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](http:///https://www.nowpublishers.com/article/Details/CIT-001))


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}



How to compute the first moment of the distribution of the convolution of Marcenko-Pastur law with a not iid matrix?
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 $\beta>0$.


