Suppose $a, n>0$ and $B_t$ is a Brownian Motion, define $$\sigma=\inf\{t:B_t\in\{-a,a\}\}.$$ I want to find $\mathbb{E}_x[e^{n\sigma}]$. (Notice since $n>0$ it is $\textbf{not}$ Laplace transform!)
I used standard methods from stochastic analysis (Ito formula + Optional Stopping Time theorem) to connect it with bounded value problem
Here is my answer
\begin{align*}
\mathbb{E}_xe^{n \sigma}=\frac{\sin(\sqrt{2n}(a+x))+\sin(\sqrt{2n}(a-x))}{\sin\left(2\sqrt{2n}a\right)}
\end{align*}
which does not make sense, because it is not monotone with respect to $n$... Any help/questions/comments would be appreciated!
Edit: $x\in(-a,b)$ and $\mathbb{E}_x$ is the expectation with respect to law of Brownian Motion starting at $x$.