Verify whether $\mathbb{E}\int_{0}^{\infty}\frac{|B_t|}{(1+B_t^2)^2}\mathrm{d}t < \infty$
Solution 1:
$$\mathbb E\int_0^\infty \frac{|B_t|}{(1+B_t^2)^2}dt = \int_0^\infty \mathbb E\frac{|B_t|}{(1+B_t^2)^2}dt = \int_0^\infty \mathbb E\frac{\sqrt t|B_1|}{(1+tB_1^2)^2}dt = \mathbb E\int_0^\infty \frac{\sqrt t|B_1|}{(1+tB_1^2)^2}dt = $$ $$ = \mathbb E\frac1{B_1^2}\int_0^\infty \frac{\sqrt{x}}{(1+x)^2}dx = C\mathbb E \frac1{B_1^2} = \infty$$