An integration both involves improper and change-of-variable $\int_{\Bbb R^2}\frac{e^{-x^2}}{1+x^2+2xy+y^2}dA$

Question

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.

Answer 1

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

Answer 2

Let $u=x+y$ and so that $\mathrm{d}u\,\mathrm{d}x=\mathrm{d}y\,\mathrm{d}x$. Thus, $$ \begin{align} \int_{\mathbb{R}^2}\frac{e^{-x^2}}{1+x^2+2xy+y^2}\,\mathrm{d}y\,\mathrm{d}x &=\int_{\mathbb{R}^2}\frac{e^{-x^2}}{1+u^2}\,\mathrm{d}u\,\mathrm{d}x\\ &=\int_{\mathbb{R}}e^{-x^2}\,\mathrm{d}x\int_{\mathbb{R}}\frac1{1+u^2}\,\mathrm{d}u\\[3pt] &=\sqrt\pi\,\pi\\[9pt] &=\pi^{3/2} \end{align} $$