 Laplace transform of symmetric Brownian interval-exit time

For standard Brownian motion started at zero and first exit ${\tau }_x$ from $(-x,x)$, the Laplace transform is $1/\cosh (x\sqrt{2\lambda })$ for $\lambda >0$. The stochastic process $e^{-\lambda t}\cosh (\sqrt{2\lambda }{B}_t)$ is the average of two copies of the Exponential martingale for Brownian motion. Its values stopped at $t\wedge {\tau }_x$ are bounded, so bounded-time application of the optional stopping theorem followed by dominated convergence evaluates the transform without assuming uniform integrability of an unstopped exponential martingale on the infinite horizon.