= Solution
For $n\geq1$, the <convolution> equals the length of the overlap of two intervals:
$$
g_n(\xi)=\begin{cases}2,&|\xi|\leq n-1,\\n+1-|\xi|,&n-1<|\xi|<n+1,\\0,&|\xi|\geq n+1.\end{cases}
$$
It is <continuous> and compactly supported, with $\|g_n\|_\infty=2$. Inverse <Fourier transform> turns this <convolution> into a product, so
$$
\boxed{h_n(x)=\frac{\sin(2\pi n x)\sin(2\pi x)}{\pi^2x^2}\ (x\ne0),\qquad h_n(0)=4n.}
$$
This function is bounded near zero and $O(x^{-2})$ at infinity, hence lies in $L^1$. <Fourier inversion>, or the <convolution theorem> applied first to the integrable interval indicators, gives $\widehat h_n=g_n$.
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