Does $\iiint_{\mathbb{R}^3} \frac{1}{1+x^2y^2z^2}$ converge?
My try
from symmetry it's enought to look when $x,y,z\geq 0$ and changing variables to $u=xyz,v=y,t=z$
$\frac{\partial (u,v,t)}{\partial(x,y,z)}=\begin{pmatrix}yz&xz&yx\\0&1&0\\0&0&1\end{pmatrix}$ and the det is $yz$ then we get $$\iiint_{x,y,z\geq0} \frac{1}{1+x^2y^2z^2}=\iiint_{u,v,t\geq0} \frac{1}{1+u^2}\frac{1}{vt}dudvdt=\int_0^{\infty}\frac{1}{v}dv\int_0^{\infty}\frac{1}{t}dt\int_0^{\infty}\frac{1}{1+u^2}du=(\ln(v)\big|_0^{\infty})(\ln(t)\big|_0^{\infty})(\arctan(u)\big|_0^{\infty})=\infty$$
therefore the integral diverges. I'm not sure if what i did is right$$$$