$\iint_{\mathbb R^2} \frac{dx \, dy}{1+x^{10}y^{10}}$ diverges or converges? Question I'm trying to solve to prepare for an exam.
I need to find out if $\displaystyle\iint_{\mathbb R^2} \frac{dx\,dy}{1+x^{10}y^{10}}$ diverges or converges.
What I did:
I switched to polar coordinates, $x=r\cos \theta$, $y=r\sin\theta$, $J=r$
and so
$$
\begin{align}
\iint_{\mathbb R^2} \frac{dx\,dy}{1+x^{10}y^{10}} & =\int_0^{2\pi} \int_0^\infty \frac{r}{1+r^{20} \sin \theta \cos \theta} \, dr \, d\theta \\[8pt]
& < \int_0^{2\pi} \int_0^\infty \frac{1}{r^{19}\sin \theta \cos \theta} \, dr \, d\theta \\[8pt]
& = \frac{1}{2} \int_0^{2\pi} \int_0^\infty \frac{1}{r^{19}\sin 2\theta} \, dr \, d\theta \\[8pt]
& =\frac{1}{2} \int_0^{2\pi} \frac{1}{\sin 2\theta} \, d\theta \int_0^\infty \frac{1}{r^{19}}\,dr
\end{align}
$$
But note that both $\int_0^\infty \frac{1}{r^{19}}dr$ and $\int_0^{2\pi} \frac{1}{\sin 2\theta} \, d\theta$ do not converge, so we proved nothing.
So that is not the way.
Does this integral converge/ diverge? why?
And also, I'd like to know why $\int_0^{2\pi} \frac{1}{\sin 2\theta} \, d\theta$ does not converge. I know it doesn't because of wolfram, not because i showed it on paper.
 A: Draw the shape around the axes inside the two hyperbolas, and four arcs, given by $xy=1$ and $xy=-1.$ Sort of a cross shape, but infinite, with both axes in its interior. Within this, $x^{10} y^{10} \leq 1,$ then  $1 +x^{10} y^{10} \leq 2,$ finally
$$ \frac{1}{1 +x^{10} y^{10}} \geq \frac{1}{2}. $$
This region has infinite area, the integral (nonnegative integrand) on the region is infinite. Being a little careful, out to some radius $R,$ the integral on the shape is at least half the area out to that radius; as $R \rightarrow \infty,$ so do the area and the integral.
A: A trivial way to see the integral fails to converge is to observe that $$\frac{1}{1+(x y)^{2n}} > \frac{1}{2}$$ whenever $|xy| < 1$, and this occurs over a subset of $\mathbb R^2$ with obviously infinite area.
A: Using a tip a friend gave me: It is enough to work on the first quadrant due to symmetry.
We will instead use the transform $v=xy, u=x$, the jacobian of said transform is $\frac{1}{u}$ which can be demonstrated.
$$\iint_{\mathbb R^2} \frac{1}{1+x^{10}y^{10}}dx\,dy = \int_0^\infty \int_{0}^{\infty} \frac{1}{1+v^{10}} \frac{1}{u} \, dv \, du =\int_0^\infty \frac{1}{1+v^{10}} \, dv \int_0^\infty \frac{1}{u} \, du  $$
Notice that $\frac{1}{1+v^{10}}>0$ and so $\int_0^\infty \frac{1}{1+v^{10}} \, dv>0$, and that $\int_0^\infty \frac{1}{u} \, du$ diverges.
We can infer that the original integral diverges as well.
A: $\newcommand{\+}{^{\dagger}}
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Scale $\quad\ds{\verts{x}y}\quad$ to $\quad\ds{y}\quad$ and you are left with a result $\quad\ds{\propto \int{\dd x \over \verts{x}}}\quad$ which 'diverges logarithmically'.
