\chapter{Integration weights for nondegenerate denominators}
\label{chap:osod}
In this appendix we will derive the equations for the integration weight
factors associated with computing the SSOD, OSDD and OSOD contributions to the
transport function (see \article\ref{sec:comp-phi}).  The integral that we
want to do is
\begin{equation}
w_i(z) = 6\int_{\Delta}
{\left[(1-a-b-c)\delta_{i1}+a\delta_{i2}+b\delta_{i3}+c\delta_{i4}\right]
da db dc
\over 
\left[z-E^p_1-aE^p_{21}-bE^p_{31}-cE^p_{41}\right]
\left[z-E^q_1-aE^q_{21}-bE^q_{31}-cE^q_{41}\right]
}
\end{equation}
where $\Delta = \{(a,b,c)|a,b,c\geq 0, a+b+c\leq 1\}$.  Note that here $E^q$
might actually be the complex conjugate of the eigenvalues of the Hamiltonian.
The only thing we assume is that the two denominators are different. To
simplify notation here we define: 
\begin{equation}
  \alpha_i = E^p_i \qquad\mbox{and}\qquad \gamma_i = E^q_i
\end{equation}
and then the integral that we need to do here is the following:
\begin{equation}
w_i(z) = 6\int_{\Delta}
{\left[(1-a-b-c)\delta_{i1}+a\delta_{i2}+b\delta_{i3}+c\delta_{i4}\right]
da db dc
\over 
\left[\alpha_1+a\alpha_{21}+b\alpha_{31}+c\alpha_{41}\right]
\left[\gamma_1+a\gamma_{21}+b\gamma_{31}+c\gamma_{41}\right]
}.
\end{equation}
Note that since we assume that the energies $\alpha_i$ and $\gamma_i$
are different we do not have to distinguish between the SS and the OS
cases.  

\section{Preliminaries}
\label{sec:prelims}
Before we start tackling these integrals let us define for later convenience
the following functions and symbols:
\begin{eqnarray}
\logdiff(x,p_i,p_j;p_k) &=& 
\log[p_i+x(p_k-p_i)]-\log[p_j+x(p_k-p_j)]
\\
\lx{n}(p_i,p_j;p_k) &=& \int_{0}^{1}dx x^n
\logdiff(x,p_i,p_j;p_k)
\\ 
\gxi(p_i,p_j;p_k,r) &=& \int_{0}^{1}dx {
\logdiff(x,p_i,p_j;p_k)
\over x-r}
\end{eqnarray}
\begin{equation}
\alpha^{i}_{jk} = {\alpha_i \over \alpha_{jk}}, \qquad
\gamma^{i}_{jk} = {\gamma_i \over \gamma_{jk}}, \qquad
D^{jk}_{\alpha\gamma} = \alpha_j\gamma_k-\alpha_k\gamma_j
\end{equation}
\begin{equation}
\Delta^{ijk}_{\alpha\gamma} = 
\begin{vmatrix}
1 & 1 & 1 \\
\alpha_{i} & \alpha_{j} & \alpha_{k} \\
\gamma_{i} & \gamma_{j} & \gamma_{k} 
\end{vmatrix} = 
\alpha_{ji}\gamma_{ki}-\alpha_{ki}\gamma_{ji}
= D_{\alpha\gamma}^{ij}+D_{\alpha\gamma}^{jk}+D_{\alpha\gamma}^{ki},
\qquad
\Gamma^{ijk}_{\alpha\gamma} = {D^{jk}_{\alpha\gamma} \over
\Delta^{ijk}_{\alpha\gamma}}.
\end{equation}
We see that both $D^{jk}_{\alpha\gamma}$ and
$\Delta^{ijk}_{\alpha\gamma}$ are antisymmetric in both $j,k$ and
$\alpha,\gamma$ and thus $\Gamma^{ijk}_{\alpha\gamma}$ is symmetric in
these indices.
We note also for future reference the following identities:
\begin{equation}
\label{eq:symbolidentities}
1+\alpha^{n}_{mn} = \alpha^{m}_{mn},
\qquad
\alpha_{ab}\left(\alpha^{b}_{ab}+\Gamma^{abc}_{\alpha\gamma}\right)
= \alpha_{bc}\Gamma^{cab}_{\alpha\gamma}
\end{equation}
Similar identities for $\gamma$ are obtained by using the symmetry
properties of the symbols.  The functions $\lx{n}$ are elementary and
easily evaluated but $\gxi$ is a rather tricky function to evaluate
and we refer the reader to appendix \ref{app:dilog} for further details.

