887. Find the spectral density of a stationary random function, given its correlation function $k_{x}(\tau)=e^{-|\tau|} \cos \tau$.
Solution. Given that $|\tau|=-\tau$ when $\tau<0$ and $|\tau|=\tau$ when $\tau>0$.

889*. Find the spectral density of a stationary function, given its correlation function $k_{x}(\tau)=$ $=D \mathrm{e}^{-\alpha |\tau|}[\cos \beta \tau + (\alpha / \beta) \sin \beta |\tau|] (\alpha>0)$.

Hint. Expand the brackets and use problem 888; express trigonometric functions in terms of exponentials using Euler's formulas.

890*. Find the spectral density of a stationary function, given its correlation function $k_{x}(\tau)=$ $=D \mathrm{e}^{-\alpha \tau^{2}}$.

Hint. Use the formula

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

complete the square in the exponent and note that the Poisson integral $\int_{-\infty}^{\infty} e^{-\left(z^{2} / 2\right)} d z=\sqrt{2 \pi}$.

891*. Find the spectral density of a stationary function, given its correlation function $\boldsymbol{k}_{x}(\tau)=$ $=D \mathrm{e}^{-\alpha |\tau|}(1+\alpha|\tau|) (\alpha>0)$.

Solution. We use the formula

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

Substituting the given correlation function, we represent the right-hand side of the equation as the sum of two integrals:

![](https://cdn.mathpix.com/cropped/2024_05_22_1d1b16277022f98e68adg-366.jpg?height=129&width=951&top_left_y=437&top_left_x=137)

Denote the first term by $I$; the derivative of this integral with respect to the parameter $\alpha$ is

$$
\frac{\partial I}{\partial \alpha}=\frac{D}{2 \pi} \int_{-\infty}^{\infty}(-|\tau|) e^{-\alpha|\tau| e^{-i \omega \tau}} d \tau
$$

Therefore,

$$
s_{x}(\omega)=I - \alpha \frac{\partial I}{\partial \alpha}
$$

Considering that (see problem 886)

$$
I=\frac{D \alpha}{\pi(\alpha^{2}+\omega^{2})}, \frac{\partial I}{\partial \alpha}=\frac{D}{\pi}\left[\frac{\omega^{2}-\alpha^{2}}{(\alpha^{2}+\omega^{2})^{2}}\right]
$$

and substituting (**) into (*), we finally obtain $s_{x}(\omega)=\frac{2 D \alpha^{3}}{\pi(\alpha^{2}+\omega^{2})^{2}}$