# $\iint_{\mathbb R^2} \frac{dx \, dy}{1+x^{10}y^{10}}$ 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.

• Recall that $\sin(x)\sim _0x$. – Git Gud Jun 18 '14 at 17:10
• Yes that is correct. that explains why $\int_{0}^{2\pi} \frac{1}{\sin 2\theta} d\theta$ does not converge. – Oria Gruber Jun 18 '14 at 17:11
• First note that there is no trouble near $(0,0)$. In the unit circle with centre the origin, our function is $\le 1$. So you can confine attention to $r\gt 1$. – André Nicolas Jun 18 '14 at 17:15
• @OriaGruber I reckon the exam must be around the corner. Good luck. – Git Gud Jun 18 '14 at 18:02
• no,no, gud help you – Will Jagy Jun 18 '14 at 18:40

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 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.

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.

• Alpha agrees that it does not converge – Ross Millikan Jun 18 '14 at 17:52
• @RossMillikan, there is an easy argument, posted answer. – Will Jagy Jun 18 '14 at 17:56
