\chapter{A few integrals on the Bethe lattice}
\label{app:bethe-lattice}
In this appendix I want to calculate a few integrals that appear
frequently when one is dealing with mean field theory on the infinite
dimensional Bethe lattice.  In what follows I use the notation
\begin{equation}
a = \omega+\mu-Re{\Sigma(\omega)}, \qquad b = Im{\Sigma(\omega)},
\qquad \sigma = a-ib.
\end{equation}
Remember that the bare spectral density for the infinite dimensional
Bethe lattice is given by
\begin{equation}
D(\epsilon) = {2 \over \pi}\sqrt{1-\epsilon^{2}}
\end{equation}

The first integral is the calculation of the local Green's function. 
\begin{eqnarray}
G_{loc}(\omega) &=& \int d\epsilon {D(\epsilon) \over \sigma-x} 
\quad = \quad
{2 \over \pi}\int_{-1}^{1}dx{\sqrt{1-x^{2}} \over \sigma-x} 
\nonumber \\
&=& {2 \over \pi}\int_{0}^{\pi}d\theta
\sin^{2}\theta{1 \over \sigma-\cos\theta}
\quad = \quad 
{1 \over \pi} \int_{0}^{2\pi}d\theta
\sin^{2}\theta{1 \over \sigma-\cos\theta}
\nonumber \\
&=&  K_{1}(\sigma)
\end{eqnarray}
Here I have defined
\begin{equation}
K_{n}(\sigma) = {1 \over \pi}\int_{0}^{2\pi}d\theta
\sin^{2}\theta{1 \over (\sigma-\cos\theta)^{n}}
\quad = \quad
{(-1)^{n-1} \over (n-1)!}{d^{n-1} \over d\sigma^{n-1}}K_{1}(\sigma).
\end{equation}
The second integral pops up in calculation of transport parameters.
\begin{eqnarray}
H(\omega) &=& \int d\epsilon {1 \over 2}D(\epsilon)
\rho^{2}(\epsilon,\omega)
\quad = \quad
{1 \over \pi^{3}}\int_{-1}^{1}dx\sqrt{1-x^{2}}
\left({b \over (a-x)^{2}+b^{2}}\right)^{2}
\nonumber \\
&=& {1 \over 2\pi^{3}}\int_{0}^{2\pi}d\theta\sin^{2}\theta
\left({b \over (\sigma-\cos\theta)(\sigma^{*}-\cos\theta)}\right)^{2}
\nonumber \\
&=& {-1 \over (2\pi)^{3}}\int_{0}^{2\pi}d\theta\sin^{2}\theta
\left({1\over (\sigma-\cos\theta)} - 
{1 \over (\sigma^{*}-\cos\theta)}\right)^{2}
\nonumber \\
&=& {-1 \over (2\pi)^{2}}Re K_{2}(\sigma)
+{1 \over (2\pi)^{2}b}Im K_{1}(\sigma).
\end{eqnarray}
Now let me turn to calculating the function $K_{1}(\sigma)$.
\begin{eqnarray}
K_{1}(\sigma) &=& {1 \over \pi}\int_{0}^{2\pi}d\theta
\sin^{2}\theta{1 \over \sigma-\cos\theta}
= {1 \over \pi}\oint {dz \over iz}{{-1 \over 4}(z-{1 \over z})^{2} \over 
\sigma - {1 \over 2}(z+{1 \over z})}
\nonumber \\
&=&
{1 \over 2\pi i}\oint {dz \over z^{2}}{(z^{2}-1)^{2} \over z^{2}-
2\sigma z+1}
\end{eqnarray}
The integrand has poles at 
$$
z_{0} = 0, \quad
z_{1} = \sigma + \sqrt{\sigma^{2}-1} \quad \mbox{and} \quad 
z_{2} = \sigma - \sqrt{\sigma^{2}-1} = {1 \over z_{2}}
$$
Only one of two latter poles is within the unit circle, let's say
it is $z_{1}$.
