I want to calculate the expectation value for the trace of the $m$-th power of the $n\times n$ adjacency matrix $A$ of a large Erdos-Renyi random graph (without self-coupling, i.e., all diagonal elements of $A$ are equal to zero). I was planning to use the invariance of trace under a change of basis and write
$\forall m<n:\:\:\:tr(A^m)=\sum_{i=1}^n\nu_i^m.$
Wigner's semicircle law states that in the asymptotic limit, the $(n-1)$ eigenvalues $\nu_1,\dots\,\nu_{n-1}$ have the semicircle probability distribution function
$f(\nu) = \frac{1}{2\pi \sigma^2}\sqrt{4\sigma^2-\frac{\nu^2}{n}}$
with second moment $\sigma^2$ of the distribution of the non-diagonal elements of $A$.
Since $tr(A)=0$ (no self-coupling), I know that $\nu_n=-\sum_{i=1}^{n-1}\nu_i$. My plan was to calculate the pdfs for $\nu_i^m$ and $tr(A^m)$ via multiple convolutions of $f(\nu_i)$ with itself. However, I already struggle with calculating the convolution of two semicircle pdfs,
$f(\nu_i)\star f(\nu_j):=\int_{-\infty}^\infty f(\nu_i)f(\nu_j-\nu_i)d \nu_i$.
How can I calculate this convolution? Or is there a better way to calculate the expectation value of $tr(A^m)$, that is, the expectation value of a non-linear function of $(n-1)$ iid random variables with semicircle pdf?
EDIT:
Since I am only interested in the expectation value of $tr(A^m)$, I do not need a pdf for $tr(A^m)$, because
$\langle tr(A^m)\rangle=\sum_{i=1}^{n}\langle \nu_i^m\rangle=(n-1)\int_{-\infty}^\infty f(\nu)\nu^md\nu+\langle \nu_n^m\rangle.$
However, I believe I still need the convolution of semicircle distributions for calculating
$\langle \nu_n^m\rangle = (-1)^m\langle \sum_{i=1}^{n-1} \nu_i^m\rangle.$