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Let $q_0$ be the stationary point and let $H=D^2\beta (q_0)$ be its real symmetric Hessian matrix. First suppose the stationary point is interior and nondegenerate, the oscillatory integral amplitude and oscillatory integral phase are sufficiently smooth, and boundary contributions are absent or smaller than the stationary contribution. Diagonalize $H$ by an orthogonal change of coordinates, with eigenvalues $h_1,h_2$. Locally,Scaling each principal coordinate by ${\lambda }^{-1/2}$ and evaluating its Fresnel integral gives the two-dimensional stationary-phase formula:Here $\mathrm{sgn}H$ is the number of positive eigenvalues minus the number of negative ones: $2$, $0$ or $-2$. The formula gives the leading nonzero term when $\alpha (q_0)\ne 0$. A partition of unity isolates this neighbourhood; integration by parts away from stationary points makes the remaining interior contribution smaller. For a smooth compactly supported oscillatory integral amplitude the local error is $O({\lambda }^{-2})$.

The printed assumption of a single stationary point alone is insufficient for an unconditional answer. A degenerate oscillatory integral phase $\beta =x^4+y^2$ with a smooth oscillatory integral amplitude supported near the origin has only one stationary point, but scales as ${\lambda }^{-3/4}$. Boundary terms can also contribute at order ${\lambda }^{-1}$: on the unit disk with oscillatory integral amplitude one and $\beta =x^2+y^2$,whose stationary contribution is $i\pi /\lambda$ and whose boundary contribution is $-i\pi e^{i\lambda }/\lambda$. Thus the boxed formula is the standard nondegenerate interior stationary contribution, with the stated hypotheses; the data as printed do not determine all possible leading behaviours.