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Does $\Bbb{E}(X^2)$ DNE $\Rightarrow \operatorname{Var}(X)$ DNE?

Suppose you have pdf $$f(x) = \begin{cases} \frac{8}{x^3} &, \text{ if $x\ge 2$} \\ 0 &, \text{ otherwise} \end{cases}$$

I have found that $\Bbb{E}(X)=4$ and am trying to find $\operatorname{Var}(X)$ using $\Bbb{E}(X^2)-(\Bbb{E}(X))^2$.

To find $\Bbb{E}(X^2)$, I’ve been using
$$\int_{-\infty}^\infty u^2 f(u) du = \int_2^\infty \frac{8}{u} du = \lim_{t\to \infty}(8\ln t – 8\ln2)$$

However, $\lim_{t\to \infty}(\ln t)$ DNE, so does that mean that neither does $\operatorname{Var}(X)$?

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