My assignment for a linear systems course is to "Use the definition of the unilateral Laplace transform and an integral table to verify the following Laplace transforms, and specify their regions of convergence:"
For the first example: $\mathcal{L}[t\sin(bt)] = \dfrac{2bs}{(s^2 + b^2)^2}$
I found the integral for $t\sin(bt)$ is $-\dfrac{t\cos(bt)}{b} + \dfrac{\sin(bt)}{b^2}$ on integral-table.com, but I'm not sure if I'm missing something crucial.
If I do the integral by definition, it's $\int_{0}^{\infty}t\sin(bt)e^{-st}dt $
I would normally just think to do integration by parts to do this but the wording of the question is making me second guess myself ... how does using the integration table for this problem help me, or is it to do with the fact that the other function in the integral is $e^{-st}$?
Please do NOT simply solve the problem ... I don't want to be accused of cheating :)
Notice that $t\sin(bt)e^{-st} = -\frac{\partial}{\partial s}\sin(bt)e^{-st}$. Therefore, $$ \mathcal{L}\{t\sin(bt)\} = -\frac{\partial}{\partial s}\mathcal{L}\{\sin(bt)\} $$ Now, what is the Laplace transform of $\sin(bt)$? Once you have that, just differentiate with the respect to $s$ and multiple by $-1$.