Laplace inverse of $\frac{1}{(s^{2}+1)^{2}}$.
I have beaten my head about this but have not got a clue as to how to do it! Can someone help? I tried to express this as some form of a derivative, but I am really just not sure what to do since a derivative of $\frac{1}{s^{2}+1}$ would involve a 2s in the numerator. Can someone please help?
$$F(s)=\frac{1}{(s^{2}+1)^{2}}=\frac{2s}{(s^{2}+1)^{2}} \times \dfrac 1 {2s}$$ $$F(s)=- \dfrac {d}{ds}\left (\frac{1}{(s^{2}+1)} \right) \times \dfrac 1 {2s}$$ $$F(s)=G(s) H(s)$$ And their inverse Laplace Transform are: $$g(t)=t \sin t \,\, , h(t)=\dfrac 12$$ Then use the convolution theorem. Evaluate this integral: $$f(t)=\int_0^t g(\tau) h(t-\tau) d \tau$$ $$f(t)=\int_0^t \tau \sin (\tau) \times \dfrac 12 d \tau$$
For the Convolution Theorem look here formula 33 here
You should find: $$f(t)=\dfrac 12 (\sin t - t\cos t)$$