 Solution

The second moment is $E[X^2]=\alpha {\int }_1^{\infty }{x}^{1-\alpha }dx$, finite exactly for $\alpha >2$, where it equals $\alpha /(\alpha -2)$. Subtracting the square of the mean givesFor $1<\alpha \le 2$, the mean exists but the variance is infinite. For $0<\alpha \le 1$, the ordinary centered variance is not defined because the mean is not finite. Thus there is no finite variance for any $\alpha \le 2$.