I have an asymptotic expression of the following form: $$U(r) = \rho r^{-1/2}\exp\left(\frac{ibr^2}{4}\right)\exp\left(\frac{-i\ln(r)}{b}\right)(1+\mathcal{O}(b^{-3}r^{-2})).$$
I want to get an asymptotic expression for $U(b^{-2}).$ The paper where this expression is mentioned states that for small enough $b$ we can write the following: $$U(b^{-2}) = \rho b\exp\left(\frac{i}{4b^3}\right)\exp\left(\frac{-2i\ln(b)}{b}\right)(1+\mathcal{O}(b|\ln(b)|)).$$
I do not understand why we have $\mathcal{O}(b|\ln(b)|)$ instead of $\mathcal{O}(b).$ Or do we just use the fact that for small $b$ we have that $$\mathcal{O}(b) = \mathcal{O}(b|\ln(b)|)?$$
I think the latter is the case. Judging from your expressions, it suffices to prove $$ \mathcal O(b) = \mathcal O(b\lvert \ln(b) \rvert) \text{ as }b \downarrow 0. $$ If $f(b) = \mathcal O(b)$ then there are constants $\delta$ and $C$ with $$0 < b < \delta \implies \frac{\lvert f(b) \rvert}{b} < C.$$ Notice that for $0 < b < 1/e$, we have $\tfrac{1}{\lvert \ln(b) \rvert} < 1$. Hence we get $$0 < b < \min\{1/e, \delta\} \implies \frac{\lvert f(b) \rvert}{b\lvert \ln(b) \rvert} = \frac{\lvert f(b) \rvert}{b}\frac{1}{\lvert \ln(b) \rvert} < C,$$ i.e. $f(b) = \mathcal O(b\lvert \ln(b) \rvert)$.