I'm looking back at a past qualifying exam.
Let $X$ be a Pareto distributed random variable with the density $$f(x \mid \alpha, \beta) = \beta\alpha^{\beta}x^{-(\beta+1)}I(x > \alpha)$$ where $\alpha, \beta > 0$ are parameters.
Let $Y = \log(X)$. Find the mean and variance of $Y$.
Explaining how to do this question is quite simple: we have $$f_{Y}(y) = \beta\alpha^{\beta}(e^{y})^{-\beta+1}e^{y}I(y>\ln(\alpha))=\beta\alpha^{\beta}e^{-\beta y}I(y>\ln(\alpha))\text{.}$$ Hence $$\mathbb{E}[Y] = \beta\alpha^{\beta}\int_{\ln(\alpha)}^{\infty}ye^{-\beta y}\text{ d}y$$ where $$\begin{align} \int_{\ln(\alpha)}^{\infty}ye^{-\beta y}\text{ d}y &= \left.y\left(\dfrac{-1}{\beta}\right)e^{-\beta y}\right|_{\ln(\alpha)}^{\infty}-\int_{\ln(\alpha)}^{\infty}\left(\dfrac{-1}{\beta}\right)e^{-\beta y}\text{ d}y \\ &= \dfrac{\ln(\alpha)}{\beta}\alpha^{-\beta}+e^{-\beta\ln(\alpha)} \\ &= \dfrac{\ln(\alpha)}{\beta}\alpha^{-\beta}+\alpha^{-\beta} \\ &= \alpha^{-\beta}\left(1+\dfrac{\ln(\alpha)}{\beta}\right) \end{align}$$ where I used the shortcut $$\int_{x}^{\infty}\beta e^{-\beta y}\text{ d}y = e^{\beta x}$$ for $x > 0$, due to this being the survival function of an exponential distribution, and then $$\mathbb{E}[Y] = \beta^{-1}+\ln(\alpha)\text{.}$$ However, $\mathbb{E}[Y^2]$ seems to be much more difficult to find - we have two iterations of integration by parts, and then we still have to square $\mathbb{E}[Y]$ - which unfortunately is an additive term, so then we have to expand that, take $\mathbb{E}[Y^2] - (\mathbb{E}[Y])^2$, etc.
This seems to be too long of an approach for this problem for a timed exam. Are there any "probabilistic" (density function, expected values of well-known continuous distributions, etc.) shortcuts for this problem?
