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How to integrate $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)} dx \ dy$$

I need some help in substitution here. I'm a beginner to double integrals. I know about the substitution in single integral but not aware about double integral. Can anyone just help me in substitution part?

I thought of substituting $x^2+y^2=r^2$ but don't actually know how to implement.

I don't need the complete solution. But just need the substitution part. Please.

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    Have you learned about polar coordinates? – Umberto P. Feb 16 '22 at 16:49
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    You should use polar coordinates. But also, I hope you're aware this integral is equal to $\bigg( \int_{-\infty}^{\infty} e^{-x^2} dx \bigg)^2 $ – Dionel Jaime Feb 16 '22 at 16:52
  • The question has the same work as the answer you have accepted. Also if you go to the linked question, there is another question linked in comments that has some more answers. – Math Lover Feb 17 '22 at 01:33
  • When you say how to apply substitution method, I am not sure what you mean. $x = r \cos\theta, y = r\sin\theta$ gets you jacobian $r$. Then you change the bounds accordingly. Are you looking for something more? – Math Lover Feb 17 '22 at 01:35
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    See example $3$ in the link here: https://tutorial.math.lamar.edu/classes/calciii/changeofvariables.aspx – Math Lover Feb 17 '22 at 01:38
  • https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables – metamorphy Feb 17 '22 at 08:04

2 Answers2

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You should use the polar coordinate substitution: $x^2+y^2=r^2,$ $x=r\cos\theta,$ and $y=r\sin\theta.$

Computing the Jacobian gives $dxdy=rdrd\theta.$

Then your integral becomes $\int_0^{\infty} \int_0^{2\pi} re^{-r^2}drd\theta.$

Chickenmancer
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  • Just to be rigorous here, the OP should notice that polar coordinates are not well behaved in all of the plane, one needs to excise a half-ray through the origin. Then, use the fact that integrating over a measure zero subset gives zero. – Laz Feb 16 '22 at 17:15
  • That's an excellent point, @Laz. Thank you for adding that. – Chickenmancer Feb 16 '22 at 18:24
  • See example 2 https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant :) – Chickenmancer Feb 17 '22 at 05:29
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Your integral is $(\int_{\mathbb{R}} \exp(-x^2)dx)^2.$ You can quickly evaluate this if you substitute $t=\sqrt 2 x$ and use the fact the standard Gaussian density integrates to one:

$$\int_\mathbb{R}\frac{1}{\sqrt{2\pi}}\exp(-t^2/2)dt=1$$

Golden_Ratio
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