Somehow this convolution is driving me crazy. I am trying to calculate for the indicator function $f:=1_{[0,1]}$ the threefold convolution $$f*f*f$$
But honestly, it does not work somehow.
$$f*f(s)=\int_0^1 1_{[0,1]}(s-t) dt$$ which I thought to be $0$ for $s\le 0$, $s$ on $(0,1)$ and $2-s$ on $[1,2]$ and $0$ if $s>2$. So, then I would need to calculate $f*f*f(s)=\int_0^1 (f*f)(s-t) dt$, but I do not trust my calculation anymore. Does anybody here know how to do this easily or has an idea about what the result will be?
One thing that I find useful in these kinds of problems is to remember that $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\int f(x-t)g(t)\,\mathrm{d}t &=\int f'(x-t)g(t)\,\mathrm{d}t\\ &=\int(\delta(x-t)-\delta(x-1-t))g(t)\,\mathrm{d}t\\ &=g(x)-g(x-1)\tag{1} \end{align} $$
$$ f(x)=\left\{\begin{array}{} 1&\text{if }x\in[0,1]\\ 0&\text{otherwise} \end{array}\right.\tag{2} $$ $\hspace{3.5cm}$
Using $(1)$, we get $g'(x)=f(x)-f(x-1)$
$$ g(x)=\int_0^1f(x-t)f(t)\,\mathrm{d}t =\left\{\begin{array}{} x&\text{if }x\in[0,1]\\ 2-x&\text{if }x\in[1,2]\\ 0&\text{otherwise} \end{array}\right.\tag{3} $$ $\hspace{3.5cm}$
Using $(1)$, we get $h'(x)=g(x)-g(x-1)$ $$ h(x) =\int_0^1f(x-t)g(t)\,\mathrm{d}t=\left\{\begin{array}{} \frac12x^2&\text{if }x\in[0,1]\\ 3x-x^2-\frac32&\text{if }x\in[1,2]\\ \frac12x^2-3x+\frac92&\text{if }x\in[2,3]\\ 0&\text{otherwise} \end{array}\right.\tag{4} $$ $\hspace{3.5cm}$