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This question already has an answer here:

$X$ 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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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 $x\to \infty$:$\frac{x}{a^2+x^2} \sim_{x\to\infty} ?$

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