Laplace Transform of $\int_{0}^{t}u^2e^u\,du$

Question

Good day,

I am having some problems with the Laplace transform of $f(t) = \int_{0}^t{u^2e^u\textrm{d}u}$. $$\mathcal{L}\left [\int_{0}^t{u^2e^u\textrm{d}u} \right ]$$

using the property :

$$\mathcal{L} \left [\int_{0}^{t}f(u)\mathrm{d}u \right ] = \frac{F(s)}{s},s>\textrm{max}\left \{ 0,a \right \}$$

This is what I have tried so far :

• $$\int_{0}^{t}u^2e^u\textrm{d}u = e^tt^2-2te^t+2e^t-2 \textrm{ (Integration by parts)}$$
• $$\mathcal{L} \left [e^tt^2-2te^t+2e^t-2 \right ] = \frac{2}{(s-1)^3}-\frac{2}{(s-1)^2}+\frac{2}{s-1}-\frac{2}{s}$$
• $$ \frac{\mathcal{L} \left [e^tt^2-2te^t+2e^t-2 \right ]}{s} = \frac{2}{s^2(s-1)^3} $$

The solution of the problem is : $$\frac{2}{s(s-1)^3}$$

Thanks in advance,

Emanuel Camacho

Answer 1

You did not use the suggested property. Since: $$ \mathcal{L}(t^2 e^t) = \frac{2}{(s-1)^3},\tag{1}$$ it follows that: $$ \mathcal{L}\left(\int_{0}^{t}u^2 e^u\,du\right) = \frac{2}{s(s-1)^3}.\tag{2}$$ To prove $(1)$, you just have to recall that $\mathcal{L}(t^2)=\frac{2}{s^3}$.