Can the variance of a continuous random variable with known distribution be impossible to find? [duplicate]

I am solving a problem where the life expectancy of a microorganism can be modeled as having the PDF:

$f(x)= \left\{ \begin{array}{ll} kx^{-3} & x\geq 1 \\ 0 & x \lt 1 \\ \end{array} \right.$

Where $k=2$. I've calculated the average to be 2 hours, but when I try to calculate the variance as: $$\sigma^2=\int_1^\infty x^2f(x)dx - \mu^2$$

The integral doesn't converge. Am I doing something wrong? Thanks in advance!

• It's absolutely possible that an expectation exists by a variance does not. Jun 4, 2017 at 22:32
• I had no idea, thanks! The integral turns out to be $log(x)$ evaluated from 1 to $\infty$, so I think it's not possible to find the variance in this case. Thanks a lot. Jun 4, 2017 at 22:37
• This is a special case of the Pareto distribution., which has density of the form $kx^{-(\alpha+1)}$ and for which only moments strictly lower than $\alpha$ are finite. Jun 5, 2017 at 0:41