Solution: At first we recall the Gaussian integral \[ \int _{-\infty }^{\infty} e^{-x^2} \mathrm{d} x = \sqrt{\pi }. \] We can compute the Gaussian integral by using the double integral as follows: \[ \left( \int_{-\infty}^{\infty} e^{-x^2} \mathrm{d}x \right)^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} \mathrm{d}x \mathrm{d}y. \] Switching to polar coordinates, where \( x = r \cos \theta \) and \( y = r \sin \theta \), we get \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} \mathrm{d}x \mathrm{d}y = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r \mathrm{d}r \mathrm{d}\theta. \] The integral simplifies to \[ \int_{0}^{2\pi} \mathrm{d}\theta \int_{0}^{\infty} r e^{-r^2} \mathrm{d}r. \] Evaluating the inner integral, we use the substitution \( u = r^2 \), \( \mathrm{d}u = 2r \mathrm{d}r \), giving \[ \int_{0}^{\infty} r e^{-r^2} \mathrm{d}r = \frac{1}{2} \int_{0}^{\infty} e^{-u} \mathrm{d}u = \frac{1}{2}. \] Thus, the double integral becomes \[ \int_{0}^{2\pi} \mathrm{d}\theta \cdot \frac{1}{2} = \pi. \] Therefore, \[ \left( \int_{-\infty}^{\infty} e^{-x^2} \mathrm{d}x \right)^2 = \pi \implies \int_{-\infty}^{\infty} e^{-x^2} \mathrm{d}x = \sqrt{\pi}. \] This also implies (as $e^{-x^2} $ is an even function ) \[ \int_{0}^{\infty} e^{-x^2} \mathrm{d}x = \frac{\sqrt{\pi}}{2}. \]

Now, we need to compute the given integral. Note that \[ \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{-x^2 / 8} \mathrm{d}x = \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{-(x/\sqrt{8})^2} \mathrm{d}x. \] Let \( u = \frac{x}{\sqrt{8}} \), then \( \mathrm{d}x = \sqrt{8} \mathrm{d}u \), and the integral becomes \begin{align*} \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{-u^2} \sqrt{8} \mathrm{d}u & = \frac{\sqrt{8}}{\sqrt{2\pi}} \int_{0}^{\infty} e^{-u^2} \mathrm{d}u \\[1ex] & = \frac{2}{\sqrt{\pi } } \left( \frac{\sqrt{\pi } }{2} \right) \\[1ex] & = 1. \end{align*} Thus, \[ \textcolor{blue}{ \boxed{ \frac{1}{\sqrt{2\pi } } \int _0^{\infty} e^{-x^2 / 8} \mathrm{d} x = 1. } } \]