Third-order virial coefficient for hard spheres Consider the virial expansion for a system of hard spheres with diameter $\sigma$. As shown in Table 8.2 on page 407 in this reference, the third-order virial coefficient is $B_3 = 5 / 18 \pi^2\sigma^6$. 
Do you know how to prove this? Do you have a reference for this? 
 A: The calculation is given in the same reference you quote, at page 425 (section 8.9.1). The reference quotes results from Appendix A of the same book, which you find on the same website.
In particular, the third virial coefficient is given by eq. $8.10\textrm{b}$:
$$B_3(T) = -\frac{1}{3V}\int f_{12}f_{13}f_{23} \, d\textbf{r}_1 d\textbf{r}_2 d\textbf{r}_3$$
This is general, and shown in any statistical physics textbook. As for the notation, $f_{ij}=f(|\textbf{r}_i-\textbf{r}_j|)$ where  $f(r)=\exp{[-\beta U(r)]}-1$ is called the Mayer f function for the potential $U(r)$.
The non-trivial step is to reduce this integral to a much nicer form by using the method explained in the appendix, section $A.11$. You apply the convolution result once and then the remaining part is a Fourier transform (it's easy to mess up, you can warm up with the $1\textrm{D}$ case). One gets:
$$B_3(T) = -\frac{1}{3} \int \frac{d^3 \textbf{k}}{(2\pi)^3} \widetilde{f}(\textbf{k})^3 $$
where $\widetilde f$ is the Fourier Transform of $f$. 
For hard spheres, 
$$f(r)=\begin{cases}-1 & r<\sigma\\ 0 &r>\sigma\end{cases}$$
Therefore after some pain, using the result on $3\mathrm{D}$ functions depending only on the modulus (section $A.9$), one gets:
$$\widetilde f(k) = 4\pi \sigma^3 \left[\frac{\cos{k\sigma}}{(k\sigma)^2}-\frac{\sin{k\sigma}}{(k\sigma)^3}\right]$$
After pulling out factors one gets the final integral:
$$B_3(T) = -\frac{32\pi}{3}\sigma^6 \int_0^\infty dx x^2 \left[\frac{\cos{x}}{x^2}-\frac{\sin{x}}{x^3} \right]^3$$
This integral can be done analytically, but is extremly painful.
In the end the hard part is getting the form of the correction right; the rest is just tricks to compute such integrals. It's probably helpful to note however that this is one of the few cases for which the correction can be computed analytically. In general one would use Monte Carlo methods. 
