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 $\mathbb{R}^{n}$ and let $\phi:U\to\mathbb{R}$ be smooth. Then $\phi^{\prime}:U\to\mathscr{L}(\mathbb{R}^{n};\mathbb{R})=(\mathbb{R}^{n})^{*}$ . For each $x\in U$ , $\mathrm{grad}\,\phi(x)$ is the unique element of $\mathbb{R}^{n}$ satisfying¹¹ 1 http://individual.utoronto.ca/jordanbell/notes/gradienthilbert.pdf

\left\langle\mathrm{grad}\,\phi(x),y\right\rangle=\phi^{\prime}(x)(y),\qquad y% \in\mathbb{R}^{n},

and $\mathrm{grad}\,\phi:U\to\mathbb{R}^{n}$ is itself smooth. $\mathrm{Hess}\,\phi:U\to\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ is the derivative of $\mathrm{grad}\,\phi$ . One checks that

\phi^{\prime\prime}(x)(u)(v)=\left\langle\mathrm{Hess}\,\phi(x)(u),v\right% \rangle,\qquad x\in U,\quad u,v\in\mathbb{R}^{n},

and $(\mathrm{Hess}\,\phi(x))^{*}=\mathrm{Hess}\,\phi(x)$ .

We call $p\in U$ a critical point of $\phi$ when $\mathrm{grad}\,\phi(p)=0$ , and we denote the set of critical points of $\phi$ by $C_{\phi}$ . For $p\in C_{\phi}$ and $\lambda\in\mathbb{R}$ let $v(p,\lambda)$ denote the dimension of the kernel of $\mathrm{Hess}\,\phi(p)-\lambda$ , and we then define the Morse index of $p$ to be

m_{\phi}(p)=\sum_{\lambda<0}v(p,\lambda).

In other words, $m_{\phi}(p)$ is the number of negative eigenvalues of $\mathrm{Hess}\,\phi(p)$ counted according to geometric multiplicity. We say that $p\in C_{\phi}$ is nondegenerate when $\mathrm{Hess}\,\phi(p)\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ is invertible.

For $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ self-adjoint and for $\lambda\in\mathbb{R}$ , let $v(\lambda)$ be the dimension of the kernel of $A-\lambda$ . Let $\nu_{+}=\sum_{\lambda>0}v(\lambda)$ , let $\nu_{-}=\sum_{\lambda<0}v(\lambda)$ , and let $\nu_{0}=v(0)$ . Because $A$ is self-adjoint, $\nu_{+}+\nu_{-}+\nu_{0}=n$ . We define the signature of $A$ as $\mathrm{sgn}\,(A)=\nu_{+}-\nu_{-}$ . In other words, $\mathrm{sgn}\,(A)$ 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, http://individual.utoronto.ca/jordanbell/notes/principalaxis.pdf

We can connect the notions of Morse index and signature. For $p\in C_{\phi}$ , write $A=\mathrm{Hess}\,\phi(p)$ . For $p$ to be a nondegenerate critical point means that $A$ is invertible and because $\mathbb{R}^{n}$ is finite-dimensional this is equivalent to $\nu_{0}=0$ . Then $\nu_{+}=n-\nu_{-}$ which yields $\mathrm{sgn}\,(A)=n-2\nu_{-}=n-2m_{\phi}(p)$ .

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 $\phi$ then there is an open subset $V$ of $U$ with $0\in V$ and a $C^{\infty}$ -diffeomorphism $\Phi:V\to V$ , $\Phi(0)=0$ , such that

\phi(x)=\phi(0)+\frac{1}{2}\left\langle\mathrm{Hess}\,\phi(0)(\Phi(x)),\Phi(x)% \right\rangle,\qquad x\in V.

2 Stationary phase

Let $U$ be a nonempty connected open subset of $\mathbb{R}^{n}$ , and let $a,\phi:U\to\mathbb{R}$ be smooth functions such that $a$ has compact support. Suppose that each $p\in C_{\phi}\cap\mathrm{supp}\,a$ is nondegenerate.⁴⁴ 4 In particular, $\phi$ is called a Morse function if it has no degenerate critical points, and in this case of course each $p\in C_{\phi}\cap\mathrm{supp}\,a$ is nondegenerate. The stationary phase approximation states that

	$\displaystyle\int_{U}a(x)e^{it\phi(x)}dx$	$\displaystyle=\sum_{p\in C_{\phi}\cap\mathrm{supp}\,a}\left(\frac{2\pi}{t}% \right)^{n/2}\frac{e^{\frac{i\pi\mathrm{sgn}\,(\mathrm{Hess}\,\phi(p))}{4}}}{\|% \det\mathrm{Hess}\,\phi(p)\|^{1/2}}e^{it\phi(p)}a(p)$
		$\displaystyle+O(t^{-\frac{n}{2}-1})$

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

Let $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ be self-adjoint and invertible and define

\phi(x)=\frac{1}{2}\left\langle Ax,x\right\rangle,\qquad x\in U.

