exchange integral at large distances I want to ask a question about the exchange integral $\ce{H_{AB}}$ - 

If I consider an exchange integral $\ce{H_{AB}}$ the following exchange integral can be evaluated:



so $\ce{S_{AB}}$ is $0$ at $\ce{R = \infty}$

but in this presentation I was shown today, at large values of R, $\ce{H_{AB}}$ is preportional to the overap integral $\ce{S_{AB}}$. 

Surely at large distances and towards infinity, as $\ce{S_{AB}}$ goes to zero, $\ce{H_{AB}}$ also goes to zero too. 

Where does the preportionality

$$H_{AB} \propto S_{AB}$$ 

come from then using these approimations when R is large?
The author of the presentation is talking about large, but not yet very large/infinite distance here. Note the annotations:
$$
\frac{e^2}{4 \pi \epsilon R} = k/R
$$
when $R \rightarrow \infty$: slowly to zero



$$
S_{AB}
$$
when $R \rightarrow \infty$: to zero



$$
\left&lt; 1s_A\left| k/r_A \right| 1s_B\right&gt;
$$

when $R \rightarrow \infty$: rapidly to zero



So there is some regime somewhere, in which the change of the entire expression is dominated by the change in $S_{AB}$. The verification of the correctness of the different notations is apparently left to the reader ...