= Convolution of a Cauchy distribution and a uniform distribution
Let $X$ have the <Standard Cauchy distribution> and let $Y$, independently, have the <uniform distribution> on $[-h,h]$, where $h>0$. The <convolution of probability densities> gives the <probability density> of $Z=X+Y$ as
$$
p_h(z)=\frac{1}{2h\pi}\int_{-h}^{h}\frac{dy}{1+(z-y)^2}=\frac{\arctan(z+h)-\arctan(z-h)}{2h\pi}.
$$
This is an <even function>. Its <derivative> is
$$
p_h'(z)=\frac{-2z}{\pi[1+(z+h)^2][1+(z-h)^2]},
$$
so its unique <mode> is zero. Its tails satisfy $p_h(z)\sim1/(\pi z^2)$ as $|z|\to\infty$: multiplying the integral by $z^2$ makes the integrand converge uniformly to $1$ on the bounded interval $[-h,h]$. Consequently both positive and negative parts of the first moment diverge, so its <expected value> does not exist. The symmetric <Cauchy principal value> of that moment is zero, which is a different assertion. Adding a bounded independent displacement smooths the <probability density> without removing the divergent first moment of the <Cauchy distribution>.
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