# show that $\mathbb{E}[T^{\alpha}_1] < \infty$ for $\alpha < 1/2$

let $$T_1 = \inf \{t \geq 0 \,|\, B_t = 1 \}$$ where $$B = (B_t)_{t \geq 0}$$ is a standard brownian motion

Question : show that $$\mathbb{E}[T^{\alpha}_1] < \infty$$ for $$\alpha < 1/2$$

My Failed attempt :

since $$T^{\alpha}_1 > 0$$ then :

\begin{align} \mathbb{E}[T^{\alpha}_1] & = \int_0^{\infty}P(T^{\alpha}_1 > t) dt = \int_0^{\infty}P(T_1 > t^{1/\alpha}) dt \end{align}

$$u = t^{1/\alpha} \implies du = \frac{1}{\alpha}t^{{1-\alpha}/\alpha}dt \implies \alpha du = u^{1 - \alpha} dt \implies dt = \alpha u^{\alpha - 1} du$$

\begin{align} \mathbb{E}[T^{\alpha}_1] & = \alpha \int_0^{\infty}u^{\alpha - 1}P(T_1 > u) du \leq \alpha \int_0^{\infty}\frac{1}{u^{1-\alpha}} du \end{align}

but $$1 - \alpha > \frac12$$

something that would solve this problem is if $$P(T_1 > u) \sim u^{-1/2}$$

what I know is $$P(T_1 > u) = P(\max_{s \leq u} B_s \leq 1)$$

but the lack of independence makes it hard for me to compute it

The distribution of $$\max_{0\leq s\leq u}B_s$$ is well-known. See https://en.wikipedia.org/wiki/Reflection_principle_(Wiener_process).
So $$\mathbb P(T_1>u) = \mathbb P(\max_{0\leq s\leq u}B_s \leq 1) = 1-\mathbb P(\max_{0\leq s\leq u}B_s > 1) = 1-2\mathbb P(B_u>1).$$ Since $$B_u$$ have normal $$\mathcal N(0,u)$$ distribution, $$Y=B_u/\sqrt{u}$$ is standard normal and for $$u\to\infty$$ $$1-2\mathbb P(B_u>1) = 1-2\mathbb P(Y>1/\sqrt{u}) = 1-2\Phi(-1/\sqrt{u}) \sim 2 \frac{\phi(1/\sqrt{u})}{\sqrt{u}} \sim \frac{2}{\sqrt{2\pi u}}.$$ This asymptotic equivalence takes place by L'Hôpital's rule $$\lim\limits_{x\to 0}\frac{1-2\Phi(-x)}{x} = \lim\limits_{x\to 0}\frac{2\phi(-x)}{1} = \frac{2}{\sqrt{2\pi}}.$$ So, as you supposed, $$\mathbb P(T_1>u)\sim cu^{-1/2}$$.