The task at hand is now to evaluate a triple integral.  We will denote
the first one of these by $I_1$ the second one by $I_2$ and the third
one, which is equal to the desired weight factor, by $I_3$.

\section{All the corner energies are the same}
In this case the integral is trivial, all the weight factors are equal
and we get,
\begin{equation}
w_1 = {1 \over 4 \alpha_1\gamma_1}
\end{equation}

%---------- New section ----------

\section{Three corner energies are equal}
We assume here that the energies at the first three corners are equal.
The weight factors for those corners will therefore be equal to $w_1$
and the one for the last corner is $w_4$.

%---------- New subsection ----------

\subsection{Weight factor for corners 1,2 and 3}
\label{sss:3id1st}
\begin{equation}
I_1(b,c) = 6\int^{1-b-c}_{0}
{ada \over [\alpha_1+c\alpha_{41}][\gamma_1+c\gamma_{41}]}
= 3{(1-b-c)^2 \over [\alpha_1+c\alpha_{41}][\gamma_1+c\gamma_{41}]}
\end{equation}
\begin{equation}
I_2(c) = \int^{1-c}_{0}I_1(b,c)db
= {(1-c)^3 \over [\alpha_1+c\alpha_{41}][\gamma_1+c\gamma_{41}]}
= {1 \over D^{14}_{\alpha\gamma}}
\left[
{(c-1)^3 \over c+\alpha^{1}_{41}}-
{(c-1)^3 \over c+\gamma^{1}_{41}}
\right]
\end{equation}
We now use the following factorization:
\begin{equation}
(x-1)^3 = (x+a)[x^2-3x+3+a^2+a(3-x)]-(1+a)^3 
\end{equation}
and thus we have 
\begin{equation}
I_2(c) = {1 \over D^{14}_{\alpha\gamma}} 
\left\{
\alpha^{1}_{41}(\alpha^{1}_{41}+3-c)-
\gamma^{1}_{41}(\gamma^{1}_{41}+3-c)-
{(\alpha^{4}_{41})^3 \over 
c+\alpha^{1}_{41}}+ 
{(\gamma^{4}_{41})^3 \over 
c+\gamma^{1}_{41}}
\right\}.
\end{equation}
This is now easy to integrate over c and we obtain:
\begin{eqnarray}
I_3 &=& {1 \over D^{14}_{\alpha\gamma}} 
\left\{
\alpha^{1}_{41}(\alpha^{1}_{41}+5/2)-
(\alpha^{4}_{41})^3 
\log\left({\alpha_4 \over \alpha_1}\right)
\right\}
\nonumber \\
&-& {1 \over D^{14}_{\alpha\gamma}}
\left\{
\gamma^{1}_{41}(\gamma^{1}_{41}+5/2)-
(\gamma^{4}_{41})^3 
\log\left({\gamma_4 \over \gamma_1}\right)
\right\}.
\end{eqnarray}

