How do you find the Inverse Laplace Transform of $\frac{1}{(s-1)^2}$? How do you find the Inverse Laplace Transform of $\frac{1}{(s-1)^2}$?. I know the Inverse Laplace Transform of $\frac{1}{s-1}$ is $e^t$.
 A: Download a table of Laplace transform from here.
Particularly note these $\displaystyle \mathscr{L} \left( t^n\right) = \frac{n!}{s^{n+1}} $ and $\displaystyle  \mathscr{L} \left( e^{at}\right) = \frac{1}{(s-a)}$
Combine these two get get this $\displaystyle  \mathscr{L} \left( t^n e^{at}\right) = \frac{n!}{(s-a)^{n+1}}$, put  $n=1$ and $a=1$, there you have it.
A: Apply the residue theorem.  The contour integral is
$$\oint_C ds \frac{e^{s t}}{(s-1)^2}$$
where $C$ consists of the vertical line $[c-i R,c+i R]$,where $c>1$, and an arc of radius R that opens to the left.  In the limit $R \to \infty$, the integral about the arc vanishes.  Thus the ILT is simply the residue at the pole $s=1$, which is
$$\left[\frac{d}{ds} e^{s t}\right]_{s=1} = t e^{t}$$
A: A good way to solve this Laplace transform is to use the displacement property:
$$
\displaystyle \mathscr{L} \{f(t) e^{t}\} = F(s-a)
$$
here $ F(s)=\displaystyle \mathscr{L} \{f(t) \}.$ For $f(t)=t$ we know that $ \displaystyle \mathscr{L} \{ t\} = \frac{1}{s²}=F(s)$. Then
$$
\displaystyle \mathscr{L} \{ t e^{t}\} = F(s-1)=\frac{1}{(s-1)²} .
$$
