What is the solution to this week's POTW?

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Here is this week's POTW:

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If $X$ and $Y$ are independent, standard Cauchy random variables, find the density of the sum $X + Y$.
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This week's problem was solved correctly by Ackbach. You can read his solution below.

Spoiler

Because $X$ and $Y$ are both standard Cauchy distributions, they will have density functions
\begin{align*}
f(x;0,1)&=\frac{1}{\pi(1+x^2)} \quad\text{and}\\
f(y;0,1)&=\frac{1}{\pi(1+y^2)},
\end{align*}
respectively. To find the density function for $X+Y,$ we first find the characteristic function (Fourier Transform) $\varphi_{X+Y}(t),$ and then take its inverse. Because $X$ and $Y$ are independent, we may calculate
\begin{align*}
\varphi_{X+Y}(t)
:&=E\left[e^{it(X+Y)}\right] \\
&=E\left[e^{itX}\,e^{itY}\right] \\
&=E\left[e^{itX}\right]E\left[e^{itY}\right] \\
&=\varphi_X(t)\,\varphi_Y(t).
\end{align*}
So the characteristic function of the sum is the product of the characteristic functions (assuming independence). So we calculate as follows:
\begin{align*}
\varphi_X(t)\,\varphi_Y(t)&=\int_{\mathbb{R}}\frac{1}{\pi(1+x^2)}\,e^{ixt}\,dx \cdot \int_{\mathbb{R}}\frac{1}{\pi(1+y^2)}\,e^{iyt}\,dy \\
&=e^{-|t|}\,e^{-|t|} \\
&=e^{-2|t|} \\
&=\varphi_Z(t;0,2).
\end{align*}
In other words, we simply recognize this as the characteristic function of another Cauchy distribution, with $x_0=0$ and $\gamma=2$. That is, if $Z=X+Y,$ then the density function is
$$f(z;0,2)=\frac{1}{2\pi\left[1+z^2/4\right]}.$$