I'm working on this homework problem for Brownian Motion. Suppose we define a stopping time $\tau_a = inf \left\lbrace t \geq 0 : B(t) = a \right\rbrace$ for some $a>0$. I already showed in a previous part that:
$$f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}}e^{\frac{-a^2}{2t}}$$
Now I need to compute $E[\tau_a]$. So I tried doing:
$$ E[\tau_a] = \int_{0}^{\infty} tf_{\tau_a}(t) dt = \int_{0}^{\infty} \frac{at}{\sqrt{2\pi t^3}}e^{\frac{-a^2}{2t}}$$
Only problem is, I can't come up with a way to solve this integral. Plugging it in to wolfram alpha gives a very complicated expression. The answer should be $\infty$, but I can't really prove that unless I can solve this integral, which is becoming a struggle. I thought about using the comparison test but I don't see an easy way to do that.
Since
$$\exp \left( - \frac{a^2}{2t} \right) \uparrow 1 \qquad \text{as $t \to \infty$},$$
there exists $R>0$ such that
$$\exp \left( - \frac{a^2}{2t} \right) \geq \frac{1}{2}$$
for all $t \geq R$. Consequently,
$$\begin{align*} \mathbb{E}(\tau_a) &\geq \int_R^{\infty} \frac{at}{\sqrt{2\pi t^3}} \exp \left(- \frac{a^2}{2t} \right) \, dt \\ &\geq \frac{a}{\sqrt{2\pi}} \frac{1}{2} \int_R^{\infty}\frac{1}{\sqrt{t}} \, dt = \infty. \end{align*}$$
Remark: The result can be also proved using Wald's identities. Wald's identities state (in particular) that for any stopping time $\tau$ such that $\mathbb{E}\tau<\infty$, we have $\mathbb{E}(B_{\tau})=0$. Since $\mathbb{E}(B_{\tau_a}) = a \neq 0$, this means that $\tau_a$ cannot be integrable.