Let x, y > 0. Defint eh first passage time of a Brownian motion $W_t$ as $\tau_a$ = min{t $\ge$ 0: $W_t$ = a}. I need to show that E[$e^{-u\tau_x}$$1_{\tau_x < \tau_{-y}}$] = $\frac{sinh(y\sqrt{2u})}{sinh((x + y)\sqrt{2u}}$.
My method, and the only method that I will be able to understand, is to use the optional sampling theorem. I noted that $Z_t = e^{\theta W_t - \frac{1}{2}\theta^{2}t}$ is martingale and that the optional sampling theorem states that $E[Z_{\tau_{min\, {a, t}}}$] = 1. Applying this to the stopping time $ {\tau_x, \wedge \tau_{-y}}$ I have managed to show that as t --> $\infty$ $Z_(\tau_x\wedge\tau_{-y})\wedge t$ = $e^{-\theta y - \frac{1}{2} \theta^2 \tau_{-y}}$$1_{\tau_{-y} \, < \, \tau_x}$ + $e^{\theta x - \frac{1}{2} \theta^2 \tau_{x}}$$1_{\tau_{x} \, < \, \tau_{-y}}$. I can't figure where to go from here. I had a similar problem, but there was only one level involved, where here we have two: x and y.
I believe we should now take the expectation of the expression I derived and set it equal to one by optional sampling theorem, but I don't know what follows. Thanks.