Mean of a Cauchy Distribution Why is the mean of a Cauchy distribution undefined? Surely, it should be $0$ by symmetry? $$\int_{-\infty}^{\infty} {\frac{x}{\pi (1+x^2)}} dx =0?$$

The problem is that
$$
\int_0^\infty \frac{x\,dx}{1+x^2}=+\infty \text{ and }\int_{-\infty}^0 \frac{x\,dx}{1+x^2}=-\infty,
$$
and one consequence of the positive and negative parts both being infinite is that
$$
\lim_{a\to\infty,\  b\to\infty} \int_{-a}^b \frac{x\,dx}{1+x^2}
$$
actually depends on the way in which $a$ and $b$ are related.  For example, if $a=b$, then you have
$$
\int_{-b}^b \frac{x\,dx}{1+x^2}=0,
$$
but if $a=2b$ then you have
\begin{align}
\int_{-2b}^b \frac{x\,dx}{1+x^2} & = \int_0^b \frac{x\,dx}{1+x^2} + \int_{-2b}^0 \frac{x\,dx}{1+x^2} \\[10pt]
& = \int_1^{1+b^2} \frac{du/2} u + \int_{1+4b^2}^1 \frac{du/2} u \\[10pt]
& = \frac 1 2 \log(1+b^2) - \frac 1 2 \log(1+4b^2) \\[10pt]
& = \frac 1 2 \log\frac{1+b^2}{1+4b^2} \\[10pt]
& \to -\frac 1 2 \log 4 \ne 0 \text{ as }b\to\infty.
\end{align}
This sort of thing can happen only when the positive and negative parts are both infinite.
Another consequence is that things like the law of large numbers do not apply.  Neither does the central limit theorem.
