Example 6. Find the distribution function $F(x)$ of a random variable $X$ that is distributed according to the normal law.
Solution. According to formula (2.3.2) and the definition of the normal distribution (3.5.1), we obtain

$$
F(x)=\int_{-\infty}^{x} \frac{1}{\sigma \sqrt{2 \pi}} e^{-(t-a)^{2} / 2 \sigma^{2}} d t=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{x} e^{-(t-a)^{2} / 2 \sigma^{2}} d t
$$

We switch to the new variable $u=\frac{t-a}{\sigma}$, then $t=a+u \sigma, d t=\sigma d u$.

Therefore,

$$
F(x)=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{\frac{x-a}{\sigma}} e^{-u^{2} / 2} \sigma d u=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\frac{x-a}{\sigma}} e^{-u^{2} / 2} d u=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{0} e^{-u^{2} / 2} d u+
$$

$$
\begin{gathered}
+\frac{1}{\sqrt{2 \pi}} \int_{0}^{\frac{x-a}{\sigma}} e^{-u^{2} / 2} d u=\frac{1}{\sqrt{2 \pi}} \cdot \frac{\sqrt{2 \pi}}{2}+\frac{1}{\sqrt{2 \pi}} \int_{0}^{\frac{x-a}{\sigma}} e^{-u^{2} / 2} d u=\frac{1}{2}+\Phi\left(\frac{x-a}{\sigma}\right) \\
F(x)=\frac{1}{2}+\Phi\left(\frac{x-a}{\sigma}\right)
\end{gathered}
$$

where $\Phi(x)$ is the Laplace function, defined by formula (3.5.3).

Remark. Here it is taken into account that

$$
\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{0} e^{-u^{2} / 2} d u=\frac{\sqrt{2 \pi}}{2}
$$

This equality follows from equality (3.5.10). Indeed, $e^{-u^{2} / 2}$ is an even function, therefore

$$
\int_{-\infty}^{+\infty} e^{-u^{2} / 2} d u=2 \int_{-\infty}^{0} e^{-u^{2} / 2} d u, \text { but } \int_{-\infty}^{+\infty} e^{-u^{2} / 2} d u=\sqrt{2 \pi}
$$