Given the random variables $\textbf{x} \sim \mathcal{CN}(\textbf{0}_{M},\sigma_{x}^{2}\textbf{I}_{M})$ and $\textbf{y} \sim \mathcal{CN}(\textbf{0}_{M},\sigma_{y}^{2}\textbf{I}_{M})$ are independent, therefore, as I have calculated,

\begin{equation}\label{eq:lemma7} \mathbb{E} \left\lbrace \Re \left[ \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right] \right\rbrace = \int_{0}^{\infty} \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dwdu, \end{equation} where $\mathcal{CN}(.,.)$ is the complex normal random variable, $\Re(.)$ is the real part, $k \triangleq \sqrt{\frac{\sigma_{x}^{2}}{\sigma_{y}^{2}}}$, $f_{W}(w)$ and $f_{U}(u)$ are defined below.

\begin{equation} f_{W}(w) = \frac{M}{\pi} {B}\left( \frac{1}{2}, M \right) \left( 1 -w^{2} \right)^{M - \frac{1}{2}}, \;\; |w| < 1. \end{equation}

\begin{equation}\label{eq:lemma6} f_{U}(u) = \frac{2\Gamma(2M)}{(\Gamma(M))^{2}} \frac{u^{2M-1}}{(u^{2} + 1)^{2M}}, \;\; u > 0. \end{equation} where ${B}(.,.)$ is the Beta function and $M > 1$.

I have been using the integral in my calculations, however, I'd like to know if a closed form expression is possible for this double integral. I've tried Mathematica, however, it never finishes running. Follows below the Mathematica code I used.

```
fw = (M/Pi)*Beta[1/2, M]*((1 - (w^2))^(M - (1/2)));
fu = (2*Gamma[2*M])/((Gamma[M])^2)*(u^(2*M - 1))/(((u^2) + 1)^(2*M));
f = ((a*u + w)*fu*fw)/(a*u + (1/(a*u)) + 2*w);
ii = Integrate[f, {w, -1, 1}, Assumptions -> (a > 0 && M > 2 && 0 < u < \[Infinity])]
Integrate[ii, {u, 0, \[Infinity]}, Assumptions -> (a > 0 && M > 2)]
```

Follows a link to a matlab Monte Carlo simulation comparison with the proposed closed-form solution.

**UPDATE 11/02/2019**

By applying the following approximation to the expectation above: $$\mathbb{E} \left[ \frac{\textbf{w}}{\textbf{z}} \right] \approx \frac{\mathbb{E}[\textbf{w}]}{\mathbb{E}[\textbf{z}]} - \frac{\text{cov}(\textbf{w},\textbf{z})}{\mathbb{E}[\textbf{z}]^{2}} + \frac{\mathbb{E}[\textbf{w}]}{\mathbb{E}[\textbf{z}]^{3}}\text{var}(\mathbb{E}[\textbf{z}]).$$

I found

$$\mathbb{E} \left\lbrace \Re \left[ \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right] \right\rbrace \approx \frac{\sigma^2_x}{\sigma_x^2+\sigma_y^2}$$.