<s>[INST] Find the marginal distribution of $(X,Y)$ Given the next robability density function$\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty} \frac{1}{\pi^2(1+x^2+y^2+z^2)^2}\, dx dy dz  $
a)Find the marginal distribution of $(X,Y)$
The marginal distribution of (X,Y) is $\displaystyle\int_{-\infty}^{\infty} \frac{1}{\pi^2(1+x^2+y^2+z^2)^2}\,dz$ 
How can I solve this integral not using spherical coordinates ??
This integral looks like a Cauchy Distribution, then I know that $\displaystyle\int_{-\infty}^{\infty} \frac{a}{\pi((x-b)^2+a^2)}\,dx = 1$
Is the marginal of $(X,Y)$ a Cauchy distribution??
Thanks for your help :D
 [/INST] Since
$$
\int_{-\infty}^\infty \frac{1}{A^2+z^2} dz=\frac{z}{A^2+z^2}\Big|_{-\infty}^\infty+\int_{-\infty}^\infty \frac{2z^2}{(A^2+z^2)^2} dz=\int_{-\infty}^\infty \frac{2A^2+2z^2-2A^2}{(A^2+z^2)^2} dz
$$
Then
$$
\int_{-\infty}^\infty \frac{2A^2}{(A^2+z^2)^2} dz=\int_{-\infty}^\infty \frac{1}{A^2+z^2} dz=\frac{1}{A}\arctan(\frac{z}{A})\Big|_{-\infty}^\infty=\frac{\pi}{A}
$$
Therefore
$$
\int_{-\infty}^\infty \frac{1}{(A^2+z^2)^2} dz=\frac{\pi}{2A^3}
$$
</s>