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I got this integral: $\int_{\Bbb R ^2} e^{-x^2-y^2} \, dx \, dy$

The first thing that came to my mind was the Fubini theorem. But then I tried to calculate it via substitution theorem because of that $e^{v^2}$. But I am not sure about the boundaries.

As the substitution I used the polar coordinates: $x = r\cos\phi,$ $y = r\sin\phi$

$$\int_a^b \int_0^\infty e^{-r^2} r\,dr\,d\phi$$

Then I used the classic substitution: $u = r^2, dr=\frac{du}{2r}$

EDIT

So $\displaystyle \frac{1}{2} \int_0^{2\pi} \left(\int_0^{\infty}e^{-u} \, du\right) \, d\phi = \frac{1}{2} \int_0^{2\pi} \Big[ -e^{-u}\Big]_0^{\infty} \, d\phi = \frac{1}{2} \int_0^{2\pi} \, d\phi=\pi$

I am not sure what I am doing, this is new for me so I will be happy for any hint.

Michael Hardy
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Leif
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    What is $;a,b;$ ? I think it should be $;0,,2\pi;$ ....and always $;r\ge0;$ ... – DonAntonio Nov 18 '17 at 23:01
  • You are integrating a nonnegative function and getting zero? This implies that the function is zero almost everywhere, which is clearly not the case. – Xander Henderson Nov 18 '17 at 23:01
  • Think again about your polar coordinates. There is an obvious mistake you made at the change of variables step. – kimchi lover Nov 18 '17 at 23:03
  • Have a look at the answer here https://math.stackexchange.com/questions/2514319/how-to-prove-that-int-0-infty-e1-2-x-2-sqrtx-dx-sqrt2-e-pi/2514356#2514356 – Donald Splutterwit Nov 18 '17 at 23:03

2 Answers2

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Check your integrals becomes

$$\int_0^{2\pi}\int_0^\infty re^{-r^2}\,dr\,d\theta=-\pi\int_0^\infty(-2r)e^{-r^2}\,dr=\ldots$$

Michael Hardy
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DonAntonio
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Your integral is equal to $$ \left(\int_{\infty}^\infty\exp(-x^2)\, dx\right)^2=(\sqrt {\pi})^2=\pi $$ by considering the density of a standard normal. This approach assumes that the result that the standard density integrates to one is computed using a different approach than using this double integral. $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp(-x^2/2)\, dx=1\implies \sqrt {\pi}=\int_{-\infty}^\infty\exp(-x^2/2)\,dx $$

Sri-Amirthan Theivendran
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