The integral is now easily done by the residue method.
\begin{equation}
K_{1}(\sigma) = {d \over dz}
\left({(z^{2}-1)^{2} \over z^{2}-
2\sigma z+1}\right)_{z = 0}
+\left(
{(z-{1 \over z})^{2} \over 2(z-\sigma)}
\right)_{z = z_{1}}
= 2(\sigma + \sqrt{\sigma^{2}-1}) = 2z_{1}.
\end{equation}
From this I now get 
\begin{equation}
K_{2} = -2{\sigma+\sqrt{\sigma^{2}-1} \over \sqrt{\sigma^{2}-1}}
\end{equation}
The original integrals are now
\begin{equation}
G_{loc}(\omega) = 2(\sigma + \sqrt{\sigma^{2}-1})
\end{equation}
and
\begin{eqnarray}
H(\omega) &=& {1 \over 2\pi^{2}}
\left(
Re{\sigma+\sqrt{\sigma^{2}-1} \over \sqrt{\sigma^{2}-1}}
+{1 \over b}Im(\sigma + \sqrt{\sigma^{2}-1}))
\right)
\nonumber \\
&=& {1 \over 4\pi^{2}}
\left(Re{G_{loc} \over {1 \over 2}G_{loc}+\Sigma-\omega-\mu}
+{Im G_{loc} \over Im \Sigma}\right)
\end{eqnarray}

The third integral comes up in calculation of energy on the Bethe
lattice.
\begin{eqnarray}
\Gamma(\omega) &=&
\int\rho(\epsilon,\omega)D(\epsilon)\epsilon d\epsilon = 
{1 \over \pi}Im\int G(\epsilon,\omega)D(\epsilon)\epsilon d\epsilon 
\nonumber \\
&=& {-2 \over \pi^{2}}Im\int_{-1}^{1}{x\sqrt{1-x^2}dx \over x-\sigma}
\end{eqnarray}
\begin{eqnarray}
2\int_{-1}^{1}{x\sqrt{1-x^2}dx \over x-\sigma} &=&
\int_{0}^{2\pi}{\cos\theta\sin^{2}\theta d\theta \over \cos\theta-\sigma} 
\nonumber \\ &=&
\oint {dz \over iz}{{1 \over 2}(z+{1 \over z})
{-1 \over 4}(z-{1 \over z})^{2} \over {1 \over 2}(z+{1 \over z}) -
\sigma}
\nonumber \\ &=&
{-1 \over 4i}\oint {dz \over z^{3}}{(z^{2}+1)(z^{2}-1)^{2} 
\over z^{2}-2\sigma z +1}
\end{eqnarray}
The last integrand has a triple pole at 0 and a single pole at
$z_{1}$.
Thus I get
\begin{eqnarray}
2\int_{-1}^{1}{x\sqrt{1-x^2}dx \over x-\sigma} &=&
{-\pi \over 2}\left(
{1 \over 2}{d^{2} \over dz^{2}}\left[
{(z^{2}+1)(z^{2}-1)^{2} \over z^{2}-2\sigma z +1}
\right]_{z = 0} + \left[
{(z+{1 \over z})(z-{1 \over z})^{2} \over 2(z-\sigma)}
\right]_{z = z_{1}}\right)
\nonumber \\ &=& 
{-\pi \over 2}(4\sigma^{2}-2+4\sigma\sqrt{\sigma^{2}-1})
= -\pi(\sigma G_{loc}-1)
\end{eqnarray}
Hence
\begin{equation}
\Gamma(\omega) = {1 \over \pi}Im(\sigma G_{loc}) 
= 2t^{2}\rho_{loc}(\omega)ReG_{loc}(\omega).
\end{equation}
Here I used the fact that
\begin{equation}
\Sigma(\omega) = \omega+\mu-t^{2}G_{loc}(\omega)-
{1 \over G_{loc}(\omega)}.
\end{equation}

