Evaluate the Gaussian integrals \[ \int_{-\infty}^{\infty}e^{-(x-\mu)^2/(2\sigma^2)}\,dx, \qquad \sigma\ne0, \] \[ \int_{-\infty}^{\infty}e^{-x^2}\,dx, \qquad \int_{-\infty}^{\infty}e^{-x^2/2}\,dx, \] and \[ \int_{-\infty}^{\infty}e^{-x^2+6x-3}\,dx. \]
Solution
The standard Gaussian integral is \[ \int_{-\infty}^{\infty}e^{-u^2}\,du=\sqrt\pi. \] For the shifted-scaled Gaussian, put \(u=(x-\mu)/(\sqrt2\,|\sigma|)\): \[ \boxed{\int_{-\infty}^{\infty}e^{-(x-\mu)^2/(2\sigma^2)}\,dx=\sqrt{2\pi}\,|\sigma|.} \] Thus \[ \boxed{\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt\pi}, \qquad \boxed{\int_{-\infty}^{\infty}e^{-x^2/2}\,dx=\sqrt{2\pi}}. \] Finally, \[ -x^2+6x-3=-(x-3)^2+6, \] so \[ \boxed{\int_{-\infty}^{\infty}e^{-x^2+6x-3}\,dx=e^6\sqrt\pi.} \]