%---------- New subsection ----------

\subsection{Weight factor for corner 4}
\label{sss:3id4th}
\begin{equation}
    I_1(b,c) = 6\int^{1-b-c}_{0} {cda \over
      [\alpha_1+c\alpha_{41}][\gamma_1+c\gamma_{41}]} = {6c(1-b-c) \over
      [\alpha_1+c\alpha_{41}][\gamma_1+c\gamma_{41}]}
\end{equation}
\begin{eqnarray}
I_2(c) &=& \int^{1-c}_{0}I_1(b,c)db
= {3c(1-c)^2 \over [\alpha_1+c\alpha_{41}][\gamma_1+c\gamma_{41}]}
\nonumber \\
&=& {3 \over D^{14}_{\alpha\gamma}}
\left[
{c(c-1)^2 \over c+\gamma^{1}_{41}}-
{c(c-1)^2 \over c+\alpha^{1}_{41}}
\right]
\end{eqnarray}
We now use the following factorization:
\begin{equation}
x(x-1)^2 = (x+a)[x^2-2x+1+a(a+2-x)]-a(1+a)^2
\end{equation}
and thus we have 
\begin{equation}
I_2(c) = {3 \over D^{14}_{\alpha\gamma}} 
\left\{ 
\gamma^{1}_{41}(\gamma^{1}_{41}+2-c)-
\alpha^{1}_{41}(\alpha^{1}_{41}+2-c)-
{\gamma^{1}_{41}(\gamma^{4}_{41}) \over 
c+\gamma^{1}_{41}}+
{\alpha^{1}_{41}(\alpha^{4}_{41})^2 \over 
c+\alpha^{1}_{41}}
\right\}.
\end{equation}
We integrate over $c$ as before and get:
\begin{eqnarray}
I_3 &=& {3 \over D^{14}_{\alpha\gamma}} 
\left\{
\gamma^{1}_{41}(\gamma^{1}_{41}+3/2)-
\gamma^{1}_{41}(\gamma^{4}_{41})^2
\log\left({\gamma_4 \over \gamma_1}\right)
\right\}
\nonumber\\
&-& {3 \over D^{14}_{\alpha\gamma}} 
\left\{
\alpha^{1}_{41}(\alpha^{1}_{41}+3/2)-
\alpha^{1}_{41}(\alpha^{1}_{41})^2
\log\left({\alpha_4 \over \alpha_1}\right)
\right\}
\end{eqnarray}

%---------- New section ----------

\section{Two pairs of corner energies are equal}
We assume here that corner 1 and 2 have equal energies and weight
factor, $w_1$ and that corner 3 and 4 have equal energies and weight
factor $w_4$.

%---------- New subsection ----------

\subsection{Weight factor for corners 1 and 2}
\label{sss:2pairs1st}
\begin{eqnarray}
I_1(b,c) &=& 6\int^{1-b-c}_{0}
{ada \over [\alpha_1+(b+c)\alpha_{41}][\gamma_1+(b+c)\gamma_{41}]}
\nonumber \\ &=& 
3{(b+c-1)^2 \over [\alpha_1+(b+c)\alpha_{41}][\gamma_1+(b+c)\gamma_{41}]}
\end{eqnarray}
We now make the following change of variables:
\begin{equation}
u = b+c, \qquad
v = b-c, \qquad
\int^{1}_{0}dc\int^{1-c}_{0}db \longrightarrow
{1 \over 2} \int^{1}_{0}du\int^{u}_{-u}dv
\end{equation}
The integral is then:
\begin{equation}
I_1(u,v) = 3{(u-1)^2 \over
[\alpha_1+u\alpha_{41}][\gamma_1+u\gamma_{41}]}
\end{equation}
Now the integral over $v$ just multiplies $I_1$ by $u$ with the factor of 2
being canceled by the Jacobian.  Then we get
\begin{eqnarray}
I_3 &=& \int^{1}_{0}{3u(u-1)^2du \over
[\alpha_1+u\alpha_{41}][\gamma_1+u\gamma_{41}]}
\nonumber \\
&=& {3 \over D^{14}_{\alpha\gamma}} 
\left\{
\gamma^{1}_{41}(\gamma^{1}_{41}+3/2)-
\gamma^{1}_{41}(\gamma^{4}_{41})^2
\log\left({\gamma_4 \over \gamma_1}\right)
\right\}
\nonumber\\
&-& {3 \over D^{14}_{\alpha\gamma}} 
\left\{
\alpha^{1}_{41}(\alpha^{1}_{41}+3/2)-
\alpha^{1}_{41}(\alpha^{4}_{41})^2
\log\left({\alpha_4 \over \alpha_1}\right)
\right\}.
\end{eqnarray}
Here we got the answer right away since we noticed that the integral
was the same as the one being done in \article\ref{sss:3id4th}.

