1. **State the problem:** Calculate the integral $$\int_{-\infty}^{\infty} e^{-x^2} \, dx$$. 2. **Formula and important rules:** This is a Gaussian integral, a well-known integral in calculus and probability. 3. **Key fact:** The integral $$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$$. 4. **Explanation:** The function $$e^{-x^2}$$ is symmetric and rapidly decays to zero as $$x \to \pm \infty$$, making the integral convergent. 5. **Intermediate work:** The integral cannot be solved by elementary antiderivatives, but by squaring the integral and converting to polar coordinates, one obtains: $$I = \int_{-\infty}^{\infty} e^{-x^2} \, dx$$ $$I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \, dx\right) \left(\int_{-\infty}^{\infty} e^{-y^2} \, dy\right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy$$ Switching to polar coordinates $$x = r\cos\theta$$, $$y = r\sin\theta$$, the double integral becomes: $$I^2 = \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r \, dr \, d\theta$$ 6. **Evaluate the inner integral:** $$\int_0^{\infty} e^{-r^2} r \, dr$$ Let $$u = r^2$$, then $$du = 2r \, dr$$ or $$r \, dr = \frac{du}{2}$$. So, $$\int_0^{\infty} e^{-r^2} r \, dr = \frac{1}{2} \int_0^{\infty} e^{-u} \, du = \frac{1}{2} [ -e^{-u} ]_0^{\infty} = \frac{1}{2} (0 + 1) = \frac{1}{2}$$ 7. **Evaluate the outer integral:** $$\int_0^{2\pi} d\theta = 2\pi$$ 8. **Combine results:** $$I^2 = 2\pi \times \frac{1}{2} = \pi$$ 9. **Final answer:** $$I = \sqrt{\pi}$$ Thus, the value of the integral $$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$$.