Tough inverse Laplace transform I know what the solution is to this inverse Laplace transform, I just have NO idea how to get there.
$$\mathcal{L}^{-1}\left(\frac{16s}{\left(s^2+4\right)^2}\right)$$
Basically, my question is what modification do I have to do to the equation above?
 A: Using a table, note the form:
$$f(t) = t\sin(at)$$
$$F(s) = \frac{2as}{(s^2+a^2)^2}$$
Using this:
$$f(t) = \mathcal{L}^{-1}\left(\frac{16s}{\left(s^2+4\right)^2}\right) = \mathcal{L}^{-1}\left(\frac{2*2*s}{\left(s^2+2^2\right)^2}*4\right) =4t\sin(2t)$$
A: Since I am bad at memorizing the tables of the inverse Laplace transform, I prefer to use the Bromwich integral (the inverse Laplace transform).
\begin{align}
\mathcal{L}^{-1}\{f(s)\} &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{16se^{st}}{(s^2 + 4)^2}ds\\
&= \sum\text{Res}
\end{align}
The residues occur at $s = \pm 2i$ both of order two.
Then
\begin{align}
\sum\text{Res} &= \lim_{s\to 2i}\frac{d}{ds}(s - 2i)^2\frac{16se^{st}}{(s^2 + 4)^2} + \lim_{s\to -2i}\frac{d}{ds}(s + 2i)^2\frac{16se^{st}}{(s^2 + 4)^2}\\
&= -2ie^{2it}t + 2e^{-2it}t\\
&= 4t\sin(2t)
\end{align}
A: Omitting $s$ rule
Omit $s$
Then we get $\frac d{dT}[L $ inverse of $\frac 16{(s^2+4)^2}]$
Take $16$(const out)
Take $l$ inverse and differentiate with respect to $t$
