Inverse Laplace transform of $\frac 1 {(s^2+1)^2}$

Question

How to find the inverse Laplace of:

$$f(t) = 2\Bigg[\mathscr{L}^{-1}\bigg(\frac{-1}{(s^2+1)^2}\bigg) + \mathscr{L}^{-1}\bigg(\frac{3}{s^2+1}\bigg) \Bigg]$$

$$f(t) = 2\Bigg[\mathscr{L}^{-1}\bigg(\frac{-1}{(s^2+1)^2}\bigg) + 3sin(t) \Bigg]$$

$$\mathscr{L}^{-1}\bigg(\frac{-1}{(s^2+1)^2}\bigg) = ?$$

Answer 1

Since $\frac{1}{s^2+1}=\frac{1}{2i}\left(\frac{1}{s-i}-\frac{1}{s+i}\right)$,$$\begin{align}\frac{1}{(s^2+1)^2}&=-\frac14\left(\frac{1}{(s-i)^2}+\frac{1}{(s+i)^2}-\frac{2}{s^2+1}\right)\\&=-\frac14[(s-i)^{-2}-(s+i)^{-2}]+\frac{1}{4i}\left(\frac{1}{s-i}-\frac{1}{s+i}\right),\end{align}$$which has inverse Laplace transform$$-\frac14[te^{it}+te^{-it}]+\frac{e^{it}-e^{-it}}{4i}=-\frac12t\cos t+\frac12\sin t.$$

Answer 2

$$g(t) = sin(t) * \sin(t)$$

$$g(t) = \frac{1}{2}\int \limits_{0}^{t} 2\sin(\tau) \sin(t - \tau) d\tau$$

Applying trig identity:

$$2\sin(A)\sin(B)=\cos(A-B)-cos(A+B)$$

we have:

$$g(t) = \frac{1}{2}\int \limits_{0}^{t} \Big[\cos(\tau -(t-\tau)) - \cos(\tau +t - \tau)\Big]d\tau$$

$$g(t) = \frac{1}{2}\int \limits_{0}^{t} \Big[\cos(-t+2\tau) - \cos(t)\Big]d\tau$$

$$g(t) = \frac{1}{2}\int \limits_{0}^{t}\cos(2\tau-t)~d\tau - \frac{1}{2}\cos(t)\int \limits_{0}^{t} ~d\tau$$

$$g(t) = \frac{1}{4}\bigg[ \sin(2\tau-t)\bigg]^{t}_{0}- \frac{1}{2}\cos(t)\bigg[\tau\bigg]^{t}_{0}$$

$$g(t) = \frac{1}{4}\bigg[ \sin(2\tau-t)\bigg]^{t}_{0} - \frac{1}{2}\cos(t)\bigg[\tau\bigg]^{t}_{0}$$

$$g(t) = \frac{1}{4}\bigg[ \sin(2t-t) - \sin(0-t)\bigg] - \frac{1}{2}t \cos(t)$$

$$g(t) = \frac{1}{4}\bigg[ \sin(t) + \sin(t)\bigg] - \frac{1}{2}t \cos(t)$$

$$g(t) = \frac{1}{4}\bigg[ 2\sin(t)\bigg] - \frac{1}{2}t \cos(t)$$

$$g(t) = \frac{1}{2}\sin(t) - \frac{1}{2}t \cos(t)$$

Answer 3

Another way is to consider: $$\frac 1 {(s^2+1)^2}=\frac {1}{2s}\frac {2s} {(s^2+1)^2}$$ We have: $$\frac {2s} {(s^2+1)^2}=-1 \left (\frac {1} {(s^2+1)} \right )'$$ $$\frac {2s} {(s^2+1)^2}=-\frac d {ds}\{\mathcal {L}\{\sin(t)\}\}=\mathcal {L}\{t\sin(t)\}$$ Then use Convolution theorem: $$\mathcal {L^{-1}}\left\{\frac 1 {(s^2+1)^2}\right \}=\frac {1}{2}*t\sin (t)$$ The Convolution integral is easier to evaluate.

$$f(t)=\frac 1 2\int_0^t \tau \sin(\tau) d \tau$$ $$f(t)=\frac 12 (\sin(t)-t\cos(t))$$ $$\boxed {\mathcal {L^{-1}}\left\{\frac 1 {(s^2+1)^2}\right \}=\frac 12 (\sin(t)-t\cos(t))}$$