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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.

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  • $\begingroup$ Recall that $\sin(x)\sim _0x$. $\endgroup$ – Git Gud Jun 18 '14 at 17:10
  • $\begingroup$ Yes that is correct. that explains why $\int_{0}^{2\pi} \frac{1}{\sin 2\theta} d\theta$ does not converge. $\endgroup$ – Oria Gruber Jun 18 '14 at 17:11
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    $\begingroup$ 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 $r\gt 1$. $\endgroup$ – André Nicolas Jun 18 '14 at 17:15
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    $\begingroup$ @OriaGruber I reckon the exam must be around the corner. Good luck. $\endgroup$ – Git Gud Jun 18 '14 at 18:02
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    $\begingroup$ no,no, gud help you $\endgroup$ – Will Jagy Jun 18 '14 at 18:40
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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.

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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.

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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.

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  • $\begingroup$ Alpha agrees that it does not converge $\endgroup$ – Ross Millikan Jun 18 '14 at 17:52
  • $\begingroup$ @RossMillikan, there is an easy argument, posted answer. $\endgroup$ – Will Jagy Jun 18 '14 at 17:56
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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'.

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