I was reading the paper
Sengupta, Anirvan M., and Partha P. Mitra. "Distributions of singular values for some random matrices." Physical Review E 60.3 (1999): 3389.
but got stuck at equation (3):
Here $\lambda_n$ are the singular values of a matrix $M$ (eventually taken to be a random matrix), $\rho(\lambda)$ denotes the density of singular values of $M$, and the resolvent $\mathcal G(z)$ is defined by:
So how does (3) follow from (2)?
I also have problems, but let me write what I got:
Perhaps you know that $\lim_{\epsilon\to 0^+}\frac{1}{\pi}\Im(\frac{1}{t-i\epsilon})=\delta(t)$. In any case, this follows from $\Im(\frac{1}{t-i\epsilon})=\frac{1}{\epsilon((t/\epsilon)^2+1)}$. Hence, we get the right part of (3) is equal to $2\lambda\sum_n\delta(\lambda^2-\lambda_n^2)$, so we should prove that $2t\delta(t^2-a^2)=\delta(t-a)$. Unfortunately, I prove instead that $2t\delta(t^2-a^2)=-\delta(t+|a|)+\delta(t-|a|)$.
It's easier to approximate $\delta$ by $\phi_\epsilon=\epsilon^{-1}\phi(t/\epsilon)$, where $\phi$ is compactly supported and even. Now, let $\psi$ be a test function and evaluate $$ \begin{align} \int 2t\phi_\epsilon(t^2-a^2)\psi(t)\,dt &= \int_{-\infty}^0+\int_0^{\infty} \\ &=-\int_{-a^2}^\infty \phi_\epsilon(s)\psi(-\sqrt{s+a^2})\,ds +\int_{-a^2}^\infty\phi_\epsilon(s)\psi(\sqrt{s+a^2})\,ds, \end{align} $$ where I used the change of variables $t^2-a^2=s$. Therefore, taking the limit $\epsilon\to 0$ we get $2t\delta(t^2-a^2) = -\delta(t+|a|)+ \delta(t-|a|)$. I don't know if I made a mistake.