I'm trying to solve the ode $$\ddot{y}+y=5\sinh2t$$ with some set of initial values provided to me. If $Y(s)$ is the Laplace transform of $y(t)$ then I can show $$Y(s)=\dfrac{5\mathcal{L}[\sinh 2t] + 2}{s^2+1}=\dfrac{10+2(s-2)(s+2)}{(s+j)(s-j)(s-2)(s+2)}$$ I noticed that I could only compute $\mathcal{L}[\sinh2t]$ if $s<2$, otherwise the integral would blow up (coming from a Physics background here... sorry about the hand-waving).
This didn't really bother me until I tried to compute $y(t)$ via the Bromwich integral for the inverse Laplace transform:
$$y(t)=\dfrac{1}{2\pi j}\int_{c}\ ds\ Y(S)e^{st}$$
Since $c$ is some vertical line in the complex plane going from $\pm\infty$, I decided to consider the closed contour integral half-circle where the half-circle loops the left-half of the plane and is closed at the vertical line at $s=2$. My reasoning is that $s<2$ is needed for the $\sinh2t$-transform to converge and by closing it with a half-circle, I can use Jordan's lemme and the Residue theorem to evaluate.
The poles are $s=-j,j,-2,2$. The problem is that since $s<2$, then my contour doesn't enclose $s=2$ and so I don't actually use the $s=2$ residue in the residue theorem. This gives me the wrong answer; only when all poles $s=-j,j,-2,2$ are used do I get the correct answer. Why do I not ignore the $s=2$ simple-pole contribution? What am I missing?