advanced33 min studyLesson 439 in Mathematics

Stationary phase and oscillatory integrals

Locate where cancellation weakens, retain the phase of the Gaussian contribution, and distinguish stationary points from endpoints and degenerate phases.

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What you’ll understand

  • Derive nonstationary cancellation using integration by parts.
  • Apply the one-dimensional nondegenerate stationary-phase formula with its phase factor.
  • Check a leading oscillatory approximation against a complex Gaussian integral.

A rapidly oscillating integrand may have large magnitude everywhere while its integral is small because neighboring contributions cancel. Stationary phase identifies locations where the phase changes slowly enough for cancellation to weaken. The leading contribution has both a magnitude and a complex phase; dropping the latter loses information needed when several waves interfere. Endpoints and degenerate critical points can contribute on different scales. A correct calculation therefore identifies the critical set, checks its geometry, and controls the rest of the integral before quoting a Gaussian-looking expression.

Away from critical points, integrate oscillation by parts

Let the amplitude be smooth with compact support and let the phase be real and smooth near it. Where the phase derivative is bounded away from zero, express the oscillatory exponential as its derivative divided by the large parameter times that phase derivative. Integration by parts moves the derivative onto the amplitude quotient. Compact support removes endpoint terms. Repeating this argument gives decay faster than every fixed inverse power when there are no stationary points on the support. The constants depend on derivative bounds and the lower bound for the phase derivative.

On a finite interval with nonzero amplitude at its endpoints, the boundary terms do not vanish. They often produce contributions of order the inverse parameter, even if no stationary point exists. This is why an interior stationary-phase remainder cannot be used unchanged for an arbitrary abruptly truncated integral. A cutoff that vanishes smoothly near artificial boundaries separates the interior calculation from endpoint effects. Absolute integration estimates alone cannot reveal this cancellation, since the integrand's magnitude need not decrease with the oscillation parameter.

∫a(x)eiλϕ(x)dx=−1iλ∫ddx(a(x)ϕ′(x))eiλϕ(x)dx(ϕ′≠0, a∈Cc∞)\int a(x)e^{i\lambda\phi(x)}dx=-\frac1{i\lambda}\int\frac d{dx}\left(\frac{a(x)}{\phi'(x)}\right)e^{i\lambda\phi(x)}dx\qquad(\phi'\ne0,\ a\in C_c^\infty)
The boundary term vanishes because the smooth amplitude is compactly supported. Iteration requires nonvanishing phase derivative on its support and controlled derivatives.↔ Scroll wide equations horizontally.

A nondegenerate stationary point gives a complex Gaussian

Assume a smooth compactly supported amplitude and a real smooth phase with finitely many stationary points on the support, all nondegenerate and in its interior. Near each point, the phase is its stationary value plus a signed quadratic term at leading order. Rescaling by the inverse square root of the large parameter gives a Fresnel-type Gaussian. Positive phase curvature contributes a factor with argument plus one-quarter pi, and negative curvature contributes the conjugate factor. The amplitudes must be summed as complex numbers before taking a final magnitude.

Under these hypotheses, the sum of leading contributions has a remainder of order the parameter to minus three-halves. A vanishing amplitude can remove a leading contribution, so the next term may matter. If the second derivative also vanishes, the quadratic scaling fails: a cubic phase has a one-third-power scale and leads to Airy-type behavior in suitable parameterized problems. Coalescing critical points require a uniform treatment; adding separate formulas whose curvature denominators diverge does not cure their breakdown.

∫a(x)eiλϕ(x)dx=∑ja(xj)eiλϕ(xj)+iπsgn⁡(ϕ′′(xj))/42πλ∣ϕ′′(xj)∣+O(λ−3/2)\int a(x)e^{i\lambda\phi(x)}dx=\sum_j a(x_j)e^{i\lambda\phi(x_j)+i\pi\operatorname{sgn}(\phi''(x_j))/4}\sqrt{\frac{2\pi}{\lambda|\phi''(x_j)|}}+O(\lambda^{-3/2})
The stated smooth compact-support and finite nondegenerate interior critical-point assumptions control the remainder as lambda tends to positive infinity.↔ Scroll wide equations horizontally.

