91Ó°ÊÓ

We need to find that the diagrams cannot be written in terms of this basic integral.

For the diagram in the equation $8.18$,we have three dots, so first we multiply by $N$then $N-1$then $N-2$, and for the two lines connected them we multiply by $f_{12}{f}_{23}$and for each dot we have $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$V$}\right.\right)∫d^3{r}_i$, the symmetry factor is $2$, since we can number the first dot with $1$and third dot with $3$and vice versa so we have two ways, so write the second diagram as :

$\left(3\right)=\frac{1}{2}\frac{N\left(N-1\right)\left(N-2\right)}{V^3}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}$

let $N≈N-1≈N-2$

$\left(3\right)=\frac{1}{2}\frac{N^3}{V^3}∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}$

the center dot be $1$,so we can write the integer as follow

$\left(3\right)=\frac{1}{2}\frac{N^3}{V^3}∫d^3{r}_1{\left(∫d^{3}rf\left(r\right)\right)}^2$

Here,

$f\left(r\right)=e^{-\mathrm{β}u\left(r\right)}-1$

the result of the integral over $r$is the intensive quantity that is independent of $r_1$and $V$, because $f\left(r\right)$goes to zero when $r$is only a few times larger than the size of the molecule, so whatever the value of this integral, the remaining integral over $r_1$results the factor of $V$,

$\left(3\right)=\frac{1}{2}\frac{N^3}{V^3}{\left(∫d^{3}rf\left(r\right)\right)}^2$

We can not express the last two diagrams in equation $8.20$,but we can still determine the size, let $v$be the size of the molecule, so the first diagram will have a size of :

$\frac{N^4{v}^3}{V^3}$

and also the second one will have the same size.