Expected first time that $|B(t)|=1$ for a standard Brownian motion

Question

I want to calculate $\mathbb{E}[T]$ where $T = \inf \{t \geq 0 \mid |B(t)| = 1\}$ and $B(t)$ is a Brownian motion with mean $0$. I saw some similar posts but for a one-sided hitting time, and in those cases the expectation is infinite. I tried the following:

$$ \mathbb{E}[T] = \int_{0}^\infty s \mathbb{P}[T=s]ds = \int_{0}^\infty \mathbb{P}[T > s]ds. $$ By the reflection principle, $\mathbb{P}[T>s] = \mathbb{P}[\sup_{t\in [0,s]} |B(t)| <1] = 2 \mathbb{P}[|B(s)| < 1]$. So this gives $$ \mathbb{E}[T]= 2 \int_0^\infty \mathbb{P}[B(s) < 1]ds. $$ Now $B(s)$ is distributed as a normal random variable with variance $\sigma^2 = s$. I have no idea how to continue, or maybe there is a simpler way?

Thanks in advance.

EDIT: Maybe a nice way can be to view the Brownian motion as (the limit of) the sum of $\{-1,1\}$ random variables. But I am also not sure how to prove the result in this case..

EDIT: Towards an answer:
So I have come some way since I asked this question, and I think I know roughly how to do it. Here goes:

Let $T_1$ be the first time when $|B(t)|=1$, and $T_2$ be the first time after $T_1$ that $|B(T_2)-B(T_1)| = 1$, $T_3$ the first time afterwards where $|B(T_3)-B(T_2)| = 1$, etc.. Let $X_i$ correspond to $B(T_i) - B(T_{i-1})$, with $T_{0}=0$. By independent increments, all these are independent, and $\mathbb{P}[X_i=1] = \mathbb{P}[X_i=-1]=\frac12$. Also, $\mathbb{E}T_1 = \mathbb{E}T_i$ for all $i$. Now, by the LLN,

$$ \mathbb{P}[ | T_1 + \ldots + T_n - n \mathbb{E}T_1| > \epsilon n] \to 0. $$ Moreover, $B(T_1+\ldots + T_n) - B(n \mathbb{E} T_1)$ is a distributed as $\mathcal{N}(0, T_1 +\ldots + T_n - n \mathbb{E}T_1)$. With increasingly high probability (as $n\to \infty$), this variance is bounded by $\epsilon n$. Now,

\begin{align} \mathbb{P}[ |B(T_1+\ldots T_n) - B(n \mathbb{E} T_1)| > \sqrt{\epsilon n} ] & \leq 2\int_{\sqrt{\epsilon n}}^\infty \frac{1}{\sqrt{2\pi \epsilon n}}e^{-x^2/(2 \epsilon n)}dx\\ &=2 \int_{1}^\infty \frac{1}{\sqrt{2 \pi} } e^{-x^2/2} dx \\ &\leq 0.25 \label{eq:adadssa} \end{align} However, $B(T_1 + \ldots + T_n) = X_1+\ldots + X_n$, which by the Central limit theorem converges to a $\mathcal{N}(0,n)$ random variable. If it then were the case that $\alpha := \mathbb{E}T_1 \neq 1$, then $B(n \mathbb{E}T_1)$ is $\mathcal{N}(0, \alpha n)$, while $B(T_1+ \ldots + T_n)$ is roughly $\mathcal{N}(0, n)$. I think this should somehow contradict the fact that $\mathbb{P}[ |B(T_1+\ldots T_n) - B(n \mathbb{E} T_1)| > \sqrt{\epsilon n} ] < 0.25$.

Answer 1

We note that: $$B_T \in \{-1,1\}. $$

So:

$$ P(B_T=-1) + P(B_T=1) = 1.$$

We also know that $(B_t^2-t)_{t\geq 0}$ is a martingale.

Doob's optional stopping theorem gives then:

$$ E[B_T^2 - T] = E[B_0^2 -0] =0, $$

from where we have

$$ E[T]=E[B_T^2] = (-1)^2P(B_T=-1) + 1^2 P(B_T=1) = 1. $$