advanced33 min studyLesson 438 in Mathematics

Laplace's method and steepest descent

Extract a Gaussian contribution near a nondegenerate minimum and track the analytic requirements behind a complex contour deformation.

On this page

What you’ll understand

  • Derive the leading Gaussian scale at an interior minimum.
  • State nondegeneracy and localization assumptions for a Laplace expansion.
  • Justify a simple steepest-descent contour shift without crossing singularities.

A large positive parameter in a decaying exponential concentrates an integral near the lowest value of its phase. Laplace's method replaces the phase near an isolated nondegenerate minimum by a quadratic and rescales the narrow contribution to a Gaussian. Steepest descent extends the geometry to complex contours, where the path is deformed so that the real part of the exponent decays away from a saddle. Both methods depend on global information: an overlooked endpoint, singularity, or competing saddle can invalidate a locally correct calculation.

A quadratic minimum selects an inverse-square-root width

For a clear real-variable theorem, take a smooth compactly supported amplitude on the real line and a real smooth phase near its support. Assume the phase has a unique global minimum on that support at an interior point where the amplitude is nonzero, its first derivative vanishes, and its second derivative is strictly positive. Compactness and uniqueness separate the phase from its minimum outside any small neighborhood. Thus the distant contribution is exponentially suppressed, while the nearby quadratic expansion determines the leading term.

Set the distance from the minimum equal to a new variable divided by the square root of the large parameter. The quadratic phase becomes order one and the integration measure contributes the inverse square root. Smooth expansion of the amplitude and higher phase terms then produces corrections. Under the stated smooth compact-support assumptions, the leading Gaussian term has an absolute remainder of order the parameter to minus three-halves times the exponential minimum factor. Boundary minima or degenerate critical points require different powers and constants, so check the geometry before using the formula.

∫Ra(x)e−λϕ(x)dx=e−λϕ(x0)[a(x0)2πλϕ′′(x0)+O(λ−3/2)],λ→+∞\int_{\mathbb R}a(x)e^{-\lambda\phi(x)}dx=e^{-\lambda\phi(x_0)}\left[a(x_0)\sqrt{\frac{2\pi}{\lambda\phi''(x_0)}}+O(\lambda^{-3/2})\right],\qquad \lambda\to+\infty
This interior-minimum expansion uses the smoothness, compact-support localization, unique minimum, and positive second derivative hypotheses stated in the text.↔ Scroll wide equations horizontally.

A Gaussian polynomial provides an exact check

For a quadratic phase and a polynomial amplitude, some of the asymptotic calculation can be exact. The integral of a Gaussian is the square root of pi divided by the parameter. Differentiating that identity with respect to a positive parameter, justified by Gaussian domination on compact positive parameter ranges, gives its second-moment integral. Adding those two contributions evaluates an amplitude equal to one plus the squared coordinate. This is an improper-integral example with explicit tail control rather than an unproved extension of the compact-support theorem.

The correction is smaller than the leading term by the reciprocal of twice the large parameter. Relative error must specify its denominator: the omitted correction divided by the leading approximation differs from the omitted correction divided by the exact positive integral. Exact examples are valuable for checking these distinctions and normalizing constants. They do not establish that a leading-term approximation has the same finite-parameter error for every amplitude or phase with a similar-looking minimum.

∫R(1+x2)e−λx2dx=πλ(1+12λ)(λ>0)\int_{\mathbb R}(1+x^2)e^{-\lambda x^2}dx=\sqrt{\frac\pi\lambda}\left(1+\frac1{2\lambda}\right)\qquad(\lambda>0)
Both Gaussian moments are evaluated exactly. The correction-to-leading ratio is 1/(2 lambda), while the leading approximation's relative error measured against the exact integral is 1/(2 lambda + 1).↔ Scroll wide equations horizontally.