%---------- New subsection ----------

\subsection{Weight factor for corners 3 and 4}
\label{sss:2pairs4th}
To obtain the weight factors for corners 3 and 4 we just have to take
the answer from \article\ref{sss:2pairs1st} and switch the indices 1
and 4. Thus we get:
\begin{eqnarray}
I_3 &=& {3 \over D^{41}_{\alpha\gamma}} 
\left\{
\gamma^{4}_{14}(\gamma^{4}_{14}+3/2)-
\gamma^{4}_{14}(\gamma^{1}_{14})^2
\log\left({\gamma_1 \over \gamma_4}\right)
\right\}
\nonumber\\
&-& {3 \over D^{41}_{\alpha\gamma}} 
\left\{
\alpha^{4}_{14}(\alpha^{4}_{14}+3/2)-
\alpha^{4}_{14}(\alpha^{1}_{14})^2
\log\left({\alpha_1 \over \alpha_4}\right)
\right\}.
\end{eqnarray}

%---------- New section ----------

\section{One pair of corner energies are equal}
We assume here that corner 1 and 2 have equal energies and weight
factor, $w_1$.
\subsection{Weight factor for corners 1 and 2}
\label{sss:2id1st}
\begin{eqnarray}
I_1(b,c) &=& 6\int^{1-b-c}_{0}
\hspace{-3ex}
{ada \over 
[\alpha_1+b\alpha_{31}+c\alpha_{41}]
[\gamma_1+b\gamma_{31}+c\gamma_{41}]}
\nonumber \\ &=&
{3(1-b-c)^2 \over 
[\alpha_1+b\alpha_{31}+c\alpha_{41}]
[\gamma_1+b\gamma_{31}+c\gamma_{41}]}
\end{eqnarray}
We now make the following change of variables:
\begin{equation}
u = 1-b-c, \qquad
v = c, \qquad
\int^{1}_{0}dc\int^{1-c}_{0}db \longrightarrow
\int^{1}_{0}du\int^{1-u}_{0}dv
\end{equation}
and then we can write $I_1$ on the form:
\begin{eqnarray}
I_1(u,v) &=& {3u^2 \over 
[\alpha_3+u\alpha_{13}+v\alpha_{43}]
[\gamma_1+u\gamma_{13}+v\gamma_{43}]}
\nonumber \\ &=&
{3u^2 \over 
[\alpha_3+u\alpha_{13}]\gamma_{43}-[\gamma_3+u\gamma_{13}]\alpha_{43}}
\nonumber \\ && \hspace{-2ex}\times
\left(
{\gamma_{43} \over [\gamma_3+u\gamma_{13}+v\gamma_{43}]}-
{\alpha_{43} \over [\alpha_3+u\alpha_{13}+v\alpha_{43}]}
\right)
\nonumber \\ &=& 
{3 \over \Delta^{134}_{\alpha\gamma}}
{u^2 \over u-\Gamma^{134}_{\alpha\gamma}}
\left(
{\alpha_{43} \over [\alpha_3+u\alpha_{13}+v\alpha_{43}]}-
{\gamma_{43} \over [\gamma_3+u\gamma_{13}+v\gamma_{43}]}
\right)
\end{eqnarray}
We can now easily get the second integral,
\begin{eqnarray}
I_2(u) &=& \int^{1-u}_{0}I_1(u,v)dv
\nonumber \\
&=& {3 \over \Delta^{134}_{\alpha\gamma}}
{u^2 \over u-\Gamma^{134}_{\alpha\gamma}}
\left[
\logdiff(u,\gamma_3,\gamma_4;\gamma_1)-
\logdiff(u,\alpha_3,\alpha_4;\alpha_1)
\right]
\nonumber \\
&=& {3 \over \Delta^{134}_{\alpha\gamma}}
\left(
u+\Gamma^{134}_{\alpha\gamma}+
{(\Gamma^{134}_{\alpha\gamma})^2 \over u-\Gamma^{134}_{\alpha\gamma}}
\right)
\logdiff(u,\gamma_3,\gamma_4;\gamma_1)
\nonumber \\
&-& {3 \over \Delta^{134}_{\alpha\gamma}}
\left(
u+\Gamma^{134}_{\alpha\gamma}+
{(\Gamma^{134}_{\alpha\gamma})^2 \over u-\Gamma^{134}_{\alpha\gamma}}
\right)
\logdiff(u,\alpha_3,\alpha_4;\alpha_1)
\end{eqnarray}
Now we go back to our table of integrals in \article\ref{sec:prelims}
and do the integral over $u$.
\begin{eqnarray}
I_3 &=& 
{3 \over \Delta^{134}_{\alpha\gamma}}
\left[
\lx{1}(\gamma_3,\gamma_4;\gamma_1)-
\lx{1}(\alpha_3,\alpha_4;\alpha_1)
\right]
\nonumber \\ &+& 
{3 \over \Delta^{134}_{\alpha\gamma}}\Gamma^{134}_{\alpha\gamma}
\left[
\lx{1}(\gamma_3,\gamma_4;\gamma_1)-
\lx{1}(\alpha_3,\alpha_4;\alpha_1)
\right]
\nonumber \\ &+& 
{3 \over \Delta^{134}_{\alpha\gamma}}(\Gamma^{134}_{\alpha\gamma})^2
\left[
\gxi(\gamma_{3},\gamma_4;\gamma_{1},\Gamma^{134}_{\alpha\gamma})-
\gxi(\alpha_{3},\alpha_4;\alpha_{1},\Gamma^{134}_{\alpha\gamma})
\right]
\end{eqnarray}

