What will be the pdf of $X+Y$ if $X$ and $Y$ are iid from Cauchy? Suppose $X$ and $Y$ follow Cauchy distribution independent of each other. What will be the pdf of $X+Y$?
What I got by using convolution theorem is that the density $g$ of $X+Y$ is $:$
$$g(x) = \int_{-\infty}^{\infty} f(y) f(x-y)\ \mathrm {dy}$$ where $f$ is the density of the Cauchy distribution given by $f(x)=\frac {1} {\pi ({1+ x^2})},x \in \Bbb R$. Then the whole integration becomes $$\frac  {1} {\pi^2} \int_{-\infty}^{\infty} \frac {\mathrm {dy}} {(1+y^2)(1+(x-y)^2)}.$$ Now how do I solve this integral? Please help me in this regard.
Thank you very much.

You can try partial fraction decomposition:
\begin{align}\frac1{(1+y^2)(1+(x-y)^2)} &= \frac{x+2y}{x(x^2+4)(1+y^2)} + \frac{3x-2y}{x(x^2+4)(1+(x-y)^2)} \\
&= \frac{1}{(x^2+4)(1+y^2)}+ 
\frac{2y}{x(x^2+4)(1+y^2)} + \frac{2(x-y)}{x(x^2+4)(1+(x-y)^2)} + \frac{1}{(x^2+4)(1+(x-y)^2)}\\
&= \frac1{x^2+4}\left(\frac1{1+y^2} + \frac1{1+(x-y)^2}\right) + \frac1{x(x^2+4)}\left(\frac{y}{1+y^2} + \frac{x-y}{1+(x-y)^2}\right)
\end{align}
so we have
\begin{align}
\frac1{\pi^2}\int_{-\infty}^\infty \frac{dy}{(1+y^2)(1+(x-y)^2)} &=
\frac1{\pi^2(x^2+4)}\left[\int_{-\infty}^\infty\left(\frac1{1+y^2} + \frac1{1+(x-y)^2}\right)dy + \frac1x\int_{-\infty}^\infty\left(\frac{y}{1+y^2} + \frac{2(x-y)}{1+(x-y)^2}\right)dy
\right]\\
&= \frac1{\pi^2(x^2+4)}\left[\Big(\arctan(1+y^2) + \arctan(1+(x-y)^2)\Big)\Big|_{-\infty}^\infty + \frac1x\Big(\ln(1+y^2) - \ln(1+(x-y)^2)\Big)\Big|_{-\infty}^\infty
\right]\\
&= \frac1{\pi^2(x^2+4)}\left[2\pi + \frac2x \lim_{y \to \infty} \ln \left(\frac{1+y^2}{1+(x-y)^2}\right)\right]\\
&= \frac{2}{\pi(x^2+4)}
\end{align}
