I want to show that for $(B_t)_t$ being the Brownian motion and a stopping time $\tau:= \text{inf}_{t \ge 0} \{B_t= a+bt\}$ where $a,b>0$ we have that the expectation value $E(e^{-\lambda \tau}, \tau< \infty) = e^{-a(b+ \sqrt{b^2 +2 \lambda})}$ for $\lambda >0.$
There is also a hint saying I should use the martingale $M(t)=e^{c B(t)- \frac{c^2t}{2}}$ where $c \ge 0.$
I guess this can be derived somehow from the optional stopping theorem, but I don't see how all this fits together. So I would start with
$1=E(M(0))=E(M(\tau))$ but I don't see how I can evaluate the right hand-side so that I get the desired expression. Also I am not sure that the conditions of the optional stopping theorem are satisfied.