Hard Integral $\frac{1}{(1+x^2+y^2+z^2)^2}$ Prove that $\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz = \pi^2$
I tried substitution, trigonometric substitution, and partial fraction decomposition, but I can't solve this problem, I only know that
$\frac{1}{(1+x^2+y^2+z^2)^2}$ is a even function :( then
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz = $$
$$ 2 \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{0}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz $$
 A: You can use spherical coordinates. Then the integral changes to the following:
$$
\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx dy dz\\ = \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{\pi}\displaystyle\int_{0}^{\infty} \frac{1}{(1+\rho^2)^2}\,\rho^2\sin\theta d\rho d\theta d\phi \\=2\pi.2\int_{0}^{\infty} \frac{1}{(1+\rho^2)^2}\,\rho^2 d\rho
$$
Now it is enough to calculate the following:
$$
\int_{0}^{\infty} \frac{1}{(1+\rho^2)^2}\,\rho^2 d\rho.
$$
which you can find by $\rho=\tan y$ and then:
$$
\int_{0}^{\infty} \frac{1}{(1+\rho^2)^2}\,\rho^2 d\rho=\int_{0}^{\pi/2} \sin^2 y dy=\frac{\pi}{4}
$$
A: An alternative is to overkill it with some measure theory. Unfortunately I don't know the names of the theorems and objects used (not in my language and not in english). If someone does, please edit my answer as you see fit.
Firstly note that $$\displaystyle\int \limits_{-\infty}^{\infty}\displaystyle\int \limits_{-\infty}^{\infty}\displaystyle\int \limits_{-\infty}^{\infty} \dfrac{1}{(1+x^2+y^2+z^2)^2}\mathrm dx \,\mathrm dy \,\mathrm dz=\iiint \limits_{\Bbb R^3\setminus\{0_{\Bbb R^3}\}}\dfrac{1}{(1+x^2+y^2+z^2)^2}\mathrm dx \,\mathrm dy \,\mathrm dz,$$
then use change of variables and something which translates to generalized polar coordinates to get
$$\iiint \limits_{\Bbb R^3\setminus\{0_{\Bbb R^3}\}}\dfrac{1}{(1+x^2+y^2+z^2)^2}\mathrm dx \,\mathrm dy \,\mathrm dz=\iint \limits_{]0,+\infty[\times S_2} \dfrac{t^2}{\left(1+(tx)^2+(ty)^2+(tz)^2\right)^2}\mathrm dt \,\mathrm d\mu_{S_2}(x,y,z).$$
Now $$\displaystyle \begin{align} \iint \limits_{]0,+\infty[\times S_2} \dfrac{t^2\mathrm dt \,\mathrm d\mu_{S_2}(x,y,z)}{\left(1+(tx)^2+(ty)^2+(tz)^2\right)^2}&=\iint \limits_{]0,+\infty[\times S_2} \dfrac{t^2\mathrm dt \,\mathrm d\mu_{S_2}(x,y,z)}{\left(1+t^2(x^2+y^2+z)^2\right)^2}\\
&=\iint \limits_{]0,+\infty[\times S_2} \dfrac{t^2\mathrm dt \,\mathrm d\mu_{S_2}(x,y,z)}{\left(1+t^2\right)^2}\\
&=\underbrace{\mu _{S_2}(S_2)}_{\large 4\pi}\int \limits_{]0,+\infty[} \dfrac{t^2}{\left(1+t^2\right)^2}\mathrm dt\\
&=4\pi\cdot \dfrac \pi 4=\pi ^2\end{align}$$
