Problem: Find the value of the integral
\[
\frac{1}{\sqrt{2\pi } } \int _0^{\infty} e^{-x^2 / 8} \mathrm{d} x .
\]
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,
\[
\textcolor{blue}{
\boxed{
\frac{1}{\sqrt{2\pi } } \int _0^{\infty} e^{-x^2 / 8} \mathrm{d} x = 1.
}
}
\]