We calculate $\mathrm{grad}\,\phi(x)=Ax$ , so $C_{\phi}=\{0\}$ . The Hessian of $\phi$ is $\mathrm{Hess}\,\phi(x)=A$ , and because $A$ is invertible, $0$ is indeed a nondegenerate critical point of $\phi$ . Thus we have the following.

Theorem 1.

For a nonempty connected open subset of $\mathbb{R}^{n}$ and for smooth functions $a,\phi:U\to\mathbb{R}$ such that $a$ has compact support and such that each $p\in C_{\phi}$ is nondegenerate,

\int_{U}a(x)e^{\frac{1}{2}\left\langle Ax,x\right\rangle}dx=\left(\frac{2\pi}{% t}\right)^{n/2}\frac{e^{\frac{i\pi\mathrm{sgn}\,(A)}{4}}}{|\det A|^{1/2}}e^{% \frac{1}{2}it\left\langle Ap,p\right\rangle}a(p)+O(t^{-\frac{n}{2}-1})

as $t\to\infty$ .

3 The Fourier transform

For $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ self-adjoint, the spectral theorem tells us that are $\lambda_{1},\ldots,\lambda_{n}\in\mathbb{R}$ and an orthonormal basis $\{v_{1},\ldots,v_{n}\}$ for $\mathbb{R}^{n}$ such that $Av_{j}=\lambda_{j}v_{j}$ .

We call $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ positive when it is self-adjoint and satisfies $\left\langle Ax,x\right\rangle\geq 0$ for all $x\in\mathbb{R}^{n}$ . In this case, the eigenvalues of $A$ are nonnegative, thus the signature of $A$ is $\sigma(A)=n$ . Suppose furthermore that $A$ is invertible, and let $P=(v_{1},\ldots,v_{n})$ and $\Lambda=\mathrm{diag}(\lambda_{1},\ldots,\lambda_{n})$ . Then

P^{T}AP=\Lambda,\qquad\Lambda^{1/2}=\mathrm{diag}(\lambda_{1}^{1/2},\ldots,% \lambda_{n}^{1/2}),\qquad A^{1/2}=P\Lambda^{1/2}P^{T}.

For $\xi\in\mathbb{R}^{n}$ and $t>0$ , using the change of variables formula with the fact that $|\det P|=1$ and then using Fubini’s theorem,

\begin{split}&\displaystyle\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}t\left% \langle Ax,x\right\rangle-i\left\langle P\xi,x\right\rangle\right)dx\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}t\left% \langle\Lambda^{1/2}P^{T}x,\Lambda^{1/2}P^{T}x\right\rangle-i\left\langle P\xi% ,x\right\rangle\right)dx\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}t\left% \langle\Lambda^{1/2}P^{T}Py,\Lambda^{1/2}P^{T}Py\right\rangle-i\left\langle P% \xi,Py\right\rangle\right)|\det P|dy\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}t\left% \|\Lambda^{1/2}y\right\|^{2}-i\left\langle\xi,y\right\rangle\right)dy\\ \displaystyle=&\displaystyle\prod_{j=1}^{n}\int_{\mathbb{R}}\exp\left(-\frac{1% }{2}t\lambda_{j}y_{j}^{2}-i\xi_{j}y_{j}\right)dy_{j}.\end{split}

Using⁶⁶ 6 http://individual.utoronto.ca/jordanbell/notes/bochnertheorem.pdf

\int_{\mathbb{R}}e^{-ax^{2}+bx+c}dx=\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^{2}}% {4a}+c\right),\qquad\mathrm{Re}\,a>0,b,c\in\mathbb{C},

gives

\int_{\mathbb{R}}\exp\left(-\frac{1}{2}t\lambda_{j}y_{j}^{2}-i\xi_{j}y_{j}% \right)dy_{j}=\frac{1}{\lambda_{j}^{1/2}}\left(\frac{2\pi}{t}\right)^{1/2}\exp% \left(-\frac{\xi_{j}^{2}}{2t\lambda_{j}}\right),

and using $\det A=\prod_{j=1}^{n}\lambda_{j}$ we have

\begin{split}&\displaystyle\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}t\left% \langle Ax,x\right\rangle-i\left\langle P\xi,x\right\rangle\right)dx\\ \displaystyle=&\displaystyle\prod_{j=1}^{n}\frac{1}{\lambda_{j}^{1/2}}\left(% \frac{2\pi}{t}\right)^{1/2}\exp\left(-\frac{\xi_{j}^{2}}{2t\lambda_{j}}\right)% \\ \displaystyle=&\displaystyle(\det A)^{-1/2}\left(\frac{2\pi}{t}\right)^{n/2}% \exp\left(-\frac{1}{2t}\sum_{j=1}^{n}\frac{\xi_{j}^{2}}{\lambda_{j}}\right),% \end{split}

