On the distribution of a normalized Gaussian vector Let $x=(x_1,\ldots,x_n)\in\mathbb{R}^n$ be an $n$-dimensional random vector that follows the normal distribution with mean vector $\mu$ and covariance matrix $\Sigma=\operatorname{diag}\left(\sigma_1^2, \ldots, \sigma_d^2\right)$; in other words, each element of $x$ is a uni-variate normal distribution with mean and variance.
I'm interested in the distribution of the normalized vector
$$
\tilde{x} = \frac{x}{\lVert x \rVert}=\frac{x}{\sqrt{x_1^2+\ldots+x_n^2}}.
$$
Since $x_i$'s are independent, I though about studying each random variable separately; i.e.:
$$
\tilde{x}_i = \frac{x_i}{\sqrt{x_1^2+\ldots+x_n^2}},
$$
or something like this
$$
\tilde{x}_i = \frac{x_i}{\sqrt{x_i^2+a}},\quad a>0,
$$
but I cannot think of any useful strategy. Using the definition of expectation and variance would lead to painful integrations, I think. 
My goal is to normalize the mean vector, and simultaneously transform the variances so as not to "break" my set's structure; even a heuristic/approximating approach would suffice.
 A: First, the distribution of $\tilde{x}$ is concentrated on $\mathbb{S}^{n-1}$. In particular, it doesn't have a density w.r.t. the Lebesgue measure on $\mathbb{R}^n$. So instead you may work with the polar representation. Let $r=\|x\|$ and let $\theta_j$ be the angle between $\tilde{x}$ and the $x_j$-th axis, i.e. $x_j=r\cos(\theta_j)$ (denote $\theta\equiv(\theta_1,\ldots,\theta_{n-1})$). Then the density of $(r,\theta)$ is 
$$
g(r,\theta)=\left(\prod_{i=1}^{n}2\pi\sigma_i^2\right)^{-1/2}\exp\left(-\sum_{i=1}^n\frac{(r\cos(\theta_i)-\mu_i)^2}{2\sigma_i^2}\right)r^{n-1},
$$
where $r\in [0,\infty)$, $\theta\in [0,\pi]^{n-2}\times [0,2\pi)$, and $\sum_{i=1}^n \cos^2(\theta_i)=1$. Consequently, the density of $\theta$ is given by
$$
f(\theta)=\int_0^{\infty}g(r,\theta)dr.
$$
In particular, when $\mu=0$ and $\Sigma=\sigma I_n$,
$$
f(\theta)=\frac{1}{(2\pi\sigma^2)^{n/2}}\int_0^{\infty}\exp\left(-\frac{r^2}{2\sigma^2}\right)r^{n-1}dr=\frac{\Gamma(n/2)}{2\pi^{n/2}},
$$
which is $1$ over the surface area of the unit $(n-1)$-sphere. 
