Integrand e^{Nf(z)} concentrates at saddle as N→∞. Compare exact n! vs Stirling.
The saddle point z₀ satisfies f′(z₀)=0. As N grows, the integrand becomes exponentially sharper — a Gaussian of width 1/√N around z₀.
For n! = ∫₀^∞ t^n e^{-t} dt = ∫ e^{N(ln t − t/N)} dt, saddle at t*=n gives Stirling's formula:
N = 10
Exact n!: -
Stirling: -
Error: -%
The steepest descent contour passes through the saddle where Im(f) = const, so the phase doesn't oscillate.