Let $X\sim\mathcal{N}(0,I_{d})$. I would like to compute the the following quantity: \begin{equation} \mathbb{E}\bigg[\frac{XX^{\top}}{\|X\|_{2}^{2}}\bigg]. \end{equation}
Letting $B=\frac{XX^{\top}}{\|X\|_{2}^{2}}$, one can see that \begin{equation} B_{ii}=\frac{X{i}^{2}}{X_{1}^{2}+\cdots+X_{d}^{2}} \end{equation} which is a ratio of chi-squared random variables. This is a beta random variable with parameters $1/2$ and $(d-1)/2$, and the expectation is $1/d$. This takes care of the diagonal entries.
It's the off-diagonal entries where I'm stuck. We have \begin{equation} B_{ij}=\frac{X_{i}X_{j}}{X_{1}^{2}+\cdots+X_{d}^{2}}. \end{equation} For this, I thought of conditioning on $X_{i}$ and $X_{j}$ first. That basically gives me $\mathbb{E}[1/(c+Z)]$, where $c=x_{i}^{2}+x_{j}^{2}$ and $Z\sim\chi_{d-2}^{2}$. Any ideas on how to compute this expectation? Any other approaches are also welcome.