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Mathematica evaluates the integral over $w$ in terms of a hypergeometric function, $$ \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dw=$$ $$\frac{k^2 4^M u^{2 M+1}\Gamma(M+\tfrac{1}{2})}{\sqrt{\pi } \left(k^2 u^2-1\right)^3\left(u^2+1\right)^{2 M}} \left[\frac{1}{(M-1)!} \left(k^4 u^4+4 M k^2 u^2+4M+3\right)-\frac{4 M}{(M+1)!} \left(k^2 u^2+1\right) (\tfrac{3}{4}+m^2+2m)\, \, _2{F}_1\left(-\tfrac{1}{2},1;M+2;\frac{4 k^2 u^2}{\left(k^2 u^2+1\right)^2}\right)\right].$$ The integral $\int_0^\infty du$ of the first term between square brackets has a closed form expression (again involving a hypergeometric function), but the integral of the second term does not.

In the large-$M$ limit we may average numerator and denominator separately, so $$\mathbb{E} \left\lbrace \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right\rbrace \rightarrow\frac{\sigma^2_x}{\sigma_x^2+\sigma_y^2}, \;\;M\rightarrow\infty.$$

Mathematica evaluates the integral over $w$ in terms of a hypergeometric function, $$ \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dw=$$ $$\frac{k^2 4^M u^{2 M+1}\Gamma(M+\tfrac{1}{2})}{\sqrt{\pi } \left(k^2 u^2-1\right)^3\left(u^2+1\right)^{2 M}} \left[\frac{1}{(M-1)!} \left(k^4 u^4+4 M k^2 u^2+4M+3\right)-\frac{4 M}{(M+1)!} \left(k^2 u^2+1\right) (\tfrac{3}{4}+m^2+2m)\, \, _2{F}_1\left(-\tfrac{1}{2},1;M+2;\frac{4 k^2 u^2}{\left(k^2 u^2+1\right)^2}\right)\right].$$ The integral $\int_0^\infty du$ of the first term between square brackets has a closed form expression (again involving a hypergeometric function), but the integral of the second term does not.

In the large-$M$ limit we may average numerator and denominator separately, so $$\mathbb{E} \left\lbrace \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right\rbrace \rightarrow\frac{\sigma^2_x}{\sigma_x^2+\sigma_y^2}, \;\;M\rightarrow\infty.$$

In the accepted answer @student has now elegantly shown this holds actually for all $M$ (quite an impressive "student" !)

Mathematica evaluates the integral over $w$ in terms of a hypergeometric function, $$ \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dw=$$ $$\frac{k^2 4^M u^{2 M+1}\Gamma(M+\tfrac{1}{2})}{\sqrt{\pi } \left(k^2 u^2-1\right)^3\left(u^2+1\right)^{2 M}} \left[\frac{1}{(M-1)!} \left(k^4 u^4+4 M k^2 u^2+4M+3\right)-\frac{4 M}{(M+1)!} \left(k^2 u^2+1\right) (\tfrac{3}{4}+m^2+2m)\, \, _2{F}_1\left(-\tfrac{1}{2},1;M+2;\frac{4 k^2 u^2}{\left(k^2 u^2+1\right)^2}\right)\right].$$ The integral $\int_0^\infty du$ of the first term between square brackets has a closed form expression (again involving a hypergeometric function), but the integral of the second term does not.

In the large-$M$ limit we may average numerator and denominator separately, so $$\mathbb{E} \left\lbrace \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right\rbrace \rightarrow\frac{\sigma^2_x}{\sigma_x^2+\sigma_y^2}.$$

Mathematica evaluates the integral over $w$ in terms of a hypergeometric function, $$ \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dw=$$ $$\frac{k^2 4^M u^{2 M+1}\Gamma(M+\tfrac{1}{2})}{\sqrt{\pi } \left(k^2 u^2-1\right)^3\left(u^2+1\right)^{2 M}} \left[\frac{1}{(M-1)!} \left(k^4 u^4+4 M k^2 u^2+4M+3\right)-\frac{4 M}{(M+1)!} \left(k^2 u^2+1\right) (\tfrac{3}{4}+m^2+2m)\, \, _2{F}_1\left(-\tfrac{1}{2},1;M+2;\frac{4 k^2 u^2}{\left(k^2 u^2+1\right)^2}\right)\right].$$ The integral $\int_0^\infty du$ of the first term between square brackets has a closed form expression (again involving a hypergeometric function), but the integral of the second term does not.

In the large-$M$ limit we may average numerator and denominator separately, so $$\mathbb{E} \left\lbrace \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right\rbrace \rightarrow\frac{\sigma^2_x}{\sigma_x^2+\sigma_y^2}, \;\;M\rightarrow\infty.$$

Mathematica evaluates the integral over $w$ in terms of a hypergeometric function, $$ \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dw=$$ $$\frac{k^2 4^M u^{2 M+1}\Gamma(M+\tfrac{1}{2})}{\sqrt{\pi } \left(k^2 u^2-1\right)^3\left(u^2+1\right)^{2 M}} \left[\frac{1}{(M-1)!} \left(k^4 u^4+4 M k^2 u^2+4M+3\right)-\frac{4 M}{(M+1)!} \left(k^2 u^2+1\right) (\tfrac{3}{4}+m^2+2m)\, \, _2{F}_1\left(-\tfrac{1}{2},1;M+2;\frac{4 k^2 u^2}{\left(k^2 u^2+1\right)^2}\right)\right].$$ The integral $\int_0^\infty du$ of the first term between square brackets has a closed form expression (again involving a hypergeometric function), but the integral of the second term does not.

Mathematica evaluates the integral over $w$ in terms of a hypergeometric function, $$ \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dw=$$ $$\frac{k^2 4^M u^{2 M+1}\Gamma(M+\tfrac{1}{2})}{\sqrt{\pi } \left(k^2 u^2-1\right)^3\left(u^2+1\right)^{2 M}} \left[\frac{1}{(M-1)!} \left(k^4 u^4+4 M k^2 u^2+4M+3\right)-\frac{4 M}{(M+1)!} \left(k^2 u^2+1\right) (\tfrac{3}{4}+m^2+2m)\, \, _2{F}_1\left(-\tfrac{1}{2},1;M+2;\frac{4 k^2 u^2}{\left(k^2 u^2+1\right)^2}\right)\right].$$ The integral $\int_0^\infty du$ of the first term between square brackets has a closed form expression (again involving a hypergeometric function), but the integral of the second term does not.

In the large-$M$ limit we may average numerator and denominator separately, so $$\mathbb{E} \left\lbrace \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right\rbrace \rightarrow\frac{\sigma^2_x}{\sigma_x^2+\sigma_y^2}.$$