# How to get the eigenvalue density contribution $\rho_1(x)$?

I'm studying the $1/N$ expansion beyond the planar limit in matrix models. Currently I'm trying to understand and reproduce the results of:

Antisymmetric Wilson loops in $\mathcal N \geq 4$ SYM beyond the planar limit. J Gordon. arXiv:1708.05778.

At some point the author claims that we can determine $\rho_1$ from $\mathcal{W}_1$ (see page 12). Where $\rho$ is the eigenvalue density and $\mathcal{W}$ is the resolvent.

Following him, I want to get equation (3.40) (where $\mathcal{P}$ denotes the Cauchy principal value):

$$\rho_1(x)=\frac{1}{2\pi^2\sqrt{\lambda-x^2}}\int d\omega \mathcal{P}\phi(\omega)\frac{\sqrt{\lambda-\omega^2}}{(x-\omega)}$$

using (3.47), (3.48) and (3.19): \begin{align} \mathcal{W}_1(p)&=\frac{-1}{2\sqrt{p^2-\lambda}}\int_{-1}^1d\omega \frac{\sqrt{\omega^2-\lambda}}{(p-\omega)}\phi(\omega) \tag{3.19}\\ \rho(x)&=\frac{1}{2\pi i}(\mathcal{W}(x-i\epsilon)-\mathcal{W}(x+i\epsilon))\tag{3.47}\\ \frac{1}{x\pm i\epsilon}&=\mathcal{P}\left(\frac{1}{x}\right) \mp i\pi\delta(x) \tag{3.48} \end{align}

NOTE: $\phi(\omega)=(1+e^{z-\omega})^{-1}$ See equation (3.7)

So, since we have the expansions

\begin{align} \rho(x)&=\sum_{n=0}^\infty N^{-n}\rho_n(x) \\ \mathcal{W}(p)&=\sum_{n=0}^\infty N^{-n}\mathcal{W}_n(p) \end{align}

I guess that

$$\rho_1(x)=\frac{1}{2\pi i}(\mathcal{W}_1(x-i\epsilon)-\mathcal{W}_1(x+i\epsilon))$$

So now: \begin{align} \mathcal{W}_1(x\pm i\epsilon)&=\frac{-1}{2\sqrt{x^2-\lambda}}\int_{-1}^1d\omega \frac{\sqrt{\omega^2-\lambda}}{(x-\omega\pm i\epsilon)}\phi(\omega) \\&= \frac{-1}{2\sqrt{x^2-\lambda}}\int_{-1}^1d\omega \phi(\omega)\sqrt{\omega^2-\lambda}\left[\mathcal{P}\left(\frac{1}{x-\omega}\right) \mp i\pi\delta(x-\omega)\right] \\&= \frac{-1}{2\sqrt{x^2-\lambda}}\int_{-1}^1d\omega \mathcal{P}\left(\frac{\phi(\omega)\sqrt{\omega^2-\lambda}}{x-\omega}\right) \pm \frac{i\pi}{2\sqrt{x^2-\lambda}}\int_{-1}^1d\omega \phi(\omega)\sqrt{\omega^2-\lambda} \delta(x-\omega) \\&= \frac{-1}{2\sqrt{x^2-\lambda}}\int_{-1}^1d\omega \mathcal{P}\left(\frac{\phi(\omega)\sqrt{\omega^2-\lambda}}{x-\omega}\right) \pm \frac{i\pi}{2}\phi(x) \end{align}

If we now calculate $\mathcal{W}_1(x-i\epsilon)-\mathcal{W}_1(x+i\epsilon)$ we get $$\mathcal{W}_1(x-i\epsilon)-\mathcal{W}_1(x+i\epsilon)=-i\pi\phi(x)$$ and we are not getting the desired result.

But if we find $\mathcal{W}_1(x-i\epsilon)+\mathcal{W}_1(x+i\epsilon)$ instead, we get something very similar to (3.49).

What is wrong with my calculations?

• When you write the expresion for $\mathcal{W}_1(x\pm i\epsilon)$, why $p^2=x^2$? Should not $p^2$ be $x^2+\epsilon^2\pm 2i\epsilon$. – Anthonny Feb 13 '18 at 20:02
• @Anthonny Yes but in the limit $\epsilon\to 0$ you can neglect it. But under the integral things are different, and you have to use (3.49) – Pathy Feb 14 '18 at 18:07