TITLE: Laplace transforms: Convolution
QUESTION [5 upvotes]: Find $$1*1*1*\cdots*1\quad n\,\,\text{  factors}$$ that is, a function $f(t)=1$ convolution with itself for a total of $n$ factors.

Would anyone mind helping me? I have no idea what I should do.
Also, if we define $g(t) = t$, what is $g*g*g$?

REPLY [2 votes]: For two functions $f(t), g(t)$ $$\mathcal{L}((f*g)(t))=\mathcal{L}(f(t))\mathcal{L}(g(t))$$Since $\displaystyle \mathcal{L}(1(t))=\frac{1}{s}$ it follows that $$\mathcal{L}(\underbrace{1*1*\cdots*1}_{n \ \mbox{times}})=\frac{1}{s^n}$$ Now, $$\mathcal{L}(t^{n-1})=\frac{(n-1)!}{s^n}$$ hence comparing, $\displaystyle \underbrace{1*1*\cdots*1}_{n\ \mbox{times}}=\frac{t^{n-1}}{(n-1)!}$