1. The problem is to evaluate the integral $$\int_{-\infty}^{\infty} e^{-x^2} \, dx$$. 2. This is a classic Gaussian integral. The formula for the Gaussian integral is: $$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$$. 3. Important rule: The integral converges because the function $e^{-x^2}$ rapidly approaches zero as $x$ goes to $\pm \infty$. 4. To understand why, consider the square of the integral: $$I = \int_{-\infty}^{\infty} e^{-x^2} \, dx$$ $$I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \, dx\right) \left(\int_{-\infty}^{\infty} e^{-y^2} \, dy\right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy$$ 5. Change to polar coordinates where $x = r \cos \theta$, $y = r \sin \theta$, and $dx \, dy = r \, dr \, d\theta$: $$I^2 = \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r \, dr \, d\theta$$ 6. Evaluate the inner integral: $$\int_0^{\infty} r e^{-r^2} \, dr$$ Let $u = r^2$, then $du = 2r \, dr$, so $r \, dr = \frac{du}{2}$: $$= \frac{1}{2} \int_0^{\infty} e^{-u} \, du = \frac{1}{2} [ -e^{-u} ]_0^{\infty} = \frac{1}{2} (0 + 1) = \frac{1}{2}$$ 7. Now integrate over $\theta$: $$I^2 = \int_0^{2\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \times 2\pi = \pi$$ 8. Taking the square root: $$I = \sqrt{\pi}$$ Final answer: $$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$$