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

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 $$\alpha > 0$$.

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}
• And $\int_{1}^{\infty} \frac{r^2}{1 + \alpha r^4} dr\leq \int_{1}^{\infty}\frac{r^2}{\alpha r^4}dr$. Literally just thought of that trick too. – Rick Joker Jan 20 at 21:05
At 0, this is continuous. At $$\infty$$ it behaves like $$x \mapsto 1/x^2$$, which integrates finitely there. The integral converges.
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.