Taking inverse laplace transform of $\frac{1}{(s^2+1)^2}$ using residues, something's wrong

Question

I am trying to compute the inverse laplace transform of $\dfrac{1}{(s^2+1)^2}$ using residues

So $$\mathcal{L}^{-1}\dfrac{1}{(s^2+1)^2} = res(\dfrac{e^{st}}{(s^2+1)^2}, i) + res(\dfrac{e^{st}}{(s^2+1)^2}, -i) $$

I found $$res(\dfrac{e^{st}}{(s^2+1)^2}, i) = -t \dfrac{e^{it}}{4}$$

$$res(\dfrac{e^{st}}{(s^2+1)^2}, -i) = -t \dfrac{e^{-it}}{4}$$

Combined we get $\mathcal{L}^{-1}\dfrac{1}{(s^2+1)^2} = -t\cos(t)/2$

But the correct solution should be $\sin(t)/2 - t\cos(t)/2$ http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+1%2F%28s%5E2%2B1%29%5E2

Where is my $\sin(t)$ term?

Answer 1

$res(\dfrac{e^{st}}{(s^2+1)^2}, i) = \lim_{s\rightarrow i} \partial_s \dfrac{e^{st}}{(s+i)^2} = \lim_{s\rightarrow i} \dfrac{te^{st}}{(s+i)^2}-2\dfrac{te^{st}}{(s+i)^3} = \dfrac{-te^{it}}{4}+\dfrac{e^{it}}{4i}$