# $\int_{0}^{\infty } e^{-t}\cdot t^{3}\cdot \sin(t) dt$ using Laplace transform

I am currently on this question:

Use Laplace Transform to evaluate the following integral:

$$\int_{0}^{\infty } e^{-t}\cdot t^{3}\cdot \sin(t) dt$$

What I did:

1. let the $$\int_{0}^{\infty } e^{-t } dt$$ = Limit $$s\to s+1$$
2. $$t\sin(t) = (-1)^3 \frac{d^{3}}{ds^{3}} \frac{1}{s^{2}+1}$$

However, my professor answered it differently, and it seems I'm either far off or both are correct. I really have no idea what happens after the second line so If anyone could explain?

Well, we are trying the following integral:

$$\mathscr{S}_\text{n}\left(\text{s},\omega\right):=\int_0^\infty x^\text{n}\sin\left(\omega x\right)\exp\left(-\text{s}x\right)\space\text{d}x\tag1$$

Using the definition of the Laplace transform, we can see that we can write:

$$\mathscr{S}_\text{n}\left(\text{s},\omega\right)=\mathscr{L}_x\left[x^\text{n}\sin\left(\omega x\right)\right]_{\left(\text{s}\right)}\tag2$$

Using the frequency-domain general derivative property of the Laplace transform, we can write:

$$\mathscr{S}_\text{n}\left(\text{s},\omega\right)=\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathscr{L}_x\left[\sin\left(\omega x\right)\right]_{\left(\text{s}\right)}\right)\tag3$$

Using the table of selected Laplace transforms, we can see that:

$$\mathscr{S}_\text{n}\left(\text{s},\omega\right)=\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{\omega}{\text{s}^2+\omega^2}\right)=\omega\cdot\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{1}{\text{s}^2+\omega^2}\right)\tag4$$

Applying this to your integral, we get:

$$\mathscr{S}_3\left(1,1\right)=\lim_{\text{s}\to1}1\cdot\left(-1\right)^3\cdot\frac{\partial^3}{\partial\text{s}^3}\left(\frac{1}{\text{s}^2+1^2}\right)=0\tag5$$