$\newcommand{\P}{\mathbb{P}}$I would like to prove that stopping time:
$\sigma=\inf\{t\geqslant0: |B_t|=\frac{1}{2}(1+\sqrt{1+t})\}$
is integrable.
This is part of the answer to the question here.
I've tried to estimate ${\P}(\sigma\geqslant t)$ in the following way: $$\begin{array}{rcl}\P(\sigma\geqslant t) & = & \P(\forall s<t |B_s|\leqslant \frac{1}{2}(1+\sqrt{1+s} ) \\ & = & \P(\forall s<t |B_1|\sqrt{s}\leqslant \frac{1}{2}(1+\sqrt{1+s}))\\ &= &\P\left(\forall s<t |B_1|\leqslant \frac{1}{2}\left(\frac{1}{s}+\sqrt{1+\frac{1}{s}}\right)\right) \end{array}$$
I'm not sure if it is correct (I get something which doesn't converge).
Thank you in advance.
First of all, note that $\sigma$ is almost surely finite; this is e.g. a consequence of the law of the iterated logarithm.
It follows from the optional stopping theorem that $M_t := B_{t \wedge \sigma}^2 - t \wedge \sigma$ is a martingale; in particular
$$\mathbb{E}(M_t) = \mathbb{E}(M_0)=0,$$
i.e.
$$\mathbb{E}(B_{t \wedge \sigma}^2) = \mathbb{E}(t \wedge \sigma). \tag{1}$$
By the very definition of $\sigma$, we have
$$\mathbb{E}(B_{t \wedge \sigma}^2) \leq \frac{1}{4} \mathbb{E}\bigg[ (1+\sqrt{1+t \wedge \sigma})^2 \bigg]$$
and therefore
$$\mathbb{E}(t \wedge \sigma) \leq \frac{1}{4} \mathbb{E}\bigg[ (1+\sqrt{1+t \wedge \sigma})^2 \bigg] \leq \frac{1}{2} \mathbb{E}(2+t \wedge \sigma).$$
Hence,
$$\frac{1}{2} \mathbb{E}(t \wedge \sigma) \leq 1.$$
Applying Fatou's lemma we conclude
$$\mathbb{E}(\sigma) \leq \liminf_{n \to \infty} \mathbb{E}(n \wedge \sigma) \leq 2.$$