I want to check if the following integral diverges or converges.

$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x+y)^{1000}}\ dx\,dy $

I'm not sure if it diverges or converges. I attempted to prove divergence, in two ways:

  1. Substituting $ x + y = u, x = v$ we get that the abosolute value of the Jacobian is $1, $we get that $ -\infty \le x =v \le \infty $. However, I'm not sure what are the bounadries on $u$. How do I find them? let's say $ a \le u \le b$. If one of the boundaries of $u$ is infinite, then our integral is now $ \int_{-\infty}^{\infty}\int_{a}^{b}e^{-(u)^{1000}}\ dudv, $ and it diverges. However, again, I'm not sure on the boundaries.

  2. Since $e^{-(x+y)^{1000}} \ge 0 $, we can look at a small enough interval near some $x_0$, such that $ x-y$ is close to $ 0$ - so in that interval, we can say that $ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x+y)^{1000}}\ dx\,dy \ge \int_{-\infty}^{\infty}\int_{x_0 - \delta}^{x_0 + \delta}\frac{1}{2}\ dx\,dy, $ and diverges.

So my questions are: in the first attempt, how to find the boundaries? Are my attempts okay? If not, what's wrong with them?

  • 1
    $\begingroup$ In your first approach, the limits on $u$ and $v$ are $\pm \infty$. You need to cover all the points of the plane with the values of $u$ and $v$. $\endgroup$ Commented Apr 8 at 9:15
  • 1
    $\begingroup$ In your second approach, you are actually looking at a small tubular neighborhood around the line $y=-x$, but that’s enough for divergence. $\endgroup$ Commented Apr 8 at 9:17
  • $\begingroup$ Thanks! Yes, this was my thought - there's always a $y$ for each $x$ that our function is larger than half. $\endgroup$
    – FNB
    Commented Apr 8 at 9:28
  • $\begingroup$ It was in an exam in 202-something. I figured there's nothing special about that number, so I changed it to 1000. I like that number. $\endgroup$
    – FNB
    Commented Apr 8 at 14:36
  • 2
    $\begingroup$ Change of variables $u=x+y$ and $v=x-y$ is better IMO, since it represents a rotation as much as $\pi/4$ and a scaling. $\endgroup$ Commented Apr 8 at 18:57


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