Convolution of a step function with itself I have a step function that states $f(x) = 1$ for $|x| < 1$ and 0 everywhere else. So far I've found this to be $$\int_{-\infty}^{\infty}\, dy\, f(y)\,f(x-y)$$ and from boundaries, $$-1 < y <1 \\ x-1 < y < x+1$$
Which now gives the function $$1\cdot\int_{-1}^{1}\, dy\, f(x-y) = \int_{t \,= \,x-1}^{t\, = \,x+1}f(t)\,dt$$ where $t = x-y$. From here I've found that when $x < -2$ or $x + 2$ that $f(t) = 0$ and between these boundaries $f(t) = 1$. I'm stuck with what to do from here, however. I'm unsure how to use the boundary conditions and find a final answer here.
The question also asks for the Fourier transform of this convolution. Is this straightforward or is there I trick to it that I need to be aware of? Thanks.
 A: First of all, the convolution of the function $f$ with itself is given by 
$$\begin{align}
(f*f)(x)&=\int_{-\infty}^\infty f(y)f(x-y)\,dy\\\\
&=\begin{cases}
0&,x\le -2\\\\
\int_{-1}^{x+1}(1)\,dx=(x+2)&,-2<x<0\\\\
\int_{x-1}^1(1)\,dx=2-x&,x<2\\\\
0&,x\ge 2
\end{cases}
\end{align}$$
We can write $(f*f)(x)=2-|x|$ for $|x|<2$ and $0$ otherwise.

The Fourier Transform of $f(x)$ is 
$$\begin{align}
\mathscr{F}\{f\}(k)&=\int_{-\infty}^\infty f(x)e^{ikx}\,dx\\\\
&=\int_{-1}^1 e^{ikx}\,dx\\\\
&=2\frac{\sin(k)}{k}
\end{align}$$

Finally, invoking the Convolution Theorem yields
$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{f*f\}(k)=\frac{4\sin^2(k)}{k^2}}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets
  $\ds{\,\mc{F}\pars{x} \equiv
\int_{-\infty}^{\infty}\mrm{f}\pars{y}\mrm{f}\pars{x - y}\,\dd y =
\int_{-\infty}^{\infty}\hat{\mc{F}}\pars{k}\expo{\ic kx}{\dd k \over 2\pi}}$ such that

\begin{align}
\hat{\mc{F}}\pars{k} & =
\int_{-\infty}^{\infty}\mc{F}\pars{x}\expo{-\ic kx}\dd x =
\int_{-\infty}^{\infty}\bracks{%
\int_{-\infty}^{\infty}\mrm{f}\pars{y}\mrm{f}\pars{x - y}\,\dd y}
\expo{-\ic kx}\dd x
\\[5mm] & =
\int_{-\infty}^{\infty}\bracks{%
\int_{-\infty}^{\infty}\mrm{f}\pars{y}\mrm{f}\pars{x}\,\dd y}
\expo{-\ic k\pars{x + y}}\dd x =
\bracks{\int_{-\infty}^{\infty}\mrm{f}\pars{y}\expo{-\ic ky}\dd y}
\bracks{\int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{-\ic kx}\dd x}
\\[5mm] & =
\hat{\mrm{f}}^{\,2}\pars{k} =
\pars{\int_{-1}^{1}\expo{-\ic kx}\dd x}^{2} =
4\pars{\int_{0}^{1}\cos\pars{kx}\,\dd x}^{2} =
\bbx{\ds{4\sin^{2}\pars{k} \over k^{2}}}
\end{align}
