convergence of $\int_{\mathbb{R}^3}\frac{dx\,dy\,dz}{1+x^4+y^4+z^4}$ I need to determine whether the integral
$$\int_{\mathbb{R}^3}\frac{dx\,dy\,dz}{1+x^4+y^4+z^4}$$
converges or diverges. 
I have no good idea for a substitution, but I tend to believe that the way to solve it is to find some upper bound of the function that converges, or vice versa:
a lower bound that diverges.
Any ideas?
 A: $$\iiint_{\mathbb{R}^3}\frac{d\mu}{1+(x^2+y^2+z^2)^2} = \int_{0}^{+\infty}\frac{4\pi\rho^2}{1+\rho^4}\,d\rho = \pi^2\sqrt{2}$$
and in a similar way
$$\iiint_{\mathbb{R}^3}\frac{d\mu}{1+\frac{1}{3}(x^2+y^2+z^2)^2} = \int_{0}^{+\infty}\frac{4\pi\rho^2}{1+\frac{1}{3}\rho^4}\,d\rho = \pi^2 3^{3/4}\sqrt{2} $$
hence the given integral is convergent since $x^4+y^4+z^4\geq \frac{1}{3}\left(x^2+y^2+z^2\right)^2$.

The exact value of the integral is given by
$$ \iiint_{\mathbb{R}^3}\int_{0}^{+\infty}\exp\left(-t(1+x^4+y^4+z^4)\right)\,dt\,d\mu = \int_{0}^{+\infty}\left(\frac{2\,\Gamma\left(\frac{5}{4}\right)}{t^{1/4}}\right)^3 e^{-t}\,dt$$
that is:
$$ 8\,\Gamma\left(\frac{1}{4}\right)\,\Gamma\left(\frac{5}{4}\right)^3 = \color{red}{\frac{1}{8}\,\Gamma\left(\frac{1}{4}\right)^4}.$$

The second approach also shows that given $\alpha,\beta,\gamma>0$,
$$ \iiint_{(0,+\infty)}\frac{dx\,dy\,dz}{1+x^{\alpha}+y^{\beta}+z^{\gamma}}$$
is convergent iff $\color{red}{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}<1}.$
