Example 1. The density function of a random variable $X$ is given by

106

$$
p(x)=\frac{c}{1+x^{2}}
$$

Find the value of the parameter $c$.
Solution. The density function must satisfy condition $(2.3 .6)$, i.e., the equality

$$
\int_{-\infty}^{+\infty} \frac{c}{1+x^{2}} d x=c \int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}=1
$$

must hold, from which

$$
c=1 / \int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}
$$

The indefinite integral is a standard one:

$$
\int \frac{d x}{1+x^{2}}=\operatorname{arctg} x
$$

Let's compute the improper integral:

$$
\begin{aligned}
& \int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}=\int_{-\infty}^{0} \frac{d x}{1+x^{2}}+\int_{0}^{+\infty} \frac{d x}{1+x^{2}}=\left.\operatorname{arctg} x\right|_{-\infty} ^{0}+\left.\operatorname{arctg} x\right|_{0} ^{+\infty}= \\
& =\operatorname{arctg} 0-\operatorname{arctg}(-\infty)+\operatorname{arctg}(+\infty)-\operatorname{arctg} 0=0-\left(-\frac{\pi}{2}\right)+\frac{\pi}{2}-0=\pi
\end{aligned}
$$

Therefore, $c=1 / \pi$; the density function is

$$
p(x)=\frac{1}{\pi} \cdot \frac{1}{1+x^{2}}, \quad p(x)=\frac{1}{\pi\left(1+x^{2}\right)}
$$