%---------- New subsection ----------

\subsection{Weight factor for corner 4}
\label{sss:2id4th}
\begin{eqnarray}
I_1(b,c) &=& 6\int^{1-b-c}_{0}
{cda \over 
[\alpha_1+b\alpha_{31}+c\alpha_{41}]
[\gamma_1+b\gamma_{31}+c\gamma_{41}]}
\nonumber \\ &=&
{c(1-b-c) \over 
[\alpha_1+b\alpha_{31}+c\alpha_{41}]
[\gamma_1+b\gamma_{31}+c\gamma_{41}]}
\nonumber \\ &=&
{6uv \over 
[\alpha_3+u\alpha_{13}+v\alpha_{43}]
[\gamma_1+u\gamma_{13}+v\gamma_{43}]}
\nonumber \\ &=&
{6uv \over 
[\alpha_3+u\alpha_{13}]\gamma_{43}-[\gamma_3+u\gamma_{13}]\alpha_{43}}
\nonumber \\ && \times
\left(
{\gamma_{43} \over [\gamma_3+u\gamma_{13}+v\gamma_{43}]}-
{\alpha_{43} \over [\alpha_3+u\alpha_{13}+v\alpha_{43}]}
\right)
\nonumber \\ &=& 
{6 \over \Delta^{134}_{\alpha\gamma}}
{u \over u-\Gamma^{134}_{\alpha\gamma}}
\left(
{v\alpha_{43} \over [\alpha_3+u\alpha_{13}+v\alpha_{43}]}-
{v\gamma_{43} \over [\gamma_3+u\gamma_{13}+v\gamma_{43}]}
\right)
\nonumber \\ &=& 
{6 \over \Delta^{134}_{\alpha\gamma}}
{u \over u-\Gamma^{134}_{\alpha\gamma}}
\left(
{\gamma_1+u\gamma_{13} \over [\gamma_3+u\gamma_{13}+v\gamma_{43}]}-
{\alpha_3+u\alpha_{13} \over [\alpha_3+u\alpha_{13}+v\alpha_{43}]}
\right)
\nonumber \\ &=& 
{6 \over \Delta^{134}_{\alpha\gamma}}
\left(
1+{\Gamma^{134}_{\alpha\gamma}\over u-\Gamma^{134}_{\alpha\gamma}}
\right)
{\gamma_{13} \over \gamma_{43}}
{\gamma_{43}(u+\gamma^{3}_{13}) \over [\gamma_3+u\gamma_{13}+v\gamma_{43}]}
\nonumber \\ &-& 
{6 \over \Delta^{134}_{\alpha\gamma}}
\left(
1+{\Gamma^{134}_{\alpha\gamma}\over u-\Gamma^{134}_{\alpha\gamma}}
\right)
{\alpha_{13} \over \alpha_{43}}
{\alpha_{43}(u+\alpha^{3}_{13}) \over [\alpha_3+u\alpha_{13}+v\alpha_{43}]}
\end{eqnarray}
We can now easily get the second integral,
\begin{eqnarray}
I_2(u) &=& \int^{1-u}_{0}I_1(u,v)dv
\nonumber \\ &=& %--------------------------------------------------
{6 \over \Delta^{134}_{\alpha\gamma}}
\left(
1+{\Gamma^{134}_{\alpha\gamma}\over u-\Gamma^{134}_{\alpha\gamma}}
\right)
{\alpha_{13} \over \alpha_{43}}(u+\alpha^{3}_{13})
\logdiff(u,\alpha_3,\alpha_4;\alpha_1)
