Laplace Transform of an integral

Question

Find the Laplace transform of $$f(t)=t\int_0^{t} \tau e^{-\tau}$$ $L(f)(s)$= ??

My thought is that I can change the $\tau$ to $t$ by Transforming the integral to get $$t/s*L[t*e^{-t}]$$

But i'm stuck from there

Answer 1

First write the function as

$$ f(t)=t\int_0^{t} \tau e^{-\tau} = t g(t), $$

then you can use the following properties of the Laplace transform :

1)

$$ \mathcal{L} \left(\int_{0}^{t} h(\tau) d\tau \right)(s) = \frac{H(s)}{s}, $$

where $H(s)$ is the Laplace transform of $h$.

2)

$$ \mathcal{L}( t^n g(t) ) = (-1)^n G^{(n)}(s), $$

where $G(s)$ is the Laplace transform of $g(t)$.

Answer 2

$$\int_0^t d\tau \, \tau \, e^{-\tau} = \left [-\tau \, e^{-\tau} \right ]_0^t + \int_0^t d\tau \, e^{-\tau} = 1-(1+t) e^{-t}$$

$$\begin{align}\int_0^{\infty} dt \, f(t) \, e^{-s t} &= \int_0^{\infty} dt \, t \, e^{-s t} - \int_0^{\infty} dt \,(t+t^2) e^{-(s+1) t}\\ &= \frac1{s^2} - \frac1{(s+1)^2}-\frac{2}{(s+1)^3}\end{align}$$