Integrate $e^{-x^2}$ from $0$ to $\infty$ and explain properties of $\mathrm{erf}(x)$

Question

I know evaluating the improper integral
$$\int_0^\infty e^{-x^2}\,dx$$ involves the use of a change to polar coordinates and the Jacobian, which provides with the $r$ variable which is all that is missing to be able to easily do the integral. This question differs (it is not a duplicate) as I also ask what exactly does the Jacobian entail (i.e. why does it work and why is it allowed). Thanks

Answer 1

This method works because of Fubini's theorem. Firstly, one can show $$ \int_0^\infty |e^{-x^2}| dx < \infty$$ so we know the integral exists. Squaring it, the integral will remain finite. But the square is also equal to $$\int_0^\infty e^{-x^2} dx \int_0^\infty e^{-y^2} dy$$ Since the integrals here are finite we're allowed combine the two integrals and change variables. The Jacobian (which is essentially the change in volume after a change of basis) then makes it easy to integrate.

If you're looking for another example of such a trick, notice the Jacobian for spherical coordinates is $\rho^2\sin \phi$. So if you cube the original integral and use Fubini's theorem you can find the value of the integral in another way.