Laplace transform of two different Heavisides multiplied

Question

The actual problem first asks me to find the laplace transform of $$f(t)=u(t-1)+u(t-2)+u(t-3)$$ which is readily obtainable by looking at a transform table to be $$F(s)=\frac{e^{-s}}{s}+\frac{e^{-2s}}{s}+\frac{e^{-3s}}{s}$$, however the problem then asks me to evaluate $$\mathcal{L}\{f(t)^2\}$$, which I am having a hard time doing. I think the minimal problem here is how to find the transform of $u(t-a)u(t-b)$. I think I should do something like $u(t-a)u((t-a)+(a-b))$ but I am both not sure of, and even if I am right, I don't know how to follow from this step.

Answer 1

Noting that $$u(t-a)=\left\{ \begin{array}{ll} 1, \space t>a\\ 0, \space t<a \end{array} \right. $$ And that $$u(t-b)=\left\{ \begin{array}{ll} 1, \space t>b\\ 0, \space t<b \end{array} \right. $$ We realise that if $ b > a $, then, when multiplying both functions, every number before the value of $b$ is multiplied by $0$, or: $$u(t-b)u(t-a)=\left\{ \begin{array}{ll} 1, \space t>b\\ 0, \space t<b \end{array} \right. \space =u(t-b) $$

which leads us to the conclusion that $$u(t-b)u(t-a)=u(t-max\{a,b\})$$