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I understand that the definition of a Laplace transform of a function $f(t)$ is $$F(s)= \int_0^{+\infty} e^{-st}f(t) \,dt$$

Is there an easy way to find the Laplace transform of $$f(t)=7t \cdot \mathrm{e}^{-3t}\cdot\sin(3t)$$

I saw some Laplace transform tables online but this does not fall into one of the scenarios. Would I have to do the entire integration?

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up vote 4 down vote accepted

Here is one approach (there are others)

$\displaystyle \mathcal{L}~~ (\sin 3t) = \frac{3}{s^2 + 3^2} = \frac{3}{s^2 + 9}$,

$\displaystyle \mathcal{L} ~~(e^{-3t} \sin 3t) = \frac{3}{((s + 3)^2 + 9)}$, via frequency shifting.

So, $\displaystyle \mathcal{L} ~~(7 e^{-3t} \sin 3t) = \frac{21}{(s + 3)^2 + 9}$.

Finally, we use frequency differentiation.

$\displaystyle \mathcal{L}~~( t~ 7 e^{-3t} \sin 3t) = -\frac{d}{ds}\left[\frac{21}{(s + 3)^2 + 9}\right] = \frac{42 (s+3)}{((s+3)^2+9)^2}$

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This is perfect. Thank you very much – Greg Harrington Apr 27 '13 at 22:35
@GregHarrington: You are very welcome! Regards – Amzoti Apr 27 '13 at 22:36
Perfect, indeed ;-) $\quad + 1 \quad$ – amWhy Apr 28 '13 at 0:13
@GregHarrington: Note that I had a sign error (now corrected in the final answer)! Regards – Amzoti Apr 28 '13 at 6:00

Hints: Use formal rules $\mathcal{L}(t f(t)) = -F^{\prime}(s)$, and combine it with the following table result: $$ \mathcal{L}\left(\mathrm{e}^{-\lambda t} \sin(\mu t) \right) = \Im \int_0^\infty \mathrm{e}^{-(s + \lambda)t} \cdot \mathrm{e}^{i \mu t} \mathrm{d} t = \Im\left( \frac{1}{s+\lambda - i \mu}\right) = \frac{\mu}{(s+\lambda)^2+\mu^2} $$

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