Minimal basis CI of H2 dissociation In Modern Quantum Chemistry by Szabo and Ostlund, they give an example of a minimal basis, full CI calculation of the dissociation of $\ce{H2}$ (on p.$241$ if you have the book). In explaining why this correctly describes the dissociation, they say:


  To see this, recall that as $\mathrm{R}\rightarrow\infty$, ... that all molecular orbital two-electron integrals tend to $\frac{1}{2}(\phi_{1}\phi_{1}|\phi_{1}\phi_{1})$, where $\phi_{1}$ is a hydrogenic orbital.


I could not find where they justified this statement. The closest I could find was in a previous section describing the dissociation using RHF (p.$166$) they state:


  [Besides $H^{\text{core}}_{11}$] All other integrals go to zero as $\mathrm{R}\rightarrow\infty$, except the one-center electron-electron repulsion integral $(\phi_{1}\phi_{1}|\phi_{1}\phi_{1})$.


What justifies this first statement? Is this unique to this case of a diatomic at infinite separation? These two comments seem to contradict rather than support each other.
Using Szabo and Ostlund's notation (page 85):

\begin{align}
J_{ij} &amp;= (\psi_i\psi_i|\psi_j\psi_j)&amp;
\text{and}&amp;&amp;
K_{ij} &amp;= (\psi_i\psi_j|\psi_i\psi_j)\tag1\label{integrals}
\end{align}

For minimal basis set $\ce{H2}$ (page 162):

\begin{align}
\psi_1 &amp;= [2(1+S_{12})]^{-1/2}(\phi_1+\phi_2)&amp;
\text{and}&amp;&amp;
\psi_2 &amp;= [2(1-S_{12})]^{-1/2}(\phi_1-\phi_2)\tag2\label{minimal}
\end{align}

Substitute equation $\eqref{minimal}$ into $\eqref{integrals}$, set $S_{12}$ and any integral involving both $\phi_1$ and $\phi_2$ to zero (large $R$) and you see that

$$J_{11} = K_{12} = \frac{1}{2}(\phi_1\phi_1|\phi_1\phi_1)$$

Remember that $(\phi_1\phi_1|\phi_1\phi_1)=(\phi_2\phi_2|\phi_2\phi_2)$ since you are dealing with a homonuclear diatomic.