# Laplace transform of stopping times

I am nearly done with a question:

Let $(B_t)$ be a Brownian motion on $\mathbb{R}$. For a fixed $x >0$, let $\tau$ be a stopping time defined by $$\tau = \inf \{t \geq 0 : B_t \not \in (-x,x) \}.$$

I wish to compute, for $\sigma>0$, $$\mathbb{E} \bigg(\exp \bigg\{ -\frac{{\sigma}^2}{2} \tau \bigg\} \bigg).$$

What I have done so far:

Noting that $\bigg\{\exp \bigg[ \sigma B_t -\frac{{\sigma}^2}{2} t \bigg] \bigg\}_{t \geq 0}$ is a martingale w.r.t the natural filtration, I applied the optional stopping theorem and the dominated convergence theorem to conclude that $$\mathbb{E} \bigg\{\exp \bigg[ \sigma B_{\tau} -\frac{{\sigma}^2}{2} \tau \bigg] \bigg\} =1.$$ Then, I get $$e^{\sigma x} \mathbb{E} \bigg( e^{-\frac{\sigma^2}{2} \tau} \mathbf{1}_{\{B_{\tau} =x\}} \bigg) + e^{- \sigma x} \mathbb{E} \bigg( e^{-\frac{\sigma^2}{2} \tau} \mathbf{1}_{\{B_{\tau} =-x\}} \bigg) =1.$$

Then I can't proceed from here. Is it true that $$\mathbb{E} \bigg( e^{-\frac{\sigma^2}{2} \tau} \mathbf{1}_{\{B_{\tau} =x\}} \bigg) = \mathbb{E} \bigg( e^{-\frac{\sigma^2}{2} \tau} \mathbf{1}_{\{B_{\tau} =-x\}} \bigg) \quad ?$$ (I am not sure whether $B_{\tau}$ has the same distribution as $-B_{\tau}$.)

If $B_t$ is a Brownian motion, then $-B_t$ is also a brownian motion (symmetry property). If you want to avoid this problem you can proceed as follows.
Since for every $\lambda \in \mathbb{R}$, $(\exp (\lambda B_t-\frac{\lambda ^2}{2}t))$ is a martingale and every finite linear combination of martingale is a martingale, $M_t= \exp (\lambda B_t-\frac{\lambda ^2}{2}t)+\exp (-\lambda B_t-\frac{\lambda ^2}{2}t)$ is a martingale. By continuity of $B_t$ we have $\tau=\inf \{ f:|B_t|=x\}$. By stopping time property : $$\mathbb{E}(M_\tau )=\mathbb{E}(M_0 )=2.$$ But we can easily see (by symmetry) that $\mathbb{E}(M_\tau )=\mathbb{E}(\exp (-\frac{\lambda ^2}{2}\tau) )2\text{ch}(\lambda x)$ so we can conclude that $\mathbb{E}(\exp (-\frac{\lambda ^2}{2}\tau) )=\text{ch}(\lambda x)^{-1}.$