find all $a\in\mathbb{R}$ such that $\int_{(0,\infty)^2}\frac{dxdy}{x^a+y^4+(xy)^2}<\infty$ Question: I want to find all $a\in\mathbb{R}$ such that $\int_{(0,\infty)^2}\frac{dxdy}{x^a+y^4+(xy)^2}<\infty$.
My thoughts: Not completely sure on where to start with this one.... I know that I need to find all $a$ so that $f(x,y)=\frac{1}{x^a+y^4+(xy)^2}$ is integrable.  I was thinking about trying to find an integrable majorant, and maybe I could put it in terms of $a$ and I would be able to see a condition on when the majorant would still be integrable, but I am not sure how to go about doing that.  I feel like I could find an integrable majorant if there was only one variable, but having two is throwing me off... maybe Fubini could be applied first?
Any thoughts, suggestions, etc. are greatly appreciated.  I am sorry I don't have more of a "start".
 A: Let me first post an answer based on an explicit computation. I am working on an idea that does not depend on an explicit computation.

We first simplify the double integral by applying the substitution $y=x^{a/4}t$:
\begin{align*}
\int_{(0,\infty)^2} \frac{\mathrm{d}x\mathrm{d}y}{x^a + y^4 + (xy)^2}
&= \int_{0}^{\infty} \frac{1}{x^{3a/4}} \int_{0}^{\infty} \frac{1}{1 + t^4 + x^{2-a/2}t^2} \, \mathrm{d}t\mathrm{d}x \\
&= \int_{0}^{\infty} \frac{1}{x^{3a/4}} \underbrace{ \int_{0}^{\infty} \frac{t^{-2}}{(t - t^{-1})^2 + 2 + x^{2-a/2}} \, \mathrm{d}t }_{=: I} \mathrm{d}x \tag{1}
\end{align*}
Now denoting the inner integral in $\text{(1)}$ by $I$ and applying the substitution $t \mapsto t^{-1}$,
$$ I
= \int_{0}^{\infty} \frac{t^{-2} \, \mathrm{d}t}{(t - t^{-1})^2 + 2 + x^{2-a/2}} \stackrel{(t\mapsto t^{-1})}{=} \int_{0}^{\infty} \frac{\mathrm{d}t}{(t - t^{-1})^2 + 2 + x^{2-a/2}}. $$
So, averaging these two representations and applying the substitution $u=t-t^{-1}$,
\begin{gather*}
I
= \frac{1}{2}\int_{0}^{\infty} \frac{(1 + t^{-2}) \, \mathrm{d}t}{(t - t^{-1})^2 + 2 + x^{2-a/2}}
\stackrel{(u=t-t^{-1})}{=} \frac{1}{2}\int_{-\infty}^{\infty} \frac{\mathrm{d}u}{u^2 + 2 + x^{2-a/2}} \\
= \frac{\pi}{2\sqrt{2+x^{2-a/2}}}.
\end{gather*}
Plugging this back to $\text{(1)}$, we end up with
\begin{align*}
\bbox[#fff8f0,5pt]{
\int_{(0,\infty)^2} \frac{\mathrm{d}x\mathrm{d}y}{x^a + y^4 + (xy)^2}
= \int_{0}^{\infty} \frac{\pi}{2x^{a/2}\sqrt{x^2 + 2x^{a/2}}} \, \mathrm{d}x.
}
\end{align*}
Writing $f(x)$ for the integrand of the integral in the right-hand side, we have
$$ f(x) \sim \frac{\text{const}}{x^{(a/2)+\min\{1,a/4\}}} \quad \text{as} \quad x \to 0^+, \qquad f(x) \sim \frac{\text{const}}{x^{a/2+\max\{1,a/4\}}} \quad \text{as} \quad x\to\infty. $$
So the integral converges if and only if
$$ (a/2)+\min\{1, a/4\} < 1 \quad\text{and}\quad (a/2)+\max\{1, a/4\} > 1$$
hold simultaneously, which occurs exactly when $0 < a < 4/3$.
