You have provided an equation and a task related to finding the characteristic function.

The equation for $f_X(x)$ is given as:

$f_X(x) = \frac{1}{\pi(1+x^2)}, \quad -\infty < x < \infty$

This is the probability density function (PDF) of the Cauchy distribution.

The task is to calculate the characteristic function $\phi_X(\omega)$ of the random variable $X$ with this PDF. The characteristic function is given by the expectation:

$\phi_X(\omega) = \mathbb{E}\left[e^{-i\omega X}\right] = \int_{-\infty}^{\infty} e^{-i\omega x} f_X(x) \, dx$

Step-by-step approach:

1. The PDF of the Cauchy distribution is $f_X(x) = \frac{1}{\pi(1+x^2)}$.

2. The characteristic function is the Fourier transform of the PDF: $\phi_X(\omega) = \int_{-\infty}^{\infty} \frac{e^{-i\omega x}}{\pi(1+x^2)} \, dx$

3. This integral is a known result for the Cauchy distribution, and the characteristic function of a standard Cauchy random variable is:

   $\phi_X(\omega) = e^{-|\omega|}$

Thus, the characteristic function of $X$ is $\phi_X(\omega) = e^{-|\omega|}$.

Do you want further details or have any questions?

Here are 5 related questions that expand on the topic:

1. What are the properties of the Cauchy distribution?
2. How does the characteristic function relate to moments of a distribution?
3. Why does the Cauchy distribution not have a finite mean or variance?
4. What is the difference between the Cauchy distribution and other symmetric distributions like the Gaussian?
5. How can we use the characteristic function to derive the PDF of the distribution?

Tip: The characteristic function provides useful insights into the distribution without the need for explicit moments, especially for distributions like the Cauchy that do not have finite moments.