An exact complex Gaussian checks magnitude and phase

A Gaussian amplitude is not compactly supported, so its use needs an independent tail argument or direct evaluation. For amplitude e−x2e^{-x^2} and phase x2/2x^2/2, combine the exponents into a complex Gaussian whose real quadratic coefficient stays positive. Its exact value is the square root of pi divided by the square root of 1−iλ/21-i\lambda/2, using the branch continued from the positive value at zero parameter. This selects a definite phase and avoids an arbitrary sign in the square root.

For large positive parameter, the coefficient's argument approaches minus one-half pi, so the reciprocal square root approaches argument plus one-quarter pi. The exact magnitude and phase therefore recover the stationary-phase constant. This example also shows why a real-valued magnitude estimate cannot reconstruct interference: two equal-sized contributions with different phase factors can reinforce or cancel. Keep phase conventions consistent across critical points and across any contour continuation used to evaluate them.

Check a stationary-phase constant at finite parameter

For J(λ)=∫Re−x2eiλx2/2dxJ(\lambda)=\int_{\mathbb R}e^{-x^2}e^{i\lambda x^2/2}dx, derive the exact value and compare its magnitude and phase with the leading stationary-phase prediction at λ=20\lambda=20.

  1. Combine the exponents to obtain J=π(1−iλ/2)−1/2J=\sqrt\pi(1-i\lambda/2)^{-1/2}, with the branch positive at λ=0\lambda=0. The positive real part of the Gaussian coefficient ensures convergence.

  2. The stationary point is zero with phase curvature one. Its leading contribution is eiπ/42π/λe^{i\pi/4}\sqrt{2\pi/\lambda}, whose magnitude at twenty is approximately 0.56049912160.5604991216.

  3. The exact magnitude is π/1011/4≈0.5591065665\sqrt\pi/101^{1/4}\approx0.5591065665 and its argument is 12arctan⁡(10)≈0.7355638372\tfrac12\arctan(10)\approx0.7355638372 radians. The leading phase is π/4≈0.7853981634\pi/4\approx0.7853981634, so the finite-parameter phase error is about 0.04983432620.0498343262 radians.

The result

Both magnitude and phase approach the stationary-phase prediction, but neither is exact at a finite parameter. The exact branch choice fixes the sign of the phase shift.

Explore the model

Interactive lab

Interactive lab · waves

Cancellation depends on relative phase

Reset and set both phase difference 00 degrees and amplitude ratio 11. Predict resultant amplitude two. Reset and set both phase difference 180180 degrees and amplitude ratio 11; predict complete cancellation, up to floating-point rounding. Compare the summed curve and amplitude readout. These two-wave sums illustrate phase-sensitive cancellation; they do not locate stationary points or evaluate an oscillatory integral.

Cancellation depends on relative phaseTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forStationary contributions must be combined with their phase factors, not only their magnitudes.

Check your understanding

For a positive large parameter and one nondegenerate stationary point with negative phase curvature, which Gaussian phase factor occurs?

Try it yourself

Let a smooth compactly supported amplitude equal one near zero, and consider ∫a(x)e−iλx2dx\int a(x)e^{-i\lambda x^2}dx as λ→+∞\lambda\to+\infty. Give the leading term and the stated remainder order.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The phase is minus x squared, so its second derivative is minus two. There are no boundary terms because the amplitude is smoothly compactly supported.

Compare with the solution

The only stationary point is zero, with curvature −2-2 and amplitude one. The leading term is e−iπ/4π/λe^{-i\pi/4}\sqrt{\pi/\lambda}, with an O(λ−3/2)O(\lambda^{-3/2}) remainder under the stated smoothness conditions. The negative curvature changes the phase sign but not the positive square-root magnitude.

Take it with you

  • Nonstationary cancellation is controlled by integration by parts and endpoint behavior.
  • A nondegenerate stationary point contributes a signed Gaussian phase.
  • Degeneracy and coalescence require new scales rather than divergent separate formulas.

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