Variance of this probability density

Question

I have the function $\rho(x) = \frac{sin^2(x)}{x^2}$ and I want to calculate its variance on $\mathbb{R}$. Does anybody know how to do this? Cause afaik the integral does not converge.

Answer 1

It was clarified in a comment that $\rho(x)$ is supposed to be the density function of a random variable $X$, presumably defined over all of $\mathbb{R}$. (If it is only for positive reals, that makes little difference to the analysis.)

The density function cannot be quite $\rho(x)$, since $\int_{-\infty}^\infty \frac{\sin^2 x}{x^2}\,dx=\pi$. But if we replace $\rho(x)$ by $\frac{\rho(x)}{\pi}$, we do get a density function.

The second moment of $X$ about the origin does not exist, for $$\int_{-\infty}^\infty x^2 \frac{\sin^2 x}{\pi x^2}\,dx$$ diverges. The divergence is easy to prove.

Indeed, $E(X)$ does not exist either.