Let $B$ be a Brownian motion and for any $a$ define $T:=\inf\{t>0: W_t\geq a\}$. I want to show that $\Bbb{E}(\exp(-\mu T))=\exp(-a\sqrt{2\mu})$.
My idea was to use the optional stopping theorem.
Let $M$ be a uniformly integrable or nonnegative right continuous martingale. Then for all stopping times $\rho, \tau$ s.t. $\rho\leq \tau$ a.s, then $M_\rho=\Bbb{E}(M_\tau|\mathcal{F}_\rho)$.
I know that $M_t:=\exp\left(\lambda B_t-\frac{\lambda^2}{2}t\right)$ is a martingale for all $\lambda$, in particular it is non negative and right continuous. Since $T$ is a stopping time we can take $\rho=0$ and $\tau=T$ in the optional stopping theorem to deduce that $$1=M_0=\Bbb{E}(M_T)$$ But we remark that after substituting $\mu=\frac{\lambda^2}{2}$ $$\begin{align}\Bbb{E}(M_T)&=\Bbb{E}\left( \exp\left(\lambda B_{M_T}-\frac{\lambda^2}{2}M_T\right)\right)\\&=\Bbb{E}\left(\exp\left(\lambda a-\frac{\lambda^2}{2}T\right)\right)\\&=\Bbb{E}\left(\exp\left(\sqrt{2\mu} a\right)\exp\left(-\mu T\right)\right)\end{align}$$ So we deduce that $\Bbb{E}(\exp(-\mu T))=\exp(-a\sqrt{2\mu})$ which is exactly what we wanted.
Does this work?