This is a straightforward application of the steepest descent method. So, we have the integral (taken from the lectures you cite)
QN=√NβJ2π∫∞−∞dμe[Nq(βJ,βb,μ)]
being
q(βJ,βb,μ)=ln{2cosh[β(Jμ+b)]}−βJμ22.
Now, you will have, by applying steepest descent method to the integral,
∂q∂μ=βJtanh[β(Jμ+b)]−βJμ=0.
Let us call μs the solution of this equation and expand the argument of the exponential around this value. You will get
q(βJ,βb,μ)=q(βJ,βb,μs)−Jβ2[1−Jβ(1−μ2s)](μ−μs)2.
This shows that
QN≈eNq(βJ,βb,μs)√NβJ2π∫∞−∞dμe−NJβ2[1−Jβ(1−μ2s)](μ−μs)2
i.e.
QN≈eNq(βJ,βb,μs)√11−Jβ(1−μ2s)
Note that the N factor is completely removed after integration and remains just into the argument of the exponential. Eq. (16-18) follow straightforwardly.
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