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I am trying to find the inverse Laplace transform of $s^{-2}(s^2 + 1)^{-1}$. I could multiply these together and use partial fraction decomposition, but, unless I am mistaken, I think there is another way by using the convolution theorem?

I would greatly appreciate it if someone could please take the time to demonstrate this.

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Let $F(s) = s^{-2}(s^2 + 1)^{-1}$ and let $\mathcal{L}(f(t)) = F(s)$ We know that $$\mathcal{L}(t) = s^{-2}$$ and $$\mathcal{L}(\sin t) = (s^2 + 1)^{-1}$$ So $$F(s) = \mathcal{L} (f(t) )= \mathcal{L}( t * \sin t) = \mathcal{L}( t )\mathcal{L}( \sin t) = s^{-2}(s^2 + 1)^{-1}$$ where $*$ stands for convolution. So, the time domain laplace inverse is really $$f(t) = t * \sin t = \int\limits_0^t \tau \sin(t - \tau) \ d \tau= \int\limits_0^t (t - \tau) \sin(\tau) \ d \tau = t - \sin (t)$$

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You are correct. The convolution theorem gives you the inverse Laplace Transform of $ s^{-2}(s^2+1)^{-1}$ as a convolution

$t*\sin t$

The convolution integral is $\int _0^t \tau \sin (t-\tau )d\tau $ which is easy to evaluate.

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