I want to find out whether this integral is convergent or not. $$\int_{\mathbb{R}^2}\frac{1}{x^4y^4+1}\ dxdy $$

I've tried to calculate it using the following variable changement, but it does'nt work i guess.$(x,y)=(r\cdot \cos(\theta),r\cdot \sin(\theta))$.

I also though of comparing the general term to another one that converge but i couldn't find.


3 Answers 3


The integral diverges.

To see this, note that the integrand is positive. Therefore, the value of the integral over $\mathbb{R}^2$ is an upper bound of the value of the integral over any subset of $\mathbb{R}^2$.

Let $S$ be the set given by $S=\{(x,y)|xy\le 1, x>0, y>0\}$. Then, we have

$$\begin{align} \iint_{\mathbb{R}^2}\frac{1}{1+x^4y^4}\,dx\,dy&\ge \iint_{S}\frac{1}{1+x^4y^4}\,dx\,dy\\\\ &\ge\frac12\iint_{S}\,dx\,dy \end{align}$$

But the area of $S$ is clearly infinite.

Therefore, we conclude the integral of interest diverges.

  • $\begingroup$ @hamzaboulahia Please let me know how I can improve my answer. I really want to give you the best answer I can. Note the answer you accepted was incorrect. $\endgroup$
    – Mark Viola
    Jul 10, 2020 at 3:31
  • $\begingroup$ @Rodrigo The integrand is bounded below by $1/2$ on a domain that has infinite area. That is how we conclude that the integral diverges. $\endgroup$
    – Mark Viola
    Feb 5 at 18:45

Another way to show divergence: The substitution $x=u/|y|$ gives

$$\int_{-\infty}^\infty\int_{-\infty}^\infty{1\over x^4y^4+1}dxdy=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty{1\over u^4+1}\cdot{du\over |y|} \right)dy=\left(\int_{-\infty}^\infty{du\over u^4+1}\right)\left(\int_{-\infty}^\infty{dy\over|y|}\right)$$

The improper integral over $u$ converges (to a nonzero value), but the one over $y$ diverges.

Remark: The use of $|y|$ instead of just $y$ in the substitution is to keep the limits of integeration running from $-\infty$ to $\infty$ for all $y$. Technically the substitution is invalid for $y=0$, but that's only a single point in the integral over $y$.

My thanks to user Simply Beautiful Art for pointing out an error in the original version of the answer.

  • $\begingroup$ Hi Barry. I hope that you're doing well and staying safe and healthy. $\endgroup$
    – Mark Viola
    Jun 20, 2020 at 19:09
  • 1
    $\begingroup$ It's not immediately clear how you managed your first step. Should it not be $$I=\int_\mathbb{R}\left(\int_\mathbb{R}\frac1{u^4+y^{-4}}\cdot\frac{\mathrm du}{y^4}\right)~\mathrm dy?$$ Or perhaps I could see you might be substituting $u=xy$, but then that gives me $$I=\int_\mathbb{R}\left(\int_\mathbb{R}\frac1{u^4+1}\cdot\frac{\mathrm du}y\right)~\mathrm dy?$$ $\endgroup$ Jun 21, 2020 at 12:02
  • $\begingroup$ @SimplyBeautifulArt, yes, you are quite right! I was thinking $x=u/y$ and somehow got $dx=du/y^4$ out of it. When I think more carefully now, it should actually be $dx=du/|y|$ to keep the $u$ limits running from $-\infty$ to $\infty$. I'll fix it a little later, when I have more time. (I was just about out the door when I saw your comment.) Thanks for bringing the error(s) to my attention! $\endgroup$ Jun 21, 2020 at 12:25
  • $\begingroup$ @SimplyBeautifulArt, fixed now. Again, thanks! $\endgroup$ Jun 21, 2020 at 14:29
  • $\begingroup$ @MarkViola, thanks for the well wishes. I am indeed doing and staying as hoped, and hope the same is true for you. $\endgroup$ Jun 21, 2020 at 14:30

This answer is wrong, as the integral diverges.

Since I missed the case of $|x|>1$ and $|y|<1$ and vice versa.


$$\int\frac1{x^4y^4+1}\ dx\ dy<\int\frac1{x^4y^4}\ dx\ dy$$

and use it to show convergence for when $|x|>1$ and $|y|>1$.

For $|\cdot|\le1$, show that it is finite.

  • $\begingroup$ Thank you very much @SimpleArt for solving my problem (y). $\endgroup$
    – Varazda
    Jan 14, 2017 at 15:36
  • $\begingroup$ @hamzaboulahia :D no problem. $\endgroup$ Jan 14, 2017 at 15:54
  • 1
    $\begingroup$ The integral diverges. Look at the region in the first quadrant bounded by $x=0$, $y=0$, and $xy=1$. In that region (of infinite area), we have $xy\le 1$ so that $\frac1{1+x^4y^4}\ge 1/2$. $\endgroup$
    – Mark Viola
    Jun 20, 2020 at 16:01
  • $\begingroup$ I posted an answer that proves that the integral diverges. The issue with the development here is that it misses the contributions from integrating in the regions $0<x<1$, $y>1$, etc. $\endgroup$
    – Mark Viola
    Jun 20, 2020 at 16:14
  • 1
    $\begingroup$ @simplybeautifulart2 You're welcome. And I hope you're staying healthy and safe. $\endgroup$
    – Mark Viola
    Jun 21, 2020 at 14:43

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