The Cauchy random variable $C(m,a)$ with center $m$ and half-width $a$ is defined by the probability density
$$p(x)=\frac{\frac{a}{\pi}}{(x-m)^2+a^2},\ -\infty\leq x\leq\infty$$

This $p(x)$ can be considered as a probability density since
\begin{align*}\int_{-\infty}^\infty p(x)dx&=\frac{a}{\pi}\int_{-\infty}^\infty\frac{dx}{(x-m)^2+a^2}\\&=\frac{1}{a\pi}\int_{-\infty}^\infty\frac{du}{\left(\frac{u}{a}\right)^2+1}\ (u=x-m)\\&=\frac{1}{\pi}\int_{-\infty}^\infty\frac{dv}{v^2+1}\ \left(v=\frac{u}{a}\right)\\&=\frac{1}{\pi}\lim_{b\to\infty}\left\{\int_{-b}^0\frac{dv}{v^2+1}+\int_0^b\frac{dv}{v^2+1}\right\}\\&=\frac{1}{\pi}\lim_{b\to\infty}\{[\tan^{-1}(v)]_{-b}^0+[\tan^{-1}(v)]_0^b\}\\&=\frac{1}{\pi}\left\{\frac{\pi}{2}+\frac{\pi}{2}\right\}\\&=1\end{align*}
Now, we want to calculate the mean and obviously, we expect it to be $m$.
\begin{align*}\int_{-\infty}^\infty xp(x)dx&=\frac{a}{\pi}\int_{-\infty}^\infty\frac{x}{(x-m)^2+a^2}dx\\
&=\frac{a}{\pi}\int_{-\infty}^\infty\frac{u}{u^2+a^2}du+\frac{am}{\pi}\int_{-\infty}^\infty\frac{du}{u^2+a^2}\ (u=x-m)\\
&=\frac{a}{\pi}\int_{-\infty}^\infty\frac{u}{u^2+a^2}du+m
\end{align*}
But,
$$\lim_{b\to\infty}\int_{-b}^0\frac{u}{u^2+1}du=\frac{1}{2}\lim_{b\to\infty}[\ln(u^2+a^2)]_{-b}^0=-\infty$$ and $$\lim_{b\to\infty}\int_0^b\frac{u}{u^2+1}du=\frac{1}{2}\lim_{b\to\infty}[\ln(u^2+a^2)]_0^b=\infty$$ This means that the mean does not exist! This result does not coincide with our intuition. What about the Cauchy principal value $$\mathrm{p. v.}\int_{-\infty}^\infty xp(x)dx?$$
Before we continue, recall that if $\int_{-\infty}^\infty f(x)dx$ exists (meaning it is finite) then $\mathrm{p. v.}\int_{-\infty}^\infty f(x)dx$ also exist and
$$\int_{-\infty}^\infty f(x)dx=\mathrm{p. v.}\int_{-\infty}^\infty f(x)dx$$
But the converse need not be true as seen below.
$$\mathrm{p. v.}\int_{-\infty}^\infty f(x)dx=\lim_{b\to\infty}\int_{-b}^b\frac{u}{u^2+1}du=0$$
since $\frac{u}{u^2+1}$ is an odd function. Hence, if we choose to use the Cauchy principal value of the improper integral $\frac{a}{\pi}\int_{-\infty}^\infty\frac{u}{u^2+a^2}du$, we obtain the mean $m$ as expected.