# Show that $X$ has no mean [duplicate]

$X$ is said to have a Cauchy distribution iff $X$ has a probability distribution of the form:

$$f(x)= \begin{cases} \frac{a}{\pi(ax^2+x^2)} & \text{,x>0} \\ \ \ \ \ 0 & \text {,otherwise} \end{cases}$$ where $a>0$

How can I show that $X$ has no mean?

## marked as duplicate by Did, user147263, Najib Idrissi, Daniel FischerOct 16 '15 at 13:22

Hint: Try to compute it: $\mathbb{E}[X]$ would be equal to $$\int_{(-\infty,\infty)} xf(x) dx$$ Does this integral exist (in a proper sense)? (i.e., is the function $x\mapsto x f(x)$ integrable on $(-\infty,\infty)$?)
In particular, look at what asymptotically $xf(x)$ becomes when $x\to \infty$:$\frac{x}{a^2+x^2} \sim_{x\to\infty} ?$