last step:
$$\mathcal L^{-1}\left\lbrace e^{-2s}\cdot {1\over {s^3}}\right\rbrace$$
$$=u(t-2)\cdot\frac 12t^2$$
That's unfortunately as far as I have gotten... If someone could help me figure out the last part that would be nice!
 A: Here's one way to approach the problem:
$$
\begin{align}
\mathcal L^{-1}\left\lbrace e^{-2s}\cdot {1\over {s^3}}\right\rbrace &=
\mathcal L^{-1}\left\{e^{-2s}\right\} * \mathcal L^{-1}\left\{1/s^3\right\}\\&=
\delta(t-2)*\frac 12 t^2u(t) \\&=
\frac 12 (t-2)^2u(t-2)
\end{align}
$$
Where $*$ represents convolution.
A: Another approach is to use the inverse Laplace transform.
$$
\mathcal{L}^{-1}\{F(s)\}(t) = \int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds
$$
For your problem, we have
$$
\mathcal{L}^{-1}\{e^{-2s}/s^3\}(t) = \int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{s(t-2)}}{s^3}ds = \sum\text{Res}\{f(s);s_j\}
$$
In order for convergence, the exponential terms needs to converge. That is, $s(t-2)<0$ or $t < 2$. We can capture this with the unit step 
$$
\mathcal{U}(t-2)=
\begin{cases}
0,&t<2\\
1,&t>2
\end{cases}
$$
We can use the series expansion to find the residue since the residue is the coefficient of the $s^{-1}$ term.
$$
\frac{e^{s(t-2)}}{s^3} = \sum_{n=0}^{\infty}\frac{[s(t-2)]^n}{n!s^3} = \frac{1}{s^3} + \frac{t-2}{s^2} + \frac{(t-2)^2}{2s} + \mathcal{O}(s^n)
$$
Therefore, the residue is $\frac{(t-2)^2}{2}$. Putting this all together, we have
$$
\mathcal{L}^{-1}\{e^{-2s}/s^3\}(t) = \frac{(t-2)^2}{2}\mathcal{U}(t-2)
$$
A: You can solve with the  Laplace's time translation property.
$$\mathscr{L}\{f(t-a)\}=e^{-sa}F(s)$$
And the transformation:
$$\mathscr{L}\{\frac{1}{s^3}\}=\frac{t^2}{2}$$
So, from: $$F(s)=\frac{t^2}{2}$$
You immediately obtain: $$\frac{(t-2)^2}{2}$$ 
