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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!

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    $\begingroup$ Have you learned the Shifting Theorem? It should be $(t-2)^2$, not $t^2$ $\endgroup$ – Amzoti Dec 9 '13 at 22:39
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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.

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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 $$ \mathcal{L}^{-1}\{e^{-2s}/s^3\}(t) = \frac{(t-2)^2}{2}\mathcal{U}(t-2) $$

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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: $$\frac{(t-2)^2}{2}$$

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