I am dealing with the following function which involves a function multiplied by its logarithm, i.e., \begin{equation} g(y) := \lvert p(y)\log p(y)\rvert, \forall y\in \mathbb{R} \end{equation}
I know that $p(y) = O(\lvert y\rvert^{-\frac{3}{2}})$, which implies that for every $\epsilon > 0$, $\exists\, \lvert y_{\epsilon}\rvert$ such that $p(y)\leq \epsilon$ for all $\lvert y\rvert > \lvert y_{\epsilon}\rvert$. Now, I want to know the growth rate of $g(y)$. My intuition tells me that $g(y) = O(\lvert y\rvert^{-r})$, where $r>1$. However, I am not able to show this rigorously and I am wondering whether my intuition is correct or not?
$$g(y) = O(|y|^{-3/2}\log|y|^{-3/2}) = O(|y|^{-3/2}\log|y|) = O(|y|^{-3/2+\epsilon})$$ for any $\epsilon > 0$. This follows by the power rule for logarithms, and the fact that $\log x = O(x^{\epsilon})$ for any $\epsilon > 0$.