Verify whether $\mathbb{E}\int_{0}^{\infty}\frac{|B_t|}{(1+B_t^2)^2}\mathrm{d}t < \infty$

Question

I need to verify if $$\mathbb{E}\int_{0}^{\infty}\frac{|B_t|}{(1+B_t^2)^2}\mathrm{d}t \ < \ +\infty$$ where $B_t$ is a standard Brownian motion. I thought about doing something like this:

1. by Fubini theorem we can change the order of integration (integrand is nonegative): $$\mathbb{E}\int_{0}^{\infty}\frac{|B_t|}{(1+B_t^2)^2}\mathrm{d}t \ = \ \int_{0}^{\infty}\mathbb{E}\frac{|B_t|}{(1+B_t^2)^2}\mathrm{d}t.$$
2. we have $B_t\sim \sqrt{t}B_1$ in distribution, hence $$\mathbb{E}\frac{|B_t|}{(1+B_t^2)^2} \ = \ \mathbb{E}\frac{\sqrt{t}|B_1|}{(1+tB_1^2)^2} \ \le \ \left\{\begin{split} \sqrt{t}\cdot C_1 \ \ \ \text{for} \ \ \ 0<t\le 1, \\ \\ ??? \ \ \ \text{for} \ \ \ 1<t<\infty, \end{split}\right.$$ where $C_1 = \mathbb{E}|B_1|$...

The problem is with $1<t<\infty$ as $1/|B_1|^k$ is not integrable for $k\ge 1$. Can this reasoning be somehow improved?

I will be glad for any suggestions...

Answer 1

$$\mathbb E\int_0^\infty \frac{|B_t|}{(1+B_t^2)^2}dt = \int_0^\infty \mathbb E\frac{|B_t|}{(1+B_t^2)^2}dt = \int_0^\infty \mathbb E\frac{\sqrt t|B_1|}{(1+tB_1^2)^2}dt = \mathbb E\int_0^\infty \frac{\sqrt t|B_1|}{(1+tB_1^2)^2}dt = $$ $$ = \mathbb E\frac1{B_1^2}\int_0^\infty \frac{\sqrt{x}}{(1+x)^2}dx = C\mathbb E \frac1{B_1^2} = \infty$$