How to compute the first moment of the distribution of the convolution of Marcenko-Pastur law with a not iid matrix?

Question

Let $\mathbf{F}$ denote an M × N matrix whose entries are independent zero-mean complex random variables, the limiting eigenvalue distribution is given by the Marchenko Pastur law $MP_{\beta}$, where $\frac{N}{M}\rightarrow \beta$.

It can be shown that the moments are giving by \begin{equation} \sum_{i=1}^{k}{1 \over k} \binom{k}{i} \binom{k}{i-1}\beta^{i}=\lim_{M,N\rightarrow\infty} {1 \over M}{\rm tr}\{({\bf FF}^{H})^{k}\} \end{equation}

Then define $\mathbf{S}=\frac{1}{M}\mathbf{C}^{1/2}\mathbf{FF}^H(\mathbf{C}^{1/2})^H$, where $\mathbf{C}$ is a positive definite matrix, whose limiting eigenvalue distribution is denoted by $\nu$. It is known that the limiting eigenvalue distribution of $\mathbf{S}$ is given by the free multiplicative convolution of $MP_{\beta}$ and $\nu$, $f_s=MP_\gamma\boxtimes\nu$.

A closed form of the resulting distribution $f_s$ is . \begin{equation} f_s(\lambda)=(1-\beta)^{+}\delta(\lambda)+\frac{\sqrt{(\lambda-a)^{+}(b-\lambda)^{+}}}{2\pi\lambda(1+\lambda\mu)} \end{equation}

where $\mu$ is a parameter >0

$a=1+\beta+2\mu \beta-2\sqrt{\beta}\sqrt{(1+\mu)(1+\mu \beta)}$

$b=1+\beta+2\mu \beta+2\sqrt{\beta}\sqrt{(1+\mu)(1+\mu \beta)}$

(see Random Matrix Theory and Wireless Communications 1)

My question is: can be found an expression of the first moment of $f_s$, i.e. the arithmetic mean as \begin{equation} \int_a^b \lambda f_s(\lambda)d\lambda=\lim_{M,N\rightarrow\infty} {1 \over M}{\rm tr}\{{\bf SS}^{H}\} \end{equation}

Answer 1

If I correct the typo's so that the MP distribution follows when $\mu=0$, the definitions should be: $$f_s(\lambda)=(1-\beta)^{+}\delta(\lambda)+{{\sqrt{(\lambda-a)^+(b-\lambda)^+}}\over{2\pi \lambda(1+\mu \lambda)}},$$ $$a=1+\beta+2\mu \beta-2\sqrt{\beta}\sqrt{(1+\mu)(1+\mu \beta)},$$ $$b=1+\beta+2\mu \beta+2\sqrt{\beta}\sqrt{(1+\mu)(1+\mu \beta)}.$$ I checked that then $\int_a^b f_s(\lambda)\,d\lambda=\min(1,\beta)$, so with the delta function contribution it is properly normalized to unity.

The desired integral evaluates to $$\int_a^b \lambda f_s(\lambda)\,d\lambda=\beta,$$ independent of $\mu>0$ for any $\beta>0$.