Find $p \in \mathbb{R}$ so that $\int_{0}^{\infty}\frac{x^2+x^{p/4}}{\sqrt{1+x^p}}dx$ converges

Question

Find the values of $p \in \mathbb{R}$ so that $\int_{0}^{\infty}\frac{x^2+x^{p/4}}{\sqrt{1+x^p}}dx$ converges.

Since there are no problems at $0$ because it's not a root of the denominator, I'll just have to worry about the case when $x \to \infty$.

I know I have to compare with $\frac{1}{x^{\alpha}}$, for some $\alpha$. But how do I choose $\alpha$? Should it be $\frac{p}{2}$? $\frac{p}{2}-2$? What if $p$ is negative?

Depending on $\alpha$, I'll get that $\frac{1}{x^\alpha}$ converges for different values of $p$, for the first one $p > 2$, for the second, $p>6$, and so on, so I have no idea what to do.

Answer 1

The effective power of the numerator is $\max\{2,p/4\}$ and the effective power of the denominator is $p/2$. You need $$\frac p2 - \max\{2 , \frac p4 \} > 1$$ for the integral to converge.

If $p \ge 8$ this reduces to $p/2 - p/4 > 1$ which is true.

If $p < 8$ this reduces to $p/2 - 2 > 1$ which is equivalent to $p > 6$.

So the integral converges for $p \in [8,\infty) \cup (6,8) = (6,\infty)$. That is, $p > 6$.

Answer 2

Hint : Around $\infty$ we have $\frac{x^2+x^{p/4}}{\sqrt{1+x^p}}\sim x^{\max(-p/2+2,-p/2+p/4)}=x^{\max(-p/2+2,-p/4)}$