Stationary phase, Laplace’s method, and the Fourier transform for Gaussian integrals

Jordan Bell
July 28, 2015

1 Critical points

Let U be a nonempty open subset of n and let ϕ:U be smooth. Then ϕ:U(n;)=(n)*. For each xU, gradϕ(x) is the unique element of n satisfying11 1


and gradϕ:Un is itself smooth. Hessϕ:U(n;n) is the derivative of gradϕ. One checks that


and (Hessϕ(x))*=Hessϕ(x).

We call pU a critical point of ϕ when gradϕ(p)=0, and we denote the set of critical points of ϕ by Cϕ. For pCϕ and λ let v(p,λ) denote the dimension of the kernel of Hessϕ(p)-λ, and we then define the Morse index of p to be


In other words, mϕ(p) is the number of negative eigenvalues of Hessϕ(p) counted according to geometric multiplicity. We say that pCϕ is nondegenerate when Hessϕ(p)(n;n) is invertible.

For A(n;n) self-adjoint and for λ, let v(λ) be the dimension of the kernel of A-λ. Let ν+=λ>0v(λ), let ν-=λ<0v(λ), and let ν0=v(0). Because A is self-adjoint, ν++ν-+ν0=n. We define the signature of A as sgn(A)=ν+-ν-. In other words, sgn(A) is the number of positive eigenvalues of A counted according to geometric multiplicity minus the number of negative eigenvalues of A counted according to geometric multiplicity.22 2 cf. Sylvester’s law of inertia,

We can connect the notions of Morse index and signature. For pCϕ, write A=Hessϕ(p). For p to be a nondegenerate critical point means that A is invertible and because n is finite-dimensional this is equivalent to ν0=0. Then ν+=n-ν- which yields sgn(A)=n-2ν-=n-2mϕ(p).

The Morse lemma33 3 Serge Lang, Differential and Riemannian Manifolds, p. 182, chapter VII, Theorem 5.1. states that if 0 is a nondegenerate critical point of ϕ then there is an open subset V of U with 0V and a C-diffeomorphism Φ:VV, Φ(0)=0, such that


2 Stationary phase

Let U be a nonempty connected open subset of n, and let a,ϕ:U be smooth functions such that a has compact support. Suppose that each pCϕsuppa is nondegenerate.44 4 In particular, ϕ is called a Morse function if it has no degenerate critical points, and in this case of course each pCϕsuppa is nondegenerate. The stationary phase approximation states that

Ua(x)eitϕ(x)𝑑x =pCϕsuppa(2πt)n/2eiπsgn(Hessϕ(p))4|detHessϕ(p)|1/2eitϕ(p)a(p)

as t.55 5 Liviu Nicolaescu, An Invitation to Morse Theory, second ed., p. 183, Proposition 3.88.

Let A(n;n) be self-adjoint and invertible and define


We calculate gradϕ(x)=Ax, so Cϕ={0}. The Hessian of ϕ is Hessϕ(x)=A, and because A is invertible, 0 is indeed a nondegenerate critical point of ϕ. Thus we have the following.

Theorem 1.

For a nonempty connected open subset of Rn and for smooth functions a,ϕ:UR such that a has compact support and such that each pCϕ is nondegenerate,


as t.

3 The Fourier transform

For A(n;n) self-adjoint, the spectral theorem tells us that are λ1,,λn and an orthonormal basis {v1,,vn} for n such that Avj=λjvj.

We call A(n;n) positive when it is self-adjoint and satisfies Ax,x0 for all xn. In this case, the eigenvalues of A are nonnegative, thus the signature of A is σ(A)=n. Suppose furthermore that A is invertible, and let P=(v1,,vn) and Λ=diag(λ1,,λn). Then


For ξn and t>0, using the change of variables formula with the fact that |detP|=1 and then using Fubini’s theorem,


Using66 6




and using detA=j=1nλj we have


and because


this becomes


and so, as P is invertible we get the following.

Theorem 2.

When AL(Rn;Rn) is positive and invertible, for t>0 and ξRn we have


4 Gaussian integrals

Let A(n;n) be positive and invertible and let bn. As above,

nexp(-12Ax,x+Pb,x)𝑑x =nexp(-12Λ1/22+b,y)𝑑y

which gives the following.77 7 cf. Gaussian measures on n:

Theorem 3.

If AL(Rn;Rn) is positive and invertible, then for bRn,


5 Laplace’s method

Let D be the open ball in n with center 0 and radius 1 and let S:D be smooth, attain its minimum value only at 0, and satisfy detHessS(x)>0 for all xD. Let g:D be smooth and for t>0 let


Laplace’s method88 8 Peter D. Miller, Applied Asymptotic Analysis, p. 92, Exercise 3.16 and R. Wong, Asymptotic Approximations of Integrals, p. 495, Theorem 3. tells us


as t.

Let A(n;n) be positive and invertible. Define S:D by


Then as above PTAP=Λ, with which S(x)=12PΛPTx,x=12Λ1/2PTx2. We get the following from according Laplace’s method.

Theorem 4.

Let AL(Rn;Rn) be positive and invertible and let g:DR be smooth. Then


as t.