Find the inverse Laplace transform of $\frac{1}{\left({s}^{2}+{1}\right)^{2}}$. What is Laplace inverse of
$$\dfrac{1}{\left({s}^{2}+{1}\right)^{2}}$$
 A: Hint:
$$\mathcal{L}^{-1}\left[\dfrac{1}{(s^2 + \omega^2)^2}\right] = \dfrac{1}{2 \omega^3}\left(\sin \omega t - \omega t \cos \omega t \right)$$
Note: $\omega$ is a real constant in this generalization.
Now, can you find the Laplace transform of $\sin \omega t$ and $\omega t \cos \omega t$ to understand what is going on with one of the shift theorems and why this is the result?
Of course you can always use the formal definitions to find this also if that is the approach required.
A: We can use the residue theorem to evaluate this inverse Laplace transform.  Consider the following contour integral in the complex $s$ plane:
$$\oint_C \frac{ds}{(s^2+1)^2}  e^{s t}$$
where $C$ is a Bromwich contour, in that the contour integral is equal to
$$\int_{c-i R}^{c+i R} ds \frac{e^{s t}}{(s^2+1)^2} + i R \int_{\pi/2}^{3 \pi/2} d\theta \, e^{i \theta} \frac{e^{R t e^{i \theta}}}{(R^2 e^{i 2 \theta}+1)^2}$$
where $c \ge 0$ (and is defined so that the vertiical line piece of $c$ is to the right of the rightmost pole).  For $t \gt 0$, the second integral is bounded by
$$\frac{2}{R^3} \int_0^{\pi/2} d\theta \, e^{-R t \sin{\theta}} \le \frac{2}{R^3} \int_0^{\pi/2} d\theta \, e^{-2 R t \theta/\pi} \le \frac{\pi}{R^4 t} $$
Thus, when $t \gt 0$, the second integral vanishes as $1/R^4$ as $R \to\infty$.  Therefore, by the residue theorem,
$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{(s^2+1)^2} = \sum_{\pm}\text{Res}_{s=\pm i} \frac{e^{s t}}{(s^2+1)^2}$$
The residues on the RHS are computed as follows because the poles are double poles:
$$\text{Res}_{s=\pm i} \frac{e^{s t}}{(s^2+1)^2} = \left [ \frac{d}{ds} \frac{e^{s t}}{(s \pm i)^2} \right ]_{z=\pm i} = \left [ \frac{(t (s \pm i)^2-2 (s \pm i)) e^{s t}}{(s \pm i)^4}\right]_{s=\pm i}$$
Note that for $t \lt 0$, we close the Bromwich contour to the right, so that there are no poles enclosed within $c$ and the ILT is identically zero.
Thus, the ILT is, for $t \gt 0$
$$\frac{(2 i)^2 t-4 i}{(2 i)^4} e^{i t} + \frac{(-2 i)^2 t + 4 i}{(-2 i)^4} e^{-i t} = -\frac{t}{2} \cos{t} + \frac12 \sin{t}$$
and $0$ otherwise.
