a question on Laplace Transform definition

**lpd** · Oct 8th 2009, 01:43 AM

Hi. I have this problem I am quite unsure of.

For which complex values of $s$ is the Laplace Transform of $e^{-2t} cos(6t)$ defined?

What I did was simplified it and had it as

$\int_0^\infty e^{(-2-s)t}cos(6t) \,dt$

And from the L.T definitions, as t approaches infitinty, the function should converge to zero.

So is it right by saying $-2-s < 0$ and hence $s > 2$ ? But then again, if I take the limit of $e^{(-2-s)t}cos(6t)$ it doesn't converge zero, rather it is undefined.

Thanks for your help.

Thanks.

**chisigma** · Oct 8th 2009, 02:04 AM

If we apply the definition of LT is...

$f(s)=\mathcal {L}\{e^{-2t}\cdot \cos 6t \} = \int_{0}^{\infty} \cos 6t \cdot e^{-(2+s)t}\cdot dt$ (1)

The integral in (1) converges for $Re(s) >-2$ and that is also the domain of $f(s)$ ...

Kind regards

$\chi$ $\sigma$

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