I am trying to get compute at least the directional component of the following expectation, where $M$ is a symmetric, invertible, PD matrix: $$\mathbb{E}_{v \sim N(0, I)}\left[\frac{vv^T}{||Mv||_2}\right]$$

(Note that the norm is not squared). My approach so far has been to write this as \begin{align} &\mathbb{E}_{v \sim N(0, M)}\left[\frac{M^{-1/2}vv^TM^{-1/2}}{||M^{1/2}v||_2^2}\|M^{1/2}v\|_2\right] \\ &= M^{-1/2}\cdot \mathbb{E}_{v \sim N(0, M)}\left[\frac{vv^T}{||v||_M^2}\|v\|_M\right]\cdot M^{-1/2} \\ &= M^{-1/2}\cdot \mathbb{E}_{v \sim N(0, M)}\left[ \left(\frac{v}{||v||_M}\right)\left(\frac{v}{||v||_M}\right)^\top \|v\|_M\right]\cdot M^{-1/2} \end{align}

Then, it seems as though (and supported by this answer) the first two components in the expectation should be independent of the last one, and since the norm is a scalar and the expectation of the normalized outer product is $I$, the directional component of this expectation is simply M^{-1}.

---- Edit to make my approach more clear ---- Going based on this answer, we can write the aforementioned expectation as \begin{align} &= M^{-1/2}\cdot \mathbb{E}_{v \sim N(0, M)}\left[ \left(\frac{v}{||v||_M}\right)\left(\frac{v}{||v||_M}\right)^\top\right] \mathbb{E}_{v \sim N(0, M)}\left[ \|v\|_M\right]\cdot M^{-1/2} \end{align}

Since I don't care about the scaling factor, and rather only the "direction" of the resulting matrix, I am willing to omit the $\mathbb{E}_{v \sim N(0, M)}\left[ \|v\|_M\right]$ term. The other expectation term should just be identity? Leaving us with $M^{-1}\cdot C$ for some scalar $C$.

Is this correct? And if so, is there a better/more elegant way to solve this problem?