From Page 177 of differential equations demystified (2005):
We must evaluate the integral I: (1)
$$ I = \int_{\infty}^{0} e^{- s^{2}} $$
observe that: (2)
$$I \cdot I = \int_{0}^{\infty} e^{- s^{2}}\, ds \cdot \int_{0}^{\infty} e^{- u^{2}}\, du = \int_{0}^{\infty} \int_{0}^{\pi / 2} e^{- r^{2}} r d\theta dr $$
Could somebody explain these steps in better detail. I'm not really sure how they get from (1) to the double integral in polar coordinates in (2)
Note that $$I = \int_{\infty}^{0} e^{- s^{2}} = \int_{\infty}^{0} e^{- u^{2}}$$ because $s$ is a dummy variable so we can as well denote it by $u$
Multiply the two integrals to get $$I \cdot I = \int_{0}^{\infty} e^{- s^{2}}\, ds \cdot \int_{0}^{\infty} e^{- u^{2}}\, du=\int_{0}^{\infty}\int_{0}^{\infty}e^{- u^{2}-s^2}dsdu $$
Use polar coordinates with the region of integration being the first quadrant.
Note that in polar coordinates element of the area is $rdrd\theta$ instead of $dsdu$
Note that $$r^2 = s^2 + u^2$$ $$= \int_{0}^{\infty} \int_{0}^{\pi / 2} e^{- r^{2}} r d\theta dr$$
