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 $$\operatorname{Var}(X)$$?

I think $$Var(X)$$ exists and $$Var(X)=+\infty$$
since $$E(X)=4$$ ,so if $$E(X^2)$$ exists , in hence $$Var(X)$$ exists.
$$E(Y)$$ exists if $$E(Y^+)<\infty$$ or $$E(Y^-)<\infty$$. on the other hands if both $$E(Y^+)=\infty$$ , $$E(Y^-)=\infty$$ so $$E(Y)$$ does not exist.
meaning-of-non-existence-of-expectation $$E((X^2)^{+})=E(X^2)=\infty$$ $$E((X^2)^{-})=E(\max(0,-X^2))=E(0)=0<\infty$$
so $$E(X^2)$$ exists and $$E(X^2)=\infty$$
This distribution is Pareto distribution that can see for $$x_m=\alpha =2$$ wikipedia say Variance $$=\infty$$ Pareto_distribution