I came across a question asking to prove $P(\max_{[0,h]}|W(s)+\mu s|\geq x)=o(h)$. The question is easy, but I want to find the value. The way I choose is by Girsanov's theorem $$P(\max_{[0,h]}|W(s)+\mu s|\geq x)=E(1_{(\max_{[0,h]}|W(s)|\geq x)} e^{\mu B(h)-\frac{1}{2}\mu^2 h}) \\ =\int_{y=-\infty}^\infty e^{\mu y-\frac{1}{2}\mu^2 h} dP(\max_{[0,h]}|W(s)|\geq x, B(h)\leq y) $$
For $|y|\geq x$, the probability can be found simply by reflection. But the $|y|\leq x$, the probability seems quite hard to evaluate, to begin with, I choose to divide it. $$P(\max_{[0,h]}|W(s)|\geq x, B(h)\leq y) \\ =P(\max_{[0,h]}W(s)\geq x, B(h)\leq y)+P(\max_{[0,h]}-W(s)\geq x, B(h)\leq y)- P(\max_{[0,h]}W(s)\geq x, \max_{[0,h]}-W(s)\geq x, B(h)\leq y) $$ The first two term can still be found by reflection, while the third term can't. The reason for not applying reflection method is that I don't know which of the two cases $\max_{[0,h]}W(s)\geq x, \max_{[0,h]}-W(s)\geq x$ happens first.
Actually evaluating $\frac{d}{dy}P(\max_{[0,h]}|W(s)|\geq x, B(h)\leq y)$ is enough for the calculate to continue. Thanks in advance.