907. Find the correlation function of a stationary random function, given its spectral density $\delta_{x}(\omega)=D \alpha / \pi\left(\alpha^{2}+\omega^{2}\right)$.
Solution. First method. From problem 886, it follows that the correlation function $k_{x}(\tau)=D e^{-\alpha|\tau|}$ corresponds to the given spectral density. Since $k_{x}(\tau)$ and $s_{x}(\omega)$ are related by mutually inverse Fourier transforms, the sought correlation function is $k_{x}(\tau)=D \mathrm{e}^{-\alpha|\tau|}$.

Second method. We use the formula

$$
k_{x}(\tau)=\int_{-\infty}^{\infty} s_{x}(\omega) e^{i \tau \omega} d \omega=\frac{D \alpha}{\pi} \int_{-\infty}^{\infty} \frac{1}{\alpha^{2}+\omega^{2}} e^{i \tau \omega} d \omega
$$

It is known that

$$
\int_{-\infty}^{\infty} \frac{1}{x^{2}+\omega^{2}} e^{i \tau \omega} \omega=(\pi / \alpha) e^{-\alpha / r_{1}}
$$

Therefore, the sought correlation function is $k_{x}(\tau)=D e^{-\alpha \mid \text { r) }}$.