I had a calculus 3 exam recently and didn't do too well, I am trying to see how I could have approached things differently.
I was asked to prove/disprove that the improper integral $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{1}{1 + x^4 + y^4 + z^4} dx dy dz$ converges.
I tried to change the coordinates to get something that resembles arcsin but I am just.. not really sure how to approach this. I'd really love a hint of some sor!
Transforming to spherical coordinates, we have $$ r^4 = (x^2 + y^2 + z^2)^2\\ = x^4+y^4+z^4 + 2x^2y^2 + 2y^2z^2 + 2x^2z^2\\ \leq 3(x^4 + y^4 + z^4) $$ which gives us $\frac13r^4\leq x^4 + y^4 + z^4 $, meaning we have $$ \frac{1}{1 + x^4 + y^4 + z^4}\leq \frac 3{3+r^4} $$ We integrate the right-hand expression over all of $\Bbb R^3$ to get $$ \int_{-\pi}^{\pi}\int_0^{2\pi}\int_0^\infty\frac{3}{3+r^4}r^2\sin\varphi\,dr\,d\varphi\,d\theta\\ $$ I don't know what $\int_0^{\infty}\frac{3r^2}{3+r^4}dr$ is, but a suitable comparison using $\frac1{r^2}$ shows that it converges. The other two are proper, bounded integrals and therefore also converge, meaning the whole triple integral also converges.