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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 $e^t$.

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Laplace transform or inverse Laplace transform? – Santosh Linkha Apr 28 '13 at 19:55
Ya sorry. That is what I meant – user74636 Apr 28 '13 at 19:55
Hints: You can use 1) Table lookup or 2) Convolution – Amzoti Apr 28 '13 at 20:27
or 3) residue theorem – Ron Gordon Apr 28 '13 at 20:34

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.

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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

$$\left[\frac{d}{ds} e^{s t}\right]_{s=1} = t e^{t}$$

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