and replace the upper limit in (24) by infinity. Thus we obtain This obviously dominates the second term 2 ρ ( τ ) on the right-hand side of (10). On the left-hand side of (10) both terms are compara- ble. The integral For the family of cooling procedures (11) we have or equivalently which leads to is gathered in the region η = O (1). Thus we can replace E ( η, τ ) by e − η and integrate up to infinity. Using Using (31) and (25) we find that the integral on the right- hand side of Eq. (29) is asymptotically we conclude that the left-hand side of (21) is asymp- totically equal to 1 − 1 / 2 = 1 / 2. Equating this to the asymptotic expression (26) for the right-hand side we ar- rive at the announced results (10) for the final density of domain walls. For a cooling scheme characterized by an arbitrary function γ ( t ), the solution is found using the same pro- cedure as in Sect. III A. Instead of the integral equation (19) we get with η = 2 ∫ t t ′ dt ′′ γ ( t $^{′′}$). The final density of domain walls satisfies Note that t ′ that appears in ρ ( t $^{′}$) and e 2( t $^{′}$− τ ) should be expressed via η by inverting Further, the upper limit in integrals in Eq. (29) is given by ζ = η ( t ′ = 0) = 2 τ dt ′′ γ ( t $^{′′}$). by ζ = η ( t ′ = 0) = 2 ∫ 0 dt ′′ γ ( t $^{′′}$). The dominant contributions to both integrals in (29) are gathered in the region where 1 − t $^{′}$/τ ≪ 1. In this region the relation between t ′ and η simplifies. We now perform the calculations for two families of cooling pro- cedures introduced in Sect. II. Computing the integral in (32) we find that the asymp- totic of the right-hand side of Eq. (29) is given by The left-hand side of Eq. (29) is asymptotically where in the last step we have used (27). Equating the RHS to the LHS we arrive at (12)–(13). For the family of cooling procedures (14) we have Computing the integral and inverting we find (keeping two leading terms) Hence the exponential factor in the integrands in (29) is and therefore the RHS of (29) is asymptotically B. Exponential Cooling IV. GENERAL COOLING PROCEDURE A. Algebraic Cooling 4 ρ ( τ ) ∫ ∞ 0 dη e − η $^{2}$/ 4 τ (2 πη ) − 1 / 2 = ρ ( τ ) Γ ( 1 4 ) τ 1 / 4 √$_{4}$$_{π}$ (26) ∫ τ 0 dη E ( η, τ ) [ I$_{0}$ ( η ) − I$_{1}$ ( η )] ∫ ∞ 0 dη e − η [ I$_{0}$ ( η ) − I$_{1}$ ( η )] = 1 (27) G$_{1}$ ( t ) = ∫ t 0 dt ′ γ ( t $^{′}$) e 2( t $^{′}$− t $^{)}$[ I$_{0}$ ( η ) − G$_{1}$ ( t $^{′}$) I$_{1}$ ( η )] (28) 1 − 1 2 ∫ ζ 0 dη e 2( t $^{′}$− τ $^{)}$[ I$_{0}$ ( η ) − I$_{1}$ ( η )] = 2 ρ ( τ ) + ∫ ζ 0 dη e 2( t $^{′}$− τ ) I$_{1}$ ( η ) ρ ( t $^{′}$) (29) η ( t $^{′}$) = 2 ∫ τ t ′ dt ′′ γ ( t $^{′′}$) (30) ρ ( τ ) ∫ ∞ 0 dη √$_{2}$$_{πη}$ exp [ − B bβ ( η 2 τ ) β η exp { − b ( η 2 τ ) − β }] e 2( t $^{′}$− τ ) = exp [ − η − B bβ ( η 2 τ ) β η exp { − b ( η 2 τ ) − β }] 1 − t ′ τ = η 2 τ + B bβ ( η 2 τ ) 1+ β exp { − b ( η 2 τ ) − β } η = 2( τ − t $^{′}$) − 2 B ∫ τ t ′ dt exp { − b (1 − t/τ ) β } LHS = 1 − 1 2 ∫ ∞ 0 dη e − η [ I$_{0}$ ( η ) − I$_{1}$ ( η )] = 1 2 RHS = √ 2 π Γ ( 3 + 2 α 2 + 2 α ) ( 1 + α A ) 1 2(1+ α ) (2 τ ) α 2(1+ α ) ρ ( τ ) ρ ( τ ) ∫ ∞ 0 dη √$_{2}$$_{πη}$ exp [ − A 1 + α η 1+ α (2 τ ) α ] (32) e 2( t $^{′}$− τ ) = exp [ − η − A 1 + α η 1+ α (2 τ ) α ] (31) 1 − t ′ τ = η 2 τ + A 1 + α ( η 2 τ ) 1+ α + . . . η = 2 ∫ τ t ′ dt ′′ γ ( t $^{′′}$) = 2( τ − t $^{′}$) − A 1 + α τ ( 1 − t ′ τ ) 1+ α + . . .