and because

\Lambda^{-1}\xi=\sum_{j=1}^{n}\frac{\xi_{j}}{\lambda_{j}}e_{j},\qquad\left% \langle\Lambda^{-1}\xi,\xi\right\rangle=\sum_{j=1}^{n}\frac{\xi_{j}^{2}}{% \lambda_{j}}

this becomes

\begin{split}&\displaystyle\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}t\left% \langle Ax,x\right\rangle-i\left\langle P\xi,x\right\rangle\right)dx\\ \displaystyle=&\displaystyle(\det A)^{-1/2}\left(\frac{2\pi}{t}\right)^{n/2}% \exp\left(-\frac{1}{2t}\left\langle\Lambda^{-1}\xi,\xi\right\rangle\right)\\ \displaystyle=&\displaystyle(\det A)^{-1/2}\left(\frac{2\pi}{t}\right)^{n/2}% \exp\left(-\frac{1}{2t}\left\langle A^{-1}P\xi,P\xi\right\rangle\right),\end{split}

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

Theorem 2.

When $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ is positive and invertible, for $t>0$ and $\xi\in\mathbb{R}^{n}$ we have

\begin{split}&\displaystyle\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}t\left% \langle Ax,x\right\rangle-i\left\langle\xi,x\right\rangle\right)dx\\ \displaystyle=&\displaystyle(\det A)^{-1/2}\left(2\pi t^{-1}\right)^{n/2}\exp% \left(-\frac{1}{2t}\left\langle A^{-1}\xi,\xi\right\rangle\right).\end{split}

4 Gaussian integrals

Let $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ be positive and invertible and let $b\in\mathbb{R}^{n}$ . As above,

	$\displaystyle\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}\left\langle Ax,x% \right\rangle+\left\langle Pb,x\right\rangle\right)dx$	$\displaystyle=\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}\left\\|\Lambda^{1/2}% \right\\|^{2}+\left\langle b,y\right\rangle\right)dy$
		$\displaystyle=\prod_{j=1}^{n}\int_{\mathbb{R}}\exp\left(-\frac{1}{2}\lambda_{j% }y_{j}^{2}+b_{j}y_{j}\right)dy_{j}$
		$\displaystyle=\prod_{j=1}^{n}\frac{(2\pi)^{1/2}}{\lambda_{j}^{1/2}}\exp\left(% \frac{b_{j}^{2}}{2\lambda_{j}}\right)$
		$\displaystyle=(\det A)^{-1/2}(2\pi)^{n/2}\exp\left(\frac{1}{2}\sum_{j=1}^{n}% \frac{b_{j}^{2}}{\lambda_{j}}\right)$
		$\displaystyle=(\det A)^{-1/2}(2\pi)^{n/2}\exp\left(\frac{1}{2}\left\langle A^{% -1}Pb,Pb\right\rangle\right),$

which gives the following.⁷⁷ 7 cf. Gaussian measures on $\mathbb{R}^{n}$ : http://individual.utoronto.ca/jordanbell/notes/gaussian.pdf

Theorem 3.

If $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ is positive and invertible, then for $b\in\mathbb{R}^{n}$ ,

\int_{\mathbb{R}^{n}}\exp\left(-\frac{1}{2}\left\langle Ax,x\right\rangle+% \left\langle b,x\right\rangle\right)dx=(\det A)^{-1/2}(2\pi)^{n/2}\exp\left(% \frac{1}{2}\left\langle A^{-1}b,b\right\rangle\right).

5 Laplace’s method

Let $D$ be the open ball in $\mathbb{R}^{n}$ with center $0$ and radius $1$ and let $S:D\to\mathbb{R}$ be smooth, attain its minimum value only at $0$ , and satisfy $\det\mathrm{Hess}\,S(x)>0$ for all $x\in D$ . Let $g:D\to\mathbb{R}$ be smooth and for $t>0$ let

J(t)=\int_{D}e^{-tS(x)}g(x)dx.

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

J(t)=(2\pi t^{-1})^{n/2}(\det\mathrm{Hess}\,S(0))^{-1/2}e^{-tS(0)}g(0)(1+O(t^{% -1}))

as $t\to\infty$ .

Let $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ be positive and invertible. Define $S:D\to\mathbb{R}$ by

S(x)=\frac{1}{2}\left\langle Ax,x\right\rangle.

Then as above $P^{T}AP=\Lambda$ , with which $S(x)=\frac{1}{2}\left\langle P\Lambda P^{T}x,x\right\rangle=\frac{1}{2}\left\|% \Lambda^{1/2}P^{T}x\right\|^{2}$ . We get the following from according Laplace’s method.

Theorem 4.

Let $A\in\mathscr{L}(\mathbb{R}^{n};\mathbb{R}^{n})$ be positive and invertible and let $g:D\to\mathbb{R}$ be smooth. Then

J(t)=(2\pi t^{-1})^{n/2}(\det A)^{-1/2}g(0)(1+O(t^{-1})),

as $t\to\infty$ .