I'm trying to understand the counterexample to the claim that every $L^2$-bounded local martingale is also a true martingale. For this I'm considering the local martingale $$ X_t:=\frac{1}{\vert B_t + x \vert}, $$ where $B_t$ is a threedimensional Brownian Motion, $x\in\mathbb{R}^3\setminus \{0\}$ and $|\cdot |$ denotes the euclidean norm.
In my calculations though, I have found $X_t$ to not even be integrable for a fixed $t\in\mathbb{R}_{\geq 0}$. My argument is as follows:
$$ \begin{align} \mathbb{E}\left[X_t\right] &=\frac{1}{\sqrt{2\pi t}^3}\int_{\mathbb{R}^3}|y-x|^{-1}\exp{(-\frac{|y|^2}{2t})}dy \\ &\geq \frac{1}{\sqrt{2\pi t}^3}\int_{|y-x|\leq c}|y-x|^{-1}\exp{(-\frac{|y|^2}{2t})}dy \\ &\geq \frac{1}{\sqrt{2\pi t}^3}\int_{|y-x|\leq c}|y-x|^{-1}\exp{(-\frac{(c+|x|)^2}{2t})}dy \\ &=k_c\cdot \frac{1}{\sqrt{2\pi t}^3}\int_{|y-x|\leq c}|y-x|^{-1}dy \\ &=k_c\cdot \frac{1}{\sqrt{2\pi t}^3}\int_{|y|\leq c}|y|^{-1}dy, \\ \end{align} $$ where $c>0$ is arbitrary and $k_c$ is a constant only depending on $c$. To my knowledge, the last integral doesn't exist, and hence $\mathbb{E}\left[X_t\right]=\infty$
But this must be wrong, since $X_t \in L^2$. Any help to find my mistake would be very appreciated!
You have to be careful about the dimensions. The integral
$$\int_{\{y \in \color{red}{\mathbb{R}}; |y| \leq c\}} \frac{1}{|y|} \, dy$$
is infinite, but the integral
$$\int_{\{y \in \color{red}{\mathbb{R}^3}; |y| \leq c\}} \frac{1}{|y|} \, dy$$
is finite (and this is the integral which appears in your calculation). This follows from the following well known statement: