Absolute value of Brownian motion I need to show that $$R_t=\frac{1}{|B_t|}$$ is bounded in $\mathcal{L^2}$ for $(t \ge 1)$, where $B_t$ is a 3-dimensional standard Brownian motion.
I am trying to find a bound for $\mathbb{E}[\int_{t=1}^{\infty}R^2_t]$.
Asymptotically $B_t^i$ is between $\sqrt{t}$ and $t$. I also know that $|B_t| \to \infty$, but the rate is not clear.
Hints would be helpful.
 A: Since $B_t$ and $\sqrt{t}B_1$ are identically distributed, $\mathrm E(R_t^2)=t^{-1}\mathrm E(R_1^2)$, hence $\mathrm E(R_t^2)\leqslant\mathrm E(R_1^2)$ for every $t\geqslant1$ and it remains to show that $\mathrm E(R_1^2)$ is finite. Now, the density of the distribution of $B_1$ is proportional to $\mathrm e^{-\|x\|^2/2}$ and $B_1$ has dimension $3$ hence the density of the distribution of $Y=\|B_1\|$ is proportional to $\varphi(y)=y^{3-1}\mathrm e^{-y^2/2}=y^2\mathrm e^{-y^2/2}$ on $y\gt0$. Since the function $y\mapsto y^{-2}\varphi(y)=\mathrm e^{-y^2/2}$ is Lebesgue integrable, the random variable $Y^{-2}=R_1^2$ is integrable.
On the other hand, $\mathrm E\left(\int\limits_1^{+\infty}R_t^2\mathrm dt\right)$ is infinite.
Edit The distribution of $B_1$ yields the distribution of $Y=\|B_1\|$ by the usual change of variables technique. To see this, note that in dimension $n$ and for every test function $u$,
$$
\mathrm E(u(Y))\propto\int_{\mathbb R^n} u(\|x\|)\mathrm e^{-\|x\|^2/2}\mathrm dx\propto\int_0^{+\infty}\int_{S^{n-1}}u(y)\mathrm e^{-y^2/2}y^{n-1}\mathrm d\sigma_{n-1}(\theta)\mathrm dy,
$$
where $\sigma_{n-1}$ denotes the uniform distribution on the unit sphere $S^{n-1}$ and $(y,\theta)\mapsto y^{n-1}$ is proportional to the Jacobian of the transformation $x\mapsto(y,\theta)$ from $\mathbb R^n\setminus\{0\}$ to $\mathbb R_+^*\times S^{n-1}$. Hence,
$$
\mathrm E(u(Y))\propto\int_0^{+\infty}u(y)y^{n-1}\mathrm e^{-y^2/2}\mathrm dy,
$$
which proves by identification that the distribution of $Y$ has a density proportional to $y^{n-1}\mathrm e^{-y^2/2}$ on $y\gt0$.
