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How to calculate $$\int_{\Bbb R^2}\frac{e^{-x^2}}{1+x^2+2xy+y^2}dA$$

This is an integration both involving "improper integral" and "change-of-variable" (also the rotation to errace the $xy$ term), which makes it become very difficult.. The problem is that, whenver dealing with such problem with double difficulty, how should I better begin?

  1. Choose a compact figure sequence $\{D_n\}_{n=1}^\infty$that covers whole $\Bbb R^2$, and evaluate $\int_{D_{n}}\cdots$, and at the last step evaluate the limit of it?
  2. Try to do the change-of-variable of "improper" integral?
  3. ...

Also, if the right thing to do is "1.", then there arises another question: what region $D_n$ should I choose? There are at least three choice. One is square: $[-n,n]\times[-n,n]$. The others are circles, and even ellipse...

Need help. It is very hard.

Eric
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  • do you want to prove the convergence of the integral or do you want the value ? – stity Feb 01 '18 at 15:05
  • @stity The exact value. I have no reference or books that mention how to solve this. So hard.. – Eric Feb 01 '18 at 15:06
  • It is even worse. If the integrand changes sign, first you should check that it converges absolutely. If you don't do that and apply Fubini's theorem to integrate one variable at a time, you might get a wrong result. Fortunately your integrand is nonnegative so you don't have to worry about convergence (meaning that the result is either a finite positive number or $+\infty$). – Giuseppe Negro Feb 01 '18 at 15:39
  • @GiuseppeNegro Can you suggest an answer that is more rigorous that suits the real-anaylsis course demanding? I find myself quite unaccustomed to the elemantary calculus way... like $\int_{\Bbb R^2}\cdots=(\int_{\Bbb R}\cdots)(\int_{\Bbb R}\cdots)$.. – Eric Feb 01 '18 at 15:41

2 Answers2

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$$\int_{\mathbb{R}^2}\frac{e^{-x^2}}{1+x^2+2xy+y^2}dA=$$ $$\int_{-\infty}^\infty e^{-x^2}\left (\int_{-\infty}^\infty\frac{1}{1+(x+y)^2}dy\right)dx=$$ $$\int_{-\infty}^\infty e^{-x^2}\left (\int_{-\infty}^\infty\frac{1}{1+y^2}dy\right)dx=$$ $$\int_{-\infty}^\infty e^{-x^2}\pi dx=\pi\sqrt{\pi}$$

stity
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  • How do you know we need to choose $u=x+y$? What if the denominator is, say $x^2+2xy+4y^2$? Is it observed by the diagonalization of quadratic forms? – Eric Feb 01 '18 at 15:23
  • @Eric I first spotted that it could be splitted in two parts ($e^{-x^2}$ and the fraction). Then I saw $x$ as a parameter of the inner integral which was no more than a translation and therefore doesn't change the value of the integral. – stity Feb 01 '18 at 15:26
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Let $u=x+y$ and so that $\mathrm{d}u\,\mathrm{d}x=\mathrm{d}y\,\mathrm{d}x$. Thus, $$ \begin{align} \int_{\mathbb{R}^2}\frac{e^{-x^2}}{1+x^2+2xy+y^2}\,\mathrm{d}y\,\mathrm{d}x &=\int_{\mathbb{R}^2}\frac{e^{-x^2}}{1+u^2}\,\mathrm{d}u\,\mathrm{d}x\\ &=\int_{\mathbb{R}}e^{-x^2}\,\mathrm{d}x\int_{\mathbb{R}}\frac1{1+u^2}\,\mathrm{d}u\\[3pt] &=\sqrt\pi\,\pi\\[9pt] &=\pi^{3/2} \end{align} $$

robjohn
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  • How do you know we need to choose $u=x+y$? – Eric Feb 01 '18 at 15:22
  • @Eric: it was suggested since the denominator is $1+(x+y)^2$. This is the same as what stity did in their answer. – robjohn Feb 01 '18 at 15:24
  • Is there a systematic way to know what the function should pick? For example, how if the denominator is $1+x^2+2xy+2y^2$? – Eric Feb 01 '18 at 15:25
  • @Eric: not really. After doing some problems, you start to notice things t that simplify the integral, like letting parts of the integrand (the denominator is many times a thing to try) be a single variable, but mainly, it is a matter of experience. – robjohn Feb 01 '18 at 15:28
  • Why does the nominator $e^{-x^2}$ does not need to change after change-of-variable? – Eric Feb 01 '18 at 15:35
  • We could say that $u=x+y$ and $v=x$, but I just left $x$ as $x$. In either case, the Jacobian $$\left|\det\begin{bmatrix}1&1\1&0\end{bmatrix}\right|=1$$ – robjohn Feb 01 '18 at 16:22
  • OK. I'm still not sure I know how to figure out the choice of transformation. For example, how to pick the proper transformation in $4x^2+4xy+5y^2$? Does diagonalizing it help? Because this is the most custom way to me.. But the eigenvalues seems to ugly.. – Eric Feb 01 '18 at 16:24
  • There may not be a transformation that gives you a simple answer. For example, $$\int_{\mathbb{R}^2}\frac{e^{-x^2}}{1+4x^2+4xy+5y^2},\mathrm{d}x,\mathrm{d}y=\frac{e^{\frac5{32}}\pi}4K_0\left(\frac5{32}\right)$$ according to Mathematica, where $K_0$ is a modified Bessel function of the second kind. – robjohn Feb 01 '18 at 16:38
  • Wow! Thanks for help. Actually, it is a question from a test, https://imgur.com/a/xIZhx. I extracted that term to ask you. So it is not in the denominator, my mistake. :) – Eric Feb 01 '18 at 16:45
  • @Eric: In the case there, diagonalizing the quadratic is definitely the way to go. Note that $4x^2+4xy+5y^2=(2x+y)^2+(2y)^2$. – robjohn Feb 01 '18 at 16:48
  • The eigenvalues are very ugly though. I have just post a new post on it. Can you help me there? – Eric Feb 01 '18 at 17:00