= n ∏ j =1 ( 2 πm$_{j}$ β ) 3 / 2 1 (2 π ) 3 ∫ d $^{3}$x exp [ − m$_{0}$ β x 2 2 ] = n ∏ j =1 ( 2 πm$_{j}$ β ) 3 / 2 / ( 2 πm$_{0}$ β ) 3 / 2 , (A9) f trans l ( β ) = − 1 β [ ln ( V$_{n}$ ( m$_{N}$A$_{l}$ 2 π ℏ $^{2}$β ) 3 / 2 ) − ln ( N$_{l}$ !) N$_{l}$ ] , (A11) F$_{n}$ ( β ) = k ∑ l =1 N$_{l}$ ( f ∗ l ( β ) + f trans l ( β ) − B$_{l}$ − E$_{cl}$ ) − ( f trans 0 ( β ) − E$_{c}$$_{0}$ ) , (A12) ω$_{n}$ = 1 2 πi ∫ c + i ∞ c − i ∞ dβ exp [ − β ( F$_{n}$ ( β ) − E$_{0}$ )] , (A13) E$_{0}$ = ε$_{0}$ − B$_{0}$ . (A14) Combining the factors resulting from the momentum integrals with those due to the fragment multiplicities, we define the translational Helmholtz free energies as where we recognize that m$_{l}$ = m$_{N}$A$_{l}$ and m$_{0}$ = m$_{N}$A$_{0}$ , with m$_{N}$ the nucleon mass. We can then define the total Helmholtz free energy as where we have replaced the sum over the fragments with a sum over distinct fragments times their multiplicities, and write the density of final states as where we have defined Up to this point, we have made several definitions but no approximations. This expression is exactly equivalent to that of Eq. (9). [1] E. Fermi, Prog. Theor. Phys. 5 (1950) 570. [2] E. Fermi, Phys. Rev. 81 (1951) 683. [3] J. J. Lord, J. Fainberg, and M. Schein, Phys. Rev. 80 (1950) 970. 16 ω$_{n}$ = k ∏ l =1 1 N$_{l}$ ! 1 2 πi ∫ c + i ∞ c − i ∞ dβ n ∏ j $_{=1}$( V$_{n}$ ( m$_{j}$ 2 π ℏ $^{2}$β ) 3 / 2 ) / ( V$_{n}$ ( m$_{0}$ 2 π ℏ $^{2}$β ) 3 / 2 ) (A10) × exp [ β ( ε$_{0}$ − B$_{0}$ − E$_{c}$$_{0}$ − n ∑ j =1 ( f ∗ j ( β ) − B$_{j}$ − E$_{cj}$ ) )] . where m$_{0}$ = ∑ n j $_{=1}$m$_{j}$ . We rewrite the full integral as k ∞ n