LaPlace transform of $t^2\sin(at)$

Question

$$\int^{\infty}_{0} e^{-st}t^2\sin(at)dt$$

I keep running into a problem when using:

$$u=e^{-st}t^2$$ $$du=2te^{-st}-st^2e^{-st}$$

$$v=\frac{1}{a} \cos(at)$$

$$dv=\sin(at)$$

Anyone have any suggestions for how to proceed? I've seen a slick substitution using Euler's formula but I'm not sure that it is necessary.

Answer 1

To expand on what @ncmathsadist means is $$ \int_0^{\infty}t^2e^{-st}\sin(at)dt $$ can be solved by using a more well known transform. Consider $$ \mathcal{L}\{\sin(at)\} = \int_0^{\infty}e^{-st}\sin(at)dt = \frac{a}{s^2 + a^2} $$ Then $$ \frac{\partial^2}{\partial s^2}\mathcal{L}\{\sin(at)\} = \text{what integral?} = \frac{\partial^2}{\partial s^2}\frac{a}{s^2 + a^2} $$

Answer 2

Try invoking the Leibnitz rule and differentiating twice under the integral sign with respect to $s$.