This is an exercise from Stochastic Calculus and Financial Applications by Michael Steele (Exercise 13.1). The goal is to use Girsanov's theorem to show that $$ \mathbb{E}\left[e^{-\mu B_T}\max_{t\in[0,T]} B_t\right]\sim e^{\mu^2 T/2}/2\mu\quad\text{and}\quad \mathbb{E}\left[e^{\mu B_T}\max_{t\in[0,T]} B_t\right]\sim \mu Te^{\mu^2 T/2}, $$ as $T\to\infty$, where $\mu>0$.
Here's my attempt for the first relation. Denoting by $\mathbb{P}$ the probability measure of the standard Brownian motion $\{B_t\}$, we define $W_t=B_t-\mu t$ and $$ d\mathbb{Q} = \exp\left(-\mu B_T-\frac{1}{2}\mu^2 T\right)\,d\mathbb{P}. $$ Then, $\{W_t\}$ is a Brownian motion with respect to $d\mathbb{Q}$, by Girsanov's theorem, and we can write $$ \mathbb{E}^{\mathbb{P}}\left[e^{-\mu B_T}\max_{t\in[0,T]} B_t\right] = e^{\frac{1}{2}\mu^2T}\mathbb{E}^{\mathbb{Q}}\left[\max_{t\in[0,T]} (W_t+\mu t)\right]. $$ This is where I'm stuck; I don't know how to deduce the asymptotic relation from the right hand side. I'd appreciate any help. Thank you!