Oscillatory Integrals

An oscillatory integral obtains meaning from cancellation together with a stated limiting prescription. Its integrand need not be absolutely integrable. For quadratic phases, Gaussian damping gives an explicit construction and fixes the phase of the square root. This page treats finite-dimensional integrals and free propagator kernels; existence of an infinite-dimensional path measure is a different question.

Required background. Distribution Theory Minimum explains testing against smooth functions. Free-Particle Hamiltonian fixes the spectral operator represented by the kernel.

Gaussian damping fixes the Fresnel phase

For real $a\ne0$, define

$$I(a)=\lim_{\varepsilon\downarrow0} \int_{\mathbb R}e^{-\varepsilon x^2}e^{iax^2/2}\,dx.$$

At every $\varepsilon>0$, the integral is absolutely convergent and equals

$$I_\varepsilon(a)=\sqrt{2\pi}\,(2\varepsilon-ia)^{-1/2}.$$

The square root is the analytic branch positive on the positive real axis. Taking the boundary value gives

$$I(a)=\sqrt{\frac{2\pi}{|a|}}\, e^{i\pi\operatorname{sgn}(a)/4}.$$

The phase follows from the regulator and cannot be chosen independently at each parameter value. Replacing the integrand by its modulus would give an infinite integral and discard precisely the cancellation that defines $I(a)$.

For a real symmetric invertible matrix $A$ in $d$ dimensions, orthogonal diagonalization and a common isotropic Gaussian regulator give

$$\operatorname{Os}\!\int_{\mathbb R^d} e^{ix^TAx/2}\,d^dx =(2\pi)^{d/2}|\det A|^{-1/2} e^{i\pi\operatorname{sgn}(A)/4}.$$

Here $\operatorname{sgn}(A)$ is the number of positive eigenvalues minus the number of negative ones, and $\operatorname{Os}$ records the oscillatory interpretation. A zero eigenvalue invalidates this scalar formula; the result may instead involve a distribution or a remaining unintegrated direction.

Free kernels are boundary values of damped operators

Use $\hbar=1$ and $H_0=-\alpha\Delta$ on $L^2(\mathbb R^d)$, with $\alpha>0$. For $z=\varepsilon+it$, $\varepsilon>0$, the Fourier multiplier of $e^{-zH_0}$ is $e^{-z\alpha|k|^2}$, giving the convolution kernel

$$K_z(r)=(4\pi\alpha z)^{-d/2} \exp\left[-\frac{|r|^2}{4\alpha z}\right].$$

For fixed nonzero $t$, its boundary value is the free real-time kernel. The branch is inherited from the right half-plane. Fourier-space dominated convergence gives

$$\|e^{-(\varepsilon+it)H_0}\psi-e^{-itH_0}\psi\|^2 =\int\left|e^{-\varepsilon\alpha|k|^2}-1\right|^2 |\widehat\psi(k)|^2\,d^dk\longrightarrow0.$$

The bound by $4|\widehat\psi|^2$ proves this for every $L^2$ vector. Pairing the Fourier multipliers with Schwartz tests also proves convergence of the kernels in tempered distributions. These are precise meanings of the real-time limit.

At $t=0$, the limiting kernel is $\delta(r)$, not a function obtained by putting zero into the displayed formula. For nonzero $t$, the kernel is an ordinary bounded function of $r$, but it is not integrable over all space in absolute value.

Composition and differentiation need their own justification

For positive regulator parameters, Gaussian integration proves the semigroup composition law. Removing the regulators in the bounded operators gives the real-time group law. This order of reasoning avoids applying Fubini’s theorem to an absolutely divergent product of undamped kernels.

There is no universal prohibition on ordinary kernel integration. For a Schwartz packet $\psi$, the expression $\int K_{it}(x-y)\psi(y)\,dy$ is absolutely convergent at fixed nonzero $t$, because the kernel is bounded and $\psi\in L^1$. The same claim does not hold for an arbitrary $L^2$ packet, whose evolution is defined by the continuous extension in Hilbert norm.

Likewise, a convergent regulated integral does not automatically justify differentiating its unregulated integrand or changing its cutoff prescription. One proves the desired operation at positive regulator and then controls the limit, or invokes an oscillatory-integral theorem with explicit phase and amplitude hypotheses.

Exercise: a Fresnel moment needs a prescription

Using the same Gaussian damping, evaluate the oscillatory moment $J(a)=\operatorname{Os}\int x^2e^{iax^2/2}\,dx$. Does the symmetric improper integral converge ordinarily?

Solution

At positive regulator, the Gaussian moment identity gives $J_\varepsilon(a)=I_\varepsilon(a)/(2\varepsilon-ia)$. Hence $J(a)=iI(a)/a$. On the other hand, integration by parts in the undamped integral over $[-R,R]$ produces the boundary term $2Re^{iaR^2/2}/(ia)$, while the remaining Fresnel integral has a finite limit. The growing boundary term prevents ordinary improper convergence. The finite oscillatory value is tied to the stated damping prescription; it is not an ordinary absolutely convergent moment.

References

• Hörmander, Lars, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, 2nd ed., Springer, 1990, Chapter 7 — oscillatory integrals and stationary phase.
• Simon, Barry, Functional Integration and Quantum Physics, 2nd ed., AMS Chelsea Publishing, 2005, doi:10.1090/chel/351 — Gaussian integration and functional-integral prescriptions.