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We begin by integrating over the excitation energies. The first integral, over ε$_{1}$ , for example, furnishes 

<img src='content_image/345084.jpg'>

The remaining integrals over ε$_{2}$$_{,}$ . . . , ε$_{n}$ can now be performed if we take ℜ [ β$_{1}$ ] > ℜ [ β$_{j}$ ] , j = 2 , . . . , n . The result is 

<img src='content_image/345085.jpg'>

We next perform the integrals over β$_{j}$ , j = 2 , . . . , n . To do this, we use the fact that the function exp [ − β$_{j}$f ∗ j ( β$_{j}$ ) ] is well behaved (no singularities) for ℜ [ β$_{j}$ ] > 0 . In particular, we note that as ℜ [ β$_{j}$ ] → ∞ , exp [ − β$_{j}$f ∗ j ( β$_{j}$ ) ] → g$_{j}$ , where g$_{j}$ is the ground-state degeneracy of fragment j . We can thus close the β$_{j}$ contour to the right, to obtain 

<img src='content_image/345086.jpg'>

since ℜ [ β$_{1}$ ] > ℜ [ β$_{j}$ ] . After performing the β$_{j}$ integrals, j = 2 , . . . , n , we have (taking β$_{1}$ → β ) 

<img src='content_image/345087.jpg'>

We next perform the integrals over the momenta. Using the integral representation of the function, we have 

<img src='content_image/345088.jpg'>

