 Square-zone cutoff of a saddle contour

For positive detuning inside a strict square region $|k_x|,|k_y|\le \Lambda$, substituting $k_x^2=k_y^2+\delta /A$ bounds $k_y$ by the displayed value, provided $0<\delta /A<{\Lambda }^2$. Integrating to $\Lambda$ instead extends the contour slightly outside that square. Either prescription has the same leading logarithmic singularity near zero detuning; it does not give the same exact finite-detuning result.