Complex saddles require permitted contour geometry

For a holomorphic amplitude and phase, a steepest-descent contour through a nondegenerate saddle keeps the phase's imaginary part constant locally and makes its real part increase away from the saddle for an integrand with negative exponential phase. Cauchy's theorem permits a deformation only through a region where the integrand is holomorphic, with endpoint and infinite-tail contributions controlled. Crossing a pole adds residue information; crossing a branch cut changes the continuation problem. Locating a saddle alone does not establish that its steepest path belongs to the original integral.

The orientation of the local contour fixes the phase of the Gaussian square root. Several accessible saddles may contribute, and their relative importance can change as parameters move. A coalescing saddle or a saddle approaching an endpoint needs a uniform canonical approximation rather than separate nondegenerate formulas with exploding coefficients. The contour-shift practice below is deliberately entire and quadratic, so the connecting vertical edges can be bounded explicitly. That gives a complete deformation argument without concealing singularities or unverified tail behavior.

Quantify a leading Laplace approximation

Evaluate I=∫R(1+x2)e−25x2dxI=\int_{\mathbb R}(1+x^2)e^{-25x^2}dx, compare it with the leading Gaussian term, and calculate the relative error using the exact integral as denominator.

  1. The phase is x2x^2, with minimum at zero and second derivative two. The leading term is π/25=π/5≈0.3544907702\sqrt{\pi/25}=\sqrt\pi/5\approx0.3544907702.

  2. The exact second-moment correction multiplies this by 1+1/50=1.021+1/50=1.02, giving I≈0.3615805856I\approx0.3615805856.

  3. The omitted amount is the leading term divided by fifty. Dividing instead by the exact integral gives relative error (1/50)/(1+1/50)=1/51≈0.01960784(1/50)/(1+1/50)=1/51\approx0.01960784.

The result

The leading term underestimates this exact integral by about 1.96%. The two-term expression is exact for this chosen amplitude and quadratic phase.

Explore the model

Interactive lab

Interactive lab · function

A quadratic phase controls concentration width

Reset and set both a=1,x=0.2a=1,x=0.2. Predict phase 0.040.04, so λ=25\lambda=25 gives exponent size one there. Reset and set both a=1,x=0.1a=1,x=0.1; the same phase threshold occurs at λ=100\lambda=100. The instrument plots the phase x2x^2, not the exponential weight or its integral. Computing e−λx2e^{-\lambda x^2} from the height shows why quadrupling the large parameter halves the dominant width.

A quadratic phase controls concentration widthThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forNear a nondegenerate minimum, the relevant region narrows like the inverse square root of the large parameter.

Check your understanding

Which situation directly violates the nondegenerate interior-minimum assumptions of the stated Laplace formula?

Try it yourself

For real bb and λ>0\lambda>0, evaluate ∫Re−λ(x−ib)2dx\int_{\mathbb R}e^{-\lambda(x-ib)^2}dx by shifting to the horizontal line through its saddle. Explain why the connecting edges vanish.

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 integrand is entire. On a vertical edge with real part ±R and bounded imaginary part between zero and b, its modulus has a factor e to minus lambda times R squared.

Compare with the solution

Shift the contour to z=s+ibz=s+ib, where the exponent becomes −λs2-\lambda s^2. On each connecting vertical edge, ∣e−λ(z−ib)2∣=e−λR2+λ(Im⁡z−b)2|e^{-\lambda(z-ib)^2}|=e^{-\lambda R^2+\lambda(\operatorname{Im}z-b)^2}, so its bounded-length integral vanishes as R→∞R\to\infty for fixed b,λb,\lambda. Cauchy's theorem gives the exact result π/λ\sqrt{\pi/\lambda}. No poles or branch cuts are crossed.

Take it with you

  • A nondegenerate minimum gives a Gaussian scale and its normalization.
  • Error order and relative-error denominators must be stated precisely.
  • Steepest descent requires an allowed analytic contour deformation and controlled tails.

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