I need to express the following improper integral as a double integral of $x$ and $y$ and then, using polar coordinates, evaluate it.


Plotting it, we find a Gaussian centered at $x=0$ which tends to infinity to both sides. We can easily express it as a double integral :


Evaluating both using Wolfram Alpha gives $\sqrt{\pi}$, so it converges.

I know that $x=rcos(\theta)$ and that $dxdy=rdrd\theta$, but substituing this in the above integral and evaluating $\theta$ from $0$ to $2\pi$ and $r$ from $0$ to $\infty$ doesn't yield the correct answer. What's wrong here?

Thanks a lot !

  • $\begingroup$ The step you made is not really helpful, it essentially makes things more complicated. Hint: let $e^{-(x^2+y^2)}$ appear. $\endgroup$ – Yves Daoust May 29 '15 at 8:49
  • $\begingroup$ $\Bbb R\times[0,1]$ is not the whole plane, it's just a strip. But in polar coordinates, if $r$ ranges in $[0,\infty)$ and $\theta$ in $[0,2\pi)$ then you're talking about the whole plane. Don't bother integrating $\iint_{\Bbb R\times[0,1]}e^{-x^2}dxdy$. The trick here is to write $I^2$ as a double integral. $\endgroup$ – anon May 29 '15 at 8:50
  • $\begingroup$ Well, if your interval is $[0,1]\times (-\infty, \infty)$, then you cannot send $\theta$ from $0$ to $2\pi$. $\endgroup$ – 5xum May 29 '15 at 8:50
  • $\begingroup$ Nominated for reopening because the question asked why this particular approach did not work. The answers so far (here and in the linked question) show other methods that do work--highly relevant but not an answer to the question asked. A couple of the comments above do address that question; they could be posted as answers. $\endgroup$ – David K May 29 '15 at 11:41

You could try: $$I^2=\left ( \int_{-\infty}^{\infty}e^{-x^2}\mathrm{d}x \right ) \left( \int_{-\infty}^{\infty}e^{-y^2}\mathrm{d}y \right) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\mathrm{d}x\ \mathrm{d}y$$ then use the polar coordinates to compute the double integral.



Compute $\int_{\mathbb R^2}e^{-x^2-y^2}d_{\lambda_2}(x,y)$ and use Fubini.


Let I denote the integral. Then Write I first in terms of x and then in terms of y. Then $I^2=4$.double integral $(e^{-(r^2)})$rdr d$\theta$ the limit of r is from 0 to $\infty$ and that of $\theta$ are from $0$ to $\frac{\pi}{2}$. The integral is easy to evaluate by separation. This is well known.


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