= Asymmetric Brownian interval-exit transform
{title2=$\mathbb E[e^{-\lambda^2T/2}\mathbf1_{\{B_T=a\}}]=\frac{\sinh(\lambda b)}{\sinh(\lambda(a+b))}$}
For standard <Brownian motion> started at zero and exit time $T$ from $(-b,a)$, $a,b>0$, stop the two <exponential Brownian martingales> with parameters $\lambda$ and $-\lambda$. Their stopped values are bounded by $e^{|\lambda|\max(a,b)}$. The <optional stopping theorem> and <dominated convergence theorem> give $e^{\lambda a}p+e^{-\lambda b}q=1$ and $e^{-\lambda a}p+e^{\lambda b}q=1$, where $p,q$ are the discounted upper and lower exit probabilities. Solving gives the displayed formula and $q=\sinh(\lambda a)/\sinh(\lambda(a+b))$. Thus
$$
\mathbb E e^{-\lambda^2T/2}=\frac{\cosh(\lambda(a-b)/2)}{\cosh(\lambda(a+b)/2)}.
$$
At $\lambda=0$, the first ratio has its removable value $b/(a+b)$, and the second equals one. Finiteness of $T$ follows from a uniformly positive one-unit-time chance to leave any fixed bounded interval.
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