$\int_{\Bbb R ^2} e^{-x^2-y^2} \, dx \, dy$

Question

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.

Answer 1

Check your integrals becomes

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

Answer 2

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. $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp(-x^2/2)\, dx=1\implies \sqrt {\pi}=\int_{-\infty}^\infty\exp(-x^2/2)\,dx $$