= Solution
Normalization uses $\int_{-\infty}^{\infty}(1+x^2)^{-1}\,dx=\pi$, hence
$$
\boxed{\gamma=\frac1\pi.}
$$
Each variable therefore has the standard <Cauchy distribution>. Let $S_n=X_1+\cdots+X_n$. We prove by induction that its <probability density function> is
$$
\boxed{f_{S_n}(x)=\frac n{\pi(n^2+x^2)}.}
$$
It holds for $n=1$. If it holds for $n=m$, <independence> and <convolution of probability densities> give
$$
f_{S_{m+1}}(x)=\frac m{\pi^2}\int_{-\infty}^{\infty}\frac{dy}{(1+y^2)[m^2+(x-y)^2]}.
$$
The supplied integral identity makes this $(m+1)/\{\pi[(m+1)^2+x^2]\}$, completing the induction. Thus addition increases the Cauchy scale to $n$.
For $\overline X_n=S_n/n$, the density transformation is $f_{\overline X_n}(x)=n f_{S_n}(nx)$, giving
$$
\boxed{f_{\overline X_n}(x)=\frac1{\pi(1+x^2)}\quad\text{for every }n\ge1.}
$$
The <Cauchy sample mean has an unchanged distribution>. In particular, for every $\delta>0$,
$$
\mathbb P(|\overline X_n|>\delta)=1-\frac2\pi\arctan\delta>0,
$$
independently of $n$, so the average does not converge in probability to zero.
There is no contradiction with the usual <weak law of large numbers>, whose finite-first-moment hypothesis is absent: $\mathbb E|X_1|=(2/\pi)\int_0^\infty x/(1+x^2)\,dx=\infty$. The positive and negative parts of $X_1$ both have infinite expectations, so $\mathbb EX_1$ is undefined. Its symmetric principal value is zero, but a principal value is not an <expected value>. The law therefore supplies no conclusion that these sample averages must approach zero.
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