Laplace Transforms and third-order derivatives

Question

The question is to calculate the Laplace transform of $(1 + t.e^{-t})^3$. I know that this can be done using a property where the problem is of the form of $t.f(t)$. However, I seem to be messing up the third order derivative.

This is a DIY exercise from the textbook and hence, no worked out solution.
I expanded the bracket $(1 + t^3e^{-3t} + 3te^{-t} + 3t^2.e^{-2t})$
and I mess up the third order derivative for $t^3.e^{-3t}$

Can someone please how me how that is done ?

Answer 1

Hints:

$$(1+t e^{-t})^3 = 1 + 3 t e^{-t} + 3 t^2 e^{-2 t} + t^3 e^{-3 t}$$

$$\int_0^{\infty} dt \, t^k \, e^{-s t} = (-1)^k \frac{d^k}{ds^k} \frac{1}{s} = \frac{1}{s^{k+1}}$$

Now work out

$$\int_0^{\infty} dt \, e^{-s t} + 3 \int_0^{\infty} dt\,t \, e^{-(s+1) t}+ 3 \int_0^{\infty} dt\,t^2 \, e^{-(s+2) t}+ \int_0^{\infty} dt\,t^3 \, e^{-(s+3) t}$$

Answer 2

A property of the Laplace transform is that if the transform of $f(t)$ is $F(s)$, then the transform of $f(t)e^{-ct}$ is $F(s+c)$. To see this property for your self find the transform for $f(t) = 1$, the constant function and then compare it with the transform of $e^{-ct}$.