Help Understanding Inverse Laplace Transforms??? I've been looking over some examples regarding inverse laplace transforms, and my textbook doesn't really go into much depth regarding examples such as this:
F(s) = (2(s-1)e^(-2s))/(s^2 -2s + 2)
I know that e^-2s is the unit step function, so that my answer will involve that, but does that mean I transform it from F(s) to F(s-2) and take the ILT of that?
Also, I know that the (s-1)/((s-1)^2 + 1) takes the form of e^(at)cosbt, so I know that will take part in the eventual answer, but I can't seem to get it to match with the book's answer.  It's as though I understand about 50% of it.
As you can see, I'm kind of confused, but would really like to understand this material better.
Thanks!
 A: We have:
$$F(s) = \dfrac{2(s-1)e^{-2s}}{s^2 -2s + 2}$$
So we do this part first:
$$\mathscr{L}^{-1}\left(\dfrac{2(s-1)}{s^2 -2s + 2}\right) = \mathscr{L}^{-1}\left(\dfrac{2(s-1)}{(s-1)^2 + 1^2}\right) = 2e^t \cos t$$
Now the $e^{-2s}$ is a time shift by two units, including a unit step function, so we shift each term that has $t$ with $t-2$, which gives:
$$\mathscr{L}^{-1} (F(s)) = 2 e^{t-2} u (t-2) \cos (2-t)$$
Would recommed reviewing the first and second shift theorems in Section 4.3
A: The inverse LT is given by
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, F(s) \, e^{s t}$$
where $c$ is greater than the real part of any pole of $F$.  This has consequences for when, in $t$, we begin to have a nonzero $f$.   
For the specific $F$ you show, we have
$$f(t) = \frac{2}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \frac{s-1}{s^2-2 s+2} \, e^{s (t-2)}$$
To compute, consider the contour integral
$$2 \oint_C dz \frac{z-1}{z^2-2 z+2} \, e^{z (t-2)}$$
where $C$ is the contour consisting of the intervals $[c-i R,c+i R]$, $[c+i R,iR]$, and $[-i R, c-i R]$, as well as the semicircle of radius $R$ centered at the origin in the left-half plane.  Thus, the contour integral is equal to
$$2 \int_{c-i \infty}^{c+i \infty} ds \, \frac{s-1}{s^2-2 s+2} \, e^{s (t-2)} +\int_c^0 dx \frac{x+i R-1}{(x+i R)^2-2 (x+i R)+2} e^{(x+i R)(t-2)} \\ + i R \int_{\pi/2}^{3 \pi/2} d\theta \, e^{i \theta} \frac{R e^{i \theta}-1}{R^2 e^{i 2 \theta}-2 R e^{i \theta}+2} e^{(t-2) R e^{i \theta}}\\ + \int_0^c dx \frac{x-i R-1}{(x-i R)^2-2 (x-i R)+2} e^{(x-i R)(t-2)} $$
Clearly, the second and fourth integrals vanish as $(e^{c (t-2)}-1)/(R (t-2))$ as $R \to \infty$.  The third integral has magnitude bounded by
$$2 \int_{\pi/2}^{3 \pi/2} d\theta \, e^{R (t-2) \cos{\theta}}$$
Note that, when $\theta \in [\pi/2, 3 \pi/2]$, $\cos{\theta} \le 0$, so that the integral vanishes as $R \to \infty$ only when $t-2 \gt 0$.  When $t-2 \lt 0$, on the other hand, it makes no sense to close the contour to the left as we have done here.  Rather, we will close to the right.  But we are getting a little ahead of ourselves; for now, assume $t-2 \gt 0$. 
Thus the contour integral is simply equal to the first integral as $R \to \infty$.  On the other hand, the contour integral is $i 2 \pi$ times the sum of the residues at the poles of $F$ inside $C$.  Since by definition, $C$ encloses all of the poles of $F$, i.e. $z_{\pm}=1 \pm i$, we have that, for $t-2 \gt 0$,
$$f(t) = e^{(1+i)(t-2)} + e^{(1-i)(t-2)} = 2 e^{t-2} \cos{(t-2)}$$
When $t-2 \lt 0$, on the other hand, we close the contour to the right, and we get that the contour integral is equal to $i 2 \pi$ times the ILT, which by the residue theorem is zero because $C$ now encloses no poles.  Therefore,
$$f(t) = 2 e^{t-2} \cos{(t-2)}\, \theta(t-2)$$
