= Solution
Use the <Fourier transform> convention $\widehat f(k)=\int_{\mathbb R}f(t)e^{-ikt}dt$, with inverse factor $1/(2\pi)$. For <functions> in <L1 space>, their <convolution> is $h(t)=\int_{\mathbb R}f(s)g(t-s)ds$. <Fubini's theorem> applies because the absolute double <integral> is $\|f\|_1\|g\|_1$. Substituting $r=t-s$ therefore proves the <convolution theorem>:
$$
\widehat h(k)=\int_{\mathbb R}\int_{\mathbb R}f(s)g(r)e^{-ik(r+s)}drds=\widehat f(k)\widehat g(k).
$$
For the specified <rectangular pulse>, the support condition is $|t|<1$. Its <Fourier transform> is
$$
\boxed{\widehat f(k)=\int_{-1}^1e^{-ikt}dt=\frac{2\sin k}k,\qquad\widehat f(0)=2.}
$$
The <convolution> $f*f$ is the length of the overlap of $(-1,1)$ and $(t-1,t+1)$. Thus it is the <triangular pulse>
$$
\boxed{(f*f)(t)=\begin{cases}2-|t|,&|t|<2,\\0,&|t|\geq2,\end{cases}\qquad\widehat{f*f}(k)=\frac{4\sin^2k}{k^2}.}
$$
The transform's value at zero is four, given by its continuous extension. In this convention, the <Parseval identity> for <square-integrable functions> is
$$
\int_{\mathbb R}|f(t)|^2dt=\frac1{2\pi}\int_{\mathbb R}|\widehat f(k)|^2dk.
$$
For the <rectangular pulse> the left side is two, giving the <sinc-squared integral>
$$
\boxed{\int_{-\infty}^{\infty}\frac{\sin^2k}{k^2}dk=\pi.}
$$
For the <triangular pulse>, $\int|f*f|^2dt=2\int_0^2(2-t)^2dt=16/3$. Apply the same <Parseval identity> with transform $4\sin^2k/k^2$ to obtain the <sinc fourth-power integral>
$$
\boxed{\int_{-\infty}^{\infty}\frac{\sin^4k}{k^4}dk=\frac{2\pi}3.}
$$
Both integrands have removable singularities at zero and integrable tails.
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