[proofplan] We evaluate $I = \int_{-\infty}^{\infty} e^{-x^2/2}\, dx$ by computing $I^2$ as a double integral over $\mathbb{R}^2$, converting to polar coordinates, and performing the resulting elementary radial integral. Taking the square root (noting $I > 0$) yields $I = \sqrt{2\pi}$. [/proofplan] [step:Square the integral and write it as a double integral over $\mathbb{R}^2$] Define $I = \int_{-\infty}^{\infty} e^{-x^2/2}\, dx$. Since the integrand is positive, $I > 0$ (and $I$ is finite because $e^{-x^2/2} \le e^{-|x|/2}$ for $|x| \ge 2$, and $e^{-|x|/2}$ is integrable). We compute $I^2$: \begin{align*} I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2/2}\, dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2/2}\, dy\right) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)/2}\, dx\, dy, \end{align*} where we combine the product of one-dimensional integrals into a double integral over $\mathbb{R}^2$. This is justified by Tonelli's theorem, since the integrand is non-negative. [guided] The difficulty with $I = \int_{-\infty}^{\infty} e^{-x^2/2}\, dx$ is that $e^{-x^2/2}$ has no elementary antiderivative. No substitution or integration by parts can reduce this to a closed form in one dimension. The key insight is that while the one-dimensional integral resists direct computation, squaring it produces a two-dimensional integral with radial symmetry — and radially symmetric integrals in $\mathbb{R}^2$ are natural candidates for polar coordinates, which separate the radial and angular parts. Concretely, we write \begin{align*} I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2/2}\, dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2/2}\, dy\right). \end{align*} Since both integrands are non-negative, the product of integrals equals the double integral (by Tonelli's theorem): \begin{align*} I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)/2}\, dx\, dy. \end{align*} The exponent $x^2 + y^2$ is the squared distance from the origin — this is what makes polar coordinates the natural choice. [/guided] [/step] [step:Convert to polar coordinates and evaluate the radial integral] We substitute $x = r\cos\theta$, $y = r\sin\theta$ with $r \in [0, \infty)$ and $\theta \in [0, 2\pi)$. The Jacobian of this change of variables is $r$, so $dx\, dy = r\, dr\, d\theta$, and $x^2 + y^2 = r^2$. The double integral becomes \begin{align*} I^2 = \int_0^{2\pi}\int_0^{\infty} e^{-r^2/2}\, r\, dr\, d\theta. \end{align*} The angular integral is $\int_0^{2\pi} d\theta = 2\pi$. For the radial integral, substitute $u = r^2/2$, so $du = r\, dr$ and the limits are $u = 0$ to $u = \infty$: \begin{align*} \int_0^{\infty} e^{-r^2/2}\, r\, dr = \int_0^{\infty} e^{-u}\, du = \left[-e^{-u}\right]_0^{\infty} = 0 - (-1) = 1. \end{align*} Therefore $I^2 = 2\pi \cdot 1 = 2\pi$. [/step] [step:Take the square root to conclude] Since $I > 0$ and $I^2 = 2\pi$, \begin{align*} I = \int_{-\infty}^{\infty} e^{-x^2/2}\, dx = \sqrt{2\pi}. \end{align*} [/step]