Solution
= Solution
Insert the two densities into the <convolution of probability densities>:
$$
h(z)=\frac1\pi\int_{-1/2}^{1/2}\frac{dy}{1+(z-y)^2}.
$$
Using the <antiderivative> $-\arctan(z-y)$ with respect to $y$ gives
$$
\boxed{h(z)=\frac{\arctan(z+1/2)-\arctan(z-1/2)}{\pi}},\qquad z\in\mathbb R.
$$
This is the <convolution of a Cauchy distribution and a uniform distribution> with half-width $1/2$.