The Gaussian Integral

Here I present the classic and very elegant proof of the value of Gaussian integral: \begin{align} I = \int\limits_{\mathbb{R}} e^{-x^2}\,\mathrm{d}x = \sqrt{\pi}\,. \end{align} The starting point is writing the square of the integral, making a change of variables, and rearranging: \begin{align} I^2 &= \left(\int\limits_{\mathbb{R}} e^{-x^2}\,\mathrm{d}x \right)^2 \\ &= \int\limits_{\mathbb{R}} e^{-x^2}\,\mathrm{d}x \cdot \int\limits_{\mathbb{R}} e^{-y^2}\,\mathrm{d}y\\ &= \iint\limits_{\mathbb{R}^2} e^{-x^2}\cdot e^{-y^2} \,\mathrm{d}x \,\mathrm{d}y \\ &= \iint\limits_{\mathbb{R}^2} e^{-\left(x^2 +y^2\right)} \,\mathrm{d}x \,\mathrm{d}y\,. \end{align} The integral can be done by shifting to polar coordinates with $r^2= x^2 + y^2$. The area element is $\mathrm{d}x\,\mathrm{d}y = r\,\mathrm{d}r\,\mathrm{d}\theta$. The squared integral is then \begin{align} I^2 &= \int\limits_{0}^{2\pi}\left.\int\limits_{0}^{\infty} \left.e^{-r^2} r \,\right.\mathrm{d} r\right.\,\mathrm{d}\theta\\ &= 2\pi\int\limits_0^{\infty} e^{-r^2}\,r\,\mathrm{d}r\\ &= 2\pi\cdot \left[-\frac{1}{2}e^{-r^2}\right]_0^{\infty}\\ &= \pi\,. \end{align} It follows immediately that $I=\sqrt{\pi}$.