Laplace transform of $f(t)=te^{-t}\sin(2t)$

Question

I was asked to find the Laplace transform of the function $\displaystyle f(t)=te^{-t}\sin(2t)$ using only the properties of Laplace transform, meaning, use clever tricks and the table shown at http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms

So it's not actually needed to calculate $\displaystyle \int_0^\infty te^{-st-t}\sin (2t)\:dt$

but rather, find shortcuts and ways to make it easier using the table and properties of the Laplace transform.

I'm really stuck. I can't find anything to make it easier. Calculating the integral seems inevitable. Anyone has any idea?

Answer 1

Here is an approach.

1. $\displaystyle \mathcal{L}(\sin 2t) = \frac{2}{s^2 + 2^2} = \frac{2}{s^2 + 4}$, using the table.

2. $\displaystyle \mathcal{L} (e^{-t} \sin 2t) = \frac{2}{(s + 1)^2 + 4}$, using frequency shifting.

3. $\displaystyle \mathcal{L}( t e^{-t} \sin 2t) = -\frac{d}{ds}\!\left(\frac{2}{(s + 1)^2 + 4}\right) = \frac{4 (s+1)}{\left((s+1)^2+4\right)^2}$, using frequency differentiation.

Answer 2

Let $f(t)=te^{−t}\sin(2t)=tg(t)$ and $g(t)=e^{−t}\sin(2t)=e^{−t}h(t)$ with $h(t)=\sin(2t)$. So $F(s)=-G'(s)$ and $G(s)=H(s+1)$ with $H(s)=\frac{2}{s^2+4}$.