Laplace Transform for a difficult function The Laplace Transform I'm having trouble with is:
$$f(t) = 6te^{-9t}\sin(6t)$$
I'm not sure what the protocol is for multiplying t into it.
The Laplace Transform for $f(t) = 6e^{-9t}\sin(6t)$ is $\dfrac 6{(s+9)^2 - 36}$.
Can't figure out how to add in the $t$.
Thanks in advance for your help.
 A: If $\displaystyle F(s) = L[f(t)](s) = \int_0^\infty f(t)e^{-st} \ dt $
and
$ \displaystyle G(s) = L[tf(t)](s) = \int_0^\infty tf(t)e^{-st} \ dt$
then
$\displaystyle {dF \over ds} = {d\ \over ds}  \int_0^\infty f(t)e^{-st} \ dt = \int_0^\infty {\partial\ \over \partial s} f(t)e^{-st} \ dt = \int_0^\infty -tf(t)e^{-st} \ dt = - \int_0^\infty tf(t)e^{-st} \ dt = -G(s)$
Hence $\displaystyle G(s) = -{dF \over ds}$.
Thus $\displaystyle L[ 6te^{-9t}\sin(6t)](s) = -{d\ \over ds}\left(\dfrac 6{(s+9)^2 - 36}\right) = \ ...$
A: Here is a partial solution:
\begin{align}
\mathcal{L}\{f(t)\}&=\mathcal{L}\{e^{-9t}[6t\sin(6t)]\} \\
&=\mathcal{L}\{6t \sin(6t)\} \vert_{s \to s+9} & \text{First Translation Theorem}\\
&= 6 \mathcal{L}\{t\sin(6t)\} \vert_{s \to s+9} & \text{linearity} \\
&=-6\frac d{ds} \mathcal{L}\{\sin(6t)\} \vert_{s \to s+9} & \text{transform derivative principle} \\
&= -6 \left. \frac d{ds} \frac {6}{s^2+36} \right\vert_{s \to s+9} 
\end{align}
Now evaluate $\frac d{ds} \frac {6}{s^2+36}$, then replace all the $s$ in your derivative with $s+9$.
