Inverse Laplace by phase shifting Can anyone help me out here?
I have to find the inverse laplace of 
$$ \frac{s (1-e^{-s/2})}{s^2+\pi^2}$$ 
Sorry it looks bad, I just don't know how to format. Here's the wolframalpha link:
http://www.wolframalpha.com/input/?i=inverse+laplace+of+%28s%281-e%5E%28-s%2F2%29%29%29%2F%28s%5E2+%2Bpi%5E2%29+
 A: $$\hat{f}(s) = \frac{s (1-e^{-s/2})}{s^2+\pi^2} e^{s t} = \frac{s }{s^2+\pi^2} e^{s t} - \frac{s\, e^{-s/2}}{s^2+\pi^2} e^{s t} $$
and sum them up.  For the first piece, from a complex analysis standpoint, you just need to evaluate the residues at the poles $s_{\pm} = \pm i \pi$ of the function  This sum of residues is
$$f_1(t) = \frac{i \pi }{2 i \pi} e^{i \pi t} + \frac{-i \pi }{-2 i \pi} e^{-i \pi t} $$
Simplifying, we get
$$f_1(t) = \cos{\pi t}$$
For the second piece, rewrite as
$$\frac{s\, e^{-s/2}}{s^2+\pi^2} e^{s t} = \frac{s }{s^2+\pi^2} e^{s (t-1/2)} $$
Note that when $t< 1/2$, the (Bromwich) integration contour must go to the right of the line $\Re{z}=c$ in order to converge.  Because there are no poles within that contour, the ILT $f(t)$ is zero here.  For $t \ge 1/2$, however, we may use the normal contour to the left of $\Re{z}=c$ and sum the residues at $s_{\pm} = \pm i \pi$:
$$f_2(t) = \frac{i \pi e^{-i \pi/2} }{2 i \pi} e^{i \pi t} + \frac{-i \pi e^{i \pi/2} }{-2 i \pi} e^{-i \pi t} = -\sin{\pi t}$$
Therefore, the ILT is
$$f(t) = \cos{\pi t} - H\left( t-\frac12 \right) \sin{\pi t}$$
where $H$ is the Heaviside step function.
A: If we set:
$$F(s) = \frac{s (1-e^{-s/2})}{s^2+\pi^2}$$ so you have $$F(s)=\frac{s}{s^2+\pi^2}-\frac{s e^{-s/2}}{s^2+\pi^2}$$ Now use the followings to get the answer:
$$\mathcal{L^{-1}}\left(\frac{s}{s^2+a^2}\right)=\cos(ax),~~\text{e}^{-as}\mathcal{L}f(t)=\mathcal{L}(u_a(t)f(t-a))$$
In fact, $$\frac{s e^{-s/2}}{s^2+\pi^2}=\text{e}^{\frac{-1}{2}s}\frac{s }{s^2+\pi^2}=\text{e}^{\frac{-1}{2}s}\mathcal{L}(\cos\pi t)=\mathcal{L}\left(u_{\left(\frac{-1}2\right)}(t)\cos\pi(t-1/2)\right)$$
