gaussian integrals

\[\begin{align} \left(\int_{-\infty}^{\infty} e^{-z^2}\right)^2 dz &= \int_{-\infty}^{\infty} e^{-x^2} dx \int_{-\infty}^{\infty} e^{-y^2} dy \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2} dxdy \\ &= \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r dr d\theta \\ &= \int_{0}^{2\pi} \int_{0}^{\infty} e^{-u} \frac{1}{2} du d\theta \\ &= \int_{0}^{2\pi} \left[ \frac{-1}{2} e^{-u} \right]_{0}^{\infty} d\theta\\ &= \int_{0}^{2\pi} \frac{1}{2} \left[ 0 + 1 \right] d\theta\\ &= \int_{0}^{2\pi} \frac{1}{2} d\theta \\ &= \pi \end{align}\]

So, \(\int_{-\infty}^{\infty} e^{-z^2} dz = \sqrt{\pi}\)