\nonumber \\ &-& %--------------------------------------------------
{6 \over \Delta^{134}_{\alpha\gamma}}
\left(
1+{\Gamma^{134}_{\alpha\gamma}\over u-\Gamma^{134}_{\alpha\gamma}}
\right)
{\gamma_{13} \over \gamma_{43}}(u+\gamma^{3}_{13})
\logdiff(u,\gamma_3,\gamma_4;\gamma_1)
\nonumber \\ &=& %--------------------------------------------------
{6 \over \Delta^{134}_{\alpha\gamma}}
{\alpha_{13} \over \alpha_{43}}\left(
[\alpha^{3}_{13}+\Gamma^{134}_{\alpha\gamma}]
+u+
{[\alpha^{3}_{13}+\Gamma^{134}_{\alpha\gamma}]  
\Gamma^{134}_{\alpha\gamma}\over u-\Gamma^{134}_{\alpha\gamma}}
\right)
\logdiff(u,\alpha_3,\alpha_4;\alpha_1)
\nonumber \\ &-& %--------------------------------------------------
{6 \over \Delta^{134}_{\alpha\gamma}}
{\gamma_{13} \over \gamma_{43}}\left(
[\gamma^{3}_{13}+\Gamma^{134}_{\alpha\gamma}]
+u+
{[\gamma^{3}_{13}+\Gamma^{134}_{\alpha\gamma}]  
\Gamma^{134}_{\alpha\gamma}\over u-\Gamma^{134}_{\alpha\gamma}}
\right)
\logdiff(u,\gamma_3,\gamma_4;\gamma_1)
\end{eqnarray}
We now use the identities (\ref{eq:symbolidentities}) and see that
\begin{equation}
{\alpha_{13} \over \alpha_{43}}
\left(\alpha^{3}_{13}+\Gamma^{134}_{\alpha\gamma}\right)
= -\Gamma^{413}_{\alpha\gamma}
\qquad\mbox{and}\qquad
{\gamma_{13} \over \gamma_{43}}
\left(\gamma^{3}_{13}+\Gamma^{134}_{\alpha\gamma}\right)
= -\Gamma^{413}_{\alpha\gamma}.
\end{equation}
Then we get,
\begin{eqnarray}
I_2(u) &=& 
{6 \over \Delta^{134}_{\alpha\gamma}}
\left({\alpha_{13} \over \alpha_{43}}u-
\Gamma^{413}_{\alpha\gamma}-
{\Gamma^{413}_{\alpha\gamma}\Gamma^{134}_{\alpha\gamma}
\over u-\Gamma^{134}_{\alpha\gamma}}
\right)
\left[
\log(\alpha_{3}+u\alpha_{13})-\log(\alpha_{4}+u\alpha_{14})
\right]
\nonumber \\ &-& %--------------------------------------------------
{6 \over \Delta^{134}_{\alpha\gamma}}
\left({\gamma_{13} \over \gamma_{43}}u-
\Gamma^{413}_{\alpha\gamma}-
{\Gamma^{413}_{\alpha\gamma}\Gamma^{134}_{\alpha\gamma}
\over u-\Gamma^{134}_{\alpha\gamma}}
\right)
\left[
\log(\gamma_{3}+u\gamma_{13})-\log(\gamma_{4}+u\gamma_{14})
\right]
\end{eqnarray}

