Does $\int_{0}^{\infty}\frac{r^2}{1+\alpha r^4}dr$ converge?

Question

I want to check that $\int_{\mathbb R^3}\frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.

I moved to spherical coordinates, the integral became $\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{r^2\sin(\theta)}{1+r^4f(\theta,\phi)}d\phi d\theta dr \leq\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{r^2\sin(\theta)}{1+\alpha r^4}d\phi d\theta dr = 4\pi \int_{0}^{\infty}\frac{r^2}{1+\alpha r^4}dr $

I don't know how to show this converges.

Edit: $f(\theta,\phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $\alpha > 0$.

Answer 1

Hint: Split the integral \begin{align} \int^\infty_0 \frac{r^2}{1+\alpha r^4}\ dr = \int^\infty_1\frac{r^2}{1+\alpha r^4}\ dr +\int^1_0 \frac{r^2}{1+\alpha r^4}\ dr \end{align}

Answer 2

There is a problem only at $\infty$. You can user equivalence:

$$\frac{r^2}{1+\alpha r^4}\sim_\infty\frac 1{\alpha r^2},$$ and $\;\displaystyle\int_1^\infty\frac 1{\alpha r^2}\,\mathrm d r\;$ converges.

Answer 3

At 0, this is continuous. At $\infty$ it behaves like $x \mapsto 1/x^2$, which integrates finitely there. The integral converges.