expectation of absolute value of cauchy distribution & general questions about expectation of absolute value of distributions The Cauchy distribution has PDF:
$$f_X(x) = \frac{1}{\pi(1 + x^2)}$$
Its expectation does not exist.
If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:
$$E(|X|) = \int_{-\infty}^{\infty} |x| \cdot \frac{1}{\pi(1 + x^2)} dx = \int_{-\infty}^{0} -x \cdot \frac{1}{\pi(1 + x^2)} dx + \int_{0}^{\infty} x \cdot \frac{1}{\pi(1 + x^2)} dx$$
If I'm not mistaken, neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?
Furthermore, is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist), or is it specific to just this distribution?
And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?
 A: 
If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:

Yes.

Neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?

Yes. (Or, to be more precise, it's $\infty$.)

Is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist)?
And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?

For any random variable, having $\mathbb E[|X|] < \infty$ and having $\mathbb E[X]$ exist as a finite number are indeed equivalent. Proof for continuous variables (discrete variables are similar): Note that
$$\mathbb E[X] = \int_{-\infty}^{\infty} x \cdot f(x) \, \textrm{d} x$$
where $f(x)$ is the associated density function. We can always rewrite this as
$$\mathbb E[X] = \int_{-\infty}^0 x \cdot f(x) \, \textrm d x + \int_0^{\infty} x \cdot f(x) \, \textrm d x.$$
Note that the left integrand is purely negative and the right integrand is purely positive. The rightmost integral therefore either converges to a finite positive number, or it diverges to $\infty$; similarly, the leftmost integral either converges to a finite negative number, or it diverges to $-\infty$. The only case in which the total expectation exists (meaning yields a finite number) is if both the integrals converge to finite numbers, in which case it is not hard to see why
$$\mathbb E[|X|] = \int_{-\infty}^0 |x| \cdot f(x) \, \textrm d x + \int_0^{\infty} |x| \cdot f(x) \, \textrm d x$$
must exist as well. The logic works just as well in reverse to show that $\mathbb E[|X|] < \infty \implies \mathbb E[X] < \infty$.