Now it is a simple matter to write down the final integral over $u$ in
terms of the functions we defined earlier,
Hence we get,
\begin{eqnarray}
I_3 &=& {6 \over \Delta^{134}_{\alpha\gamma}}
\left[
{\alpha_{13} \over \alpha_{43}}\lx{1}(\alpha_3,\alpha_4;\alpha_1)
-{\gamma_{13} \over \gamma_{43}}\lx{1}(\gamma_3,\gamma_4;\gamma_1)
\right]
\nonumber \\ &-& %--------------------------------------------------
{6\Gamma^{413}_{\alpha\gamma} \over \Delta^{134}_{\alpha\gamma}}
\left[
\lx{0}(\alpha_3,\alpha_4;\alpha_1)-\lx{0}(\gamma_3,\gamma_4;\gamma_1)
\right]
\nonumber \\ &-& %--------------------------------------------------
{6\Gamma^{413}_{\alpha\gamma}\Gamma^{134}_{\alpha\gamma}  
\over \Delta^{134}_{\alpha\gamma}}
\left[
\gxi(\alpha_{3},\alpha_4;\alpha_{1},\Gamma^{134})-
\gxi(\gamma_{3},\gamma_4;\gamma_{1},\Gamma^{134})
\right]
\end{eqnarray}

Finally to obtain the weight factor for corner 3 we just take the
result for corner 4 and switch the indices 3 and 4.
\begin{eqnarray}
I_3 &=& {6 \over \Delta^{134}_{\alpha\gamma}}
\left[
{\alpha_{14} \over \alpha_{34}}\lx{1}(\alpha_3,\alpha_4;\alpha_1)
-{\gamma_{14} \over \gamma_{34}}\lx{1}(\gamma_3,\gamma_4;\gamma_1)
\right]
\nonumber \\ &-& %--------------------------------------------------
{6\Gamma^{314}_{\alpha\gamma} \over \Delta^{134}_{\alpha\gamma}}
\left[
\lx{0}(\alpha_3,\alpha_4;\alpha_1)-\lx{0}(\gamma_3,\gamma_4;\gamma_1)
\right]
\nonumber \\ &-& %--------------------------------------------------
{6\Gamma^{314}_{\alpha\gamma}\Gamma^{134}_{\alpha\gamma}  
\over \Delta^{134}_{\alpha\gamma}}
\left[
\gxi(\alpha_{3},\alpha_4;\alpha_{1},\Gamma^{134})-
\gxi(\gamma_{3},\gamma_4;\gamma_{1},\Gamma^{134})
\right]
\end{eqnarray}


\section{All corners are different}
In this case we have not been able to derive analytical formulas for
the weight factors.  The approximate solution that we have come up
with is to express the integrals in the following fashion where
$I(a,b,c,d)$ denotes the integral in question calculated with the
energies at corners $a$, $b$, $c$ and $d$.
\begin{eqnarray}
I(1,2,3,4) &=& {1 \over 12}[
I(1,1,3,4)+I(2,2,3,4)+I(1,2,1,4)+I(3,2,3,4)
\nonumber \\ &&
+I(1,2,3,1)+I(4,2,3,4)+I(1,2,2,4)+I(1,3,3,4)
\nonumber \\ &&
+I(1,2,3,2)+I(1,4,3,4)+I(1,2,3,3)+I(1,2,4,4)].
\end{eqnarray}
The idea is to collapse two of the four corners and approximate the
integral as the weighted contribution from all such collapses.

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