How to quickly compute the inverse laplace transform of $\frac{1}{(s^2+1)^2}$

Question

I wish to find the inverse laplace transform of $\dfrac{1}{(s^2+1)^2}$

I tried using partial fraction but things do not seem to work out i.e.

$\dfrac{1}{(s^2+1)^2} = \dfrac{A}{s^2+1} + \dfrac{B}{s^2+1}$

$\Rightarrow$ $As^2 + Bs^2 = 0$, $A + B = 1$ which is inconclusive

Is there a quick way to find the inverse laplace transform of this guy?

Answer 1

I guess you know the rule $$ \mathcal{L}^{-1}[ f'(s) ](t) =- t \,\mathcal{L}^{-1} [f(s)](t)$$ and the fact that $$\mathcal{L}^{-1} \Bigl[\frac{1}{s^2+1} \Bigr](t) = \sin(t)$$ and $$\mathcal{L}^{-1} \Bigl[\frac{s}{s^2+1} \Bigr](t) = \cos(t)\quad ?$$

Now observe that $$\frac{d}{ds} \biggl(\frac{s}{s^2+1}\biggr) = -\frac{1}{(s^2+1)} +\frac{2}{(s^2+1)^2}$$ and thus $$\frac{1}{(s^2+1)^2}= \frac12\frac{d}{ds} \biggl(\frac{s}{s^2+1}\biggr) + \frac12 \frac{1}{(s^2+1)}$$ Using these results, we obtain $$\mathcal{L}^{-1}\biggl[ \frac{1}{(s^2+1)^2} \biggr](t) = -\frac{1}{2} t\,\mathcal{L}^{-1} \biggl[ \frac{s}{s^2+1}\biggr] (t)+\frac{1}{2} \sin t= \frac{-t \cos t +\sin t}{2} $$

Answer 2

$f(t)=\int_0^t \cos (t-u)\cos u du$

$=\int_0^t (\cos t \cos^2 u + \sin t \sin u \cos u)du$

$\cos t \int_0^t \frac{1+\cos 2u}{2}du+\sin t \int_0^t \sin u d\sin u$