I am looking at this answer and I can't find a way to formally justify the final conclusion.
I have to say I understand mostly everything involved in the proof. The only thing that's bugging me out is when the answerer concludes:
Therefore, we conclude the integral of interest diverges.
I understand that the area of $S$ is infinite and, therefore, $ \frac{1}{2} \iint_S dx \, dy = +\infty.$ I also know that we have the inequality $$ \frac{1}{2}\iint_S dx \, dy \leqslant \iint_{\mathbb R^2} \frac{1}{1+x^4 \, y^4} \, dx \, dy $$ Obviously my intuition tells me that from this inequality the integral must assume the value $+\infty$ and therefore, it diverges. But I don't know how to prove such thing formally.
Thanks for any help in advance.
Here is a rigorous proof.
Since the region $S = \{(x,y):x>0, y>0, xy \leqslant1\}$ is unbounded, we have an improper double integral of $(x,y) \mapsto 1$ over $S$ which can be evaluated as
$$\int\int_S \,dx\,dy = \lim_{n \to \infty}\int\int_{S_n}\, dx\,dy$$
where $(S_n)$ is any sequence of compact rectifiable sets such that $S_1 \subset S_2 \subset S_3 \subset \ldots$ and $\cup_{n=1}^\infty S_n = S$.
This is the precise definition of an improper Riemann integral over a multidimensional domain (also called the extended Riemann integral) as discussed in treatments of advanced calculus such as Analysis on Manifolds by Munkres or Calculus on Manifolds by Spivak. When the integrand is non-negative, the particular choice of the sets $S_n$ beyond the given requirements is arbitrary, and we can take for convenience
$$S_n = S \cap [1/n,n]^2$$
Hence,
$$\int\int_S \,dx\,dy = \lim_{n \to \infty}\int\int_{S_n}\, dx\,dy= \lim_{n \to \infty}\int_{1/n}^n\left(\int_0^{1/x}\, dy\right)\,dx$$
Note that we express the double integral as the iterated integral on the RHS -- which is permissible for a non-negative integrand by Fubini's theorem.
Thus,
$$\int\int_S \,dx\,dy = \lim_{n \to \infty}\int_{1/n}^n\left(\int_0^{1/x}\, dy\right)\,dx = \lim_{n \to\infty}\int_{1/n}^n \frac{dx}{x}\\ = \lim_{n \to \infty}(\log n - \log \frac{1}{n}) = +\infty$$