Expected Value Diverges? Given the probability density function $f(x) = \frac{1}{x^2}$ for $x > 1$. The expected value of this function is: $$E[X] = \int_1^\infty \frac{1}{x} dx = \infty$$
Can someone please explain this? Is $E[X]$ only finite under certain condition? Is $\infty$ really the expected value for this probability density function?
 A: This is an example of a Pareto distribution which typically has a density function of the form $$f(x) = \dfrac{\alpha x_{m}^{\alpha}}{x^{\alpha+1}} \text{ for }x \gt x_m $$ and so a cumulative distribution function of
$$F(x) = 1-\left(\dfrac{x_{m}}{x}\right)^{\alpha} \text{ for }x \gt x_m $$ where $x_m \gt 0$ is lowest value of the support (a location parameter, $x_m=1$ in your question) and $\alpha \gt 0$ is a shape parameter ($\alpha =1$ in your question).  
Trying to derive the mean would give $\dfrac{\alpha}{\alpha-1} x_m$, but this is only meaningful when $\alpha \gt 1$. For $0 \lt \alpha \le 1$ the mean is infinite, as you found with your example.
Something similar happens with the variance: you can derive $\dfrac{\alpha^2}{(\alpha-1)^2(\alpha-2)}x_m^2$ but this is only valid for $\alpha \gt 2$.  For $1 \lt \alpha \le 2$ the variance is infinite, while for $0 \lt \alpha \le 1$ the variance is undefined (since there is no finite mean) and the second moment is infinite.  
The Pareto distribution is a fat-tailed distribution. The smaller $\alpha$ is, the fatter the tail becomes.  With $\alpha = 1$ the tail is sufficiently fat to prevent the distribution having a finite mean.
It is still possible to descibe a centre of the distribution: for example in general the median of a Pareto distribution is $2^{1/\alpha} x_m$ and in your particular example the median is $2$. 
