Show that a function is not integrable show that $f(x,y) = \dfrac{2xy}{1+x^4+y^4}$ is not $\lambda_2$ integrable.
I am given the solution, and it states: $$\int f^+ \ d \lambda_2 \geq \int_{(0,\infty)^2} f^+  d \lambda_2 = \int_0^\infty \int_0^\infty \frac{2xy}{1+x^4+y^4} \ dy \ dx = \int_0^\infty \int_0^\infty \dfrac{x}{1+x^4+y^2} \ dy dx$$ then they proceed to calculate the integral.
Also, they have $$\int f^- \ d \lambda_2 \geq \int _{(0,\infty) \times (-\infty,0)} f^- \ d \lambda_2 = \int f^+ \ d \lambda_2$$
I have some questions.
1) Where did the intequalities: $\int f^+ \ d \lambda_2 \geq \int_{(0,\infty)^2} f^+  d \lambda_2$ and $\int f^- \ d \lambda_2 \geq \int _{(0,\infty) \times (-\infty,0)} f^- \ d \lambda_2$ come from?
2) in the first set of inequalities, why did the "$2x$" change to $"x"$?
3) I am having troubles justifying why $\int _{(0,\infty) \times (-\infty,0)} f^- \ d \lambda_2 = \int f^+ \ d \lambda_2$ also
thank you
 A: 1) $\int f^+ \ d \lambda_2 $ is the integral on the intervall $(-\infty,\infty)^2$ so since $f^+$ is positive, you have the inequality : $\int f^+ \ d \lambda_2 \geq \int_{(0,\infty)^2} f^+  d \lambda_2$. You have the other inequality with the same idea.
2) Actually it is not $x$ which changes, but $y$. If you make the substitution $u=y^2$, you have $du=2ydy$. And finally $\int_0^\infty \int_0^\infty \frac{2xy}{1+x^4+y^4} \ dy \ dx = \int_0^\infty \int_0^\infty \frac{x}{1+x^4+u^2} \ du dx$.
3) I have no idea at the moment...
A: Here's another approach to this problem. Let $S$ be the infinite sector $\{re^{it}: 0 \le r <\infty, t \in [\pi/8,\pi/4]\}.$ Then
$$\int_{\mathbb R^2} f^+ \ge \int_{S} f.$$
Now let's start thinking in terms of polar coordinates. In $S,$
$$f(re^{it}) \ge \frac{2r^2\cos (\pi/4)\sin(\pi/8)}{1 + 2r^4} = C\frac{r^2}{1 + 2r^4}.$$
Integrating using polar coordinate gives
$$\int_{\pi/8}^{\pi/4} \int_0^\infty \frac{r^2}{1+2r^4}r\, dr\, dt=\infty.$$
The reason is that inner integral above diverges, because the integrand is on the order of $1/r$  for large $r.$ This shows $\int_{\mathbb R^2} f^+ = \infty.$ The argument for $\int_{\mathbb R^2} f^- = \infty$ is the same by symmetry.
