Laplace transform of $t\,e^{-t}$

Question

Can someone explain why the answer is $\frac{1}{(s+1)^2}$?

My understanding was to multiply the Laplace transforms of $t$ and $e^{-t}$. So $\left(\frac{1}{s^2}\right)\left(\frac{1}{1+s}\right)$?

Thank you.

Answer 1

No, $\mathcal{L}(f\cdot g)\not=\mathcal{L}(f)\cdot\mathcal{L}(g)$ in general.

You might be thinking of the Convolution Theorem, which says $$ \mathcal{L}(f* g)=\mathcal{L}(f)\cdot\mathcal{L}(g) $$ where $*$ denotes the convolution product rather than the usual product.

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However, there are multiple, elementary ways to compute the transform you seek:

1. From Definition: You could compute the result straight from the definition of the Laplace transform: $$ \mathcal{L}(te^{-t})=\int_0^\infty te^{-t}e^{-st}\,dt=\int_0^\infty te^{-(s+1)t}\,dt=\cdots={1\over (s+1)^2}, $$ where in the dots, you integrate by parts and take the appropriate limits at $t\to\infty$.

2. Exponential Shift: However, if you know $f(t)=t\implies F(s)={1\over s^2}$ and you've established the "exponential shift" property, i.e. $$\mathcal{L}(e^{at}f(t))=F(s-a),$$ where $F(s)=\mathcal{L}(f(t))$, then $$ \mathcal{L}(e^{-t}\cdot t)=F(s+1)\text{ where }F(s)=\mathcal{L}(t)={1\over s^2}\text{ so } $$ $$ \mathcal{L}(e^{-t}\cdot t)=F(s+1)={1\over (s+1)^2}. $$

3. Differentiation in $s$ Domain: Using $\mathcal{L}(t^nf(t))=(-1)^n{d^nF\over ds^n}$, and the fact that $\mathcal{L}(e^{at})={1\over s-a}$, $$ \mathcal{L}(te^{-t})=-{dF\over ds}\text{ where }F(s)=\mathcal{L}(e^{-t})={1\over s+1}, $$ so $$ \mathcal{L}(te^{-t})=-{dF\over ds}=-{d\over ds}\left({1\over s+1}\right)=-\left({-1\over (s+1)^2}\right)={1\over (s+1)^2}. $$

These are just some of the ways you could show the result.