# Does this integral converge $\int_{\mathbb{R}^2}\frac{1}{x^4y^4+1}\ dxdy$?

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.

• Jun 20, 2020 at 15:53

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.

• @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. Jul 10, 2020 at 3:31
• @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. 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.

• Hi Barry. I hope that you're doing well and staying safe and healthy. Jun 20, 2020 at 19:09
• 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 $$I=\int_\mathbb{R}\left(\int_\mathbb{R}\frac1{u^4+1}\cdot\frac{\mathrm du}y\right)~\mathrm dy?$$ Jun 21, 2020 at 12:02
• @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! Jun 21, 2020 at 12:25
• @SimplyBeautifulArt, fixed now. Again, thanks! Jun 21, 2020 at 14:29
• @MarkViola, thanks for the well wishes. I am indeed doing and staying as hoped, and hope the same is true for you. 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.

Hint:

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

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

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

• Thank you very much @SimpleArt for solving my problem (y). Jan 14, 2017 at 15:36
• @hamzaboulahia :D no problem. Jan 14, 2017 at 15:54
• 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 $\frac1{1+x^4y^4}\ge 1/2$. Jun 20, 2020 at 16:01
• 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. Jun 20, 2020 at 16:14
• @simplybeautifulart2 You're welcome. And I hope you're staying healthy and safe. Jun 21, 2020 at 14:43