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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?

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$$\iiint_{\mathbb{R}^3}\frac{d\mu}{1+(x^2+y^2+z^2)^2} = \int_{0}^{+\infty}\frac{4\pi\rho^2}{1+\rho^4}\,d\rho = \pi^2\sqrt{2}$$ 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 $\color{red}{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}<1}.$

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    The last inequality is wrong. (Take for example $x = y = z = 1$). – 2017-01-31
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    @RobertoRastapopoulos: I noticed the issue. Now fixed. – 2017-01-31
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    The last inequality follows from Cauchy-Schwarz or $$ 3(x^4+y^4+z^4)-(x^2+y^2+z^2)^2 = (x^2-y^2)^2+(x^2-z^2)^2+(y^2-z^2)^2 \geq 0.$$ – 2017-01-31
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    Might be a stupid question, but how do you apply Cauchy-Schwarz in this case? – 2017-01-31
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    @RobertoRastapopoulos: $$( 1\cdot x^2+1\cdot y^2+1\cdot z^2)^2 \leq (1+1+1)(x^4+y^4+z^4) $$ – 2017-01-31
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    Ah, yes. Amazing, thanks! – 2017-01-31