# Gaussian integrals

Jordan Bell
August 29, 2015

## 1 One dimensional Gaussian integrals

For $p\in\mathbb{C}$, let11 1 Eberhard Zeidler, Quantum Field Theory I: Basics in Mathematics and Physics, p. 493, Problem 7.1.

 $h(p)=\int_{\mathbb{R}}e^{-x^{2}/2}e^{-ipx}dx.$

Then we check that

 $h^{\prime}(p)=-i\int_{\mathbb{R}}xe^{-x^{2}/2}e^{-ipx}dx=i\int_{\mathbb{R}}% \frac{d}{dx}\left(e^{-x^{2}/2}\right)e^{-ipx}dx.$

Integrating by parts yields

 $h^{\prime}(p)=-p\int_{\mathbb{R}}e^{-x^{2}/2}e^{-ipx}dx=-ph(p).$

Since $h^{\prime}(p)=-ph(p)$,22 2 cf. Einar Hille, Ordinary Differential Equations in the Complex Domain.

 $h(p)=h(0)e^{-p^{2}/2}.$

Now, using Fubini’s theorem and then polar coordinates,

 $\displaystyle h(0)^{2}$ $\displaystyle=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}e^{-x^{2}/2}dx\right)e^{% -y^{2}/2}dy$ $\displaystyle=\int_{\mathbb{R}^{2}}e^{-(x^{2}+y^{2})/2}dxdy$ $\displaystyle=\int_{0}^{\infty}\left(\int_{S^{1}}e^{-r^{2}/2}e^{-r^{2}/2}d% \sigma(\theta)\right)rdr$ $\displaystyle=2\pi\int_{0}^{\infty}re^{-r^{2}/2}dr$ $\displaystyle=2\pi,$

so

 $h(p)=(2\pi)^{1/2}e^{-p^{2}/2}.$

For $a>0$ and $p\in\mathbb{C}$, doing the change of variable $y=a^{1/2}x$,

 $\displaystyle(2\pi)^{-1/2}\int_{\mathbb{R}}e^{-ax^{2}/2}e^{-ipx}dx$ $\displaystyle=(2\pi)^{-1/2}a^{-1/2}\int_{\mathbb{R}}e^{-y^{2}/2}e^{-ipa^{-1/2}% y}dy$ $\displaystyle=(2\pi)^{-1/2}a^{-1/2}h(pa^{-1/2})$ $\displaystyle=a^{-1/2}e^{-a^{-1}p^{2}/2}.$

For $t>0$ and $m\in\mathbb{R}$, doing the change of variable $y=x-m$, and using the above with $a=t^{-2}$ and $p=0$,

 $\displaystyle(2\pi t^{2})^{-1/2}\int_{\mathbb{R}}\exp\left(-\frac{(x-m)^{2}}{2% t^{2}}\right)dx$ $\displaystyle=(2\pi)^{-1/2}t^{-1}\int_{\mathbb{R}}e^{-ay^{2}/2}dx$ $\displaystyle=t^{-1}\cdot a^{-1/2}$ $\displaystyle=1.$
###### Theorem 1.

For $a>0$ and $p\in\mathbb{C}$,

 $(2\pi)^{-1/2}\int_{\mathbb{R}}e^{-ax^{2}/2}e^{-ipx}dx=a^{-1/2}e^{-a^{-1}p^{2}/% 2}.$

For $t>0$ and $m\in\mathbb{R}$,

 $(2\pi t^{2})^{-1/2}\int_{\mathbb{R}}\exp\left(-\frac{(x-m)^{2}}{2t^{2}}\right)% dx=1.$

For $t>0$ and $x\in\mathbb{R}$, let

 $p_{t}(x)=(2\pi t^{2})^{-1/2}\exp\left(-\frac{x^{2}}{2t^{2}}\right).$

For $\phi\in\mathscr{S}(\mathbb{R})$, doing the change of variable $x=ty$,

 $\int_{\mathbb{R}}\phi(x)p_{t}(x)dx=(2\pi)^{-1/2}\int_{\mathbb{R}}\phi(ty)e^{-y% ^{2}/2}dy=\int_{\mathbb{R}}\phi(tx)p_{1}(x)dx.$

Then as $t\downarrow 0$, using the dominated convergence theorem,

 $\int_{\mathbb{R}}\phi(x)p_{t}(x)dx\to\int_{\mathbb{R}}\phi(0)p_{1}(x)dx=\phi(0).$

For $\phi\in L^{1}(\mathbb{R}^{N})$, let

 $\widehat{\phi}(\xi)=(2\pi)^{-N/2}\int_{\mathbb{R}^{N}}e^{-i\left\langle\xi,x% \right\rangle}\phi(x)dx,\qquad\xi\in\mathbb{R}^{N}.$

By Theorem 1, with $a=t^{-2}$,

 $\displaystyle\widehat{p}_{t}(\xi)$ $\displaystyle=(2\pi t^{2})^{-1/2}\cdot(2\pi)^{-1/2}\int_{\mathbb{R}}e^{-ax^{2}% /2}e^{-i\xi x}dx$ $\displaystyle=(2\pi t^{2})^{-1/2}a^{-1/2}e^{-a^{-1}\xi^{2}/2}$ $\displaystyle=(2\pi)^{-1/2}e^{-t^{2}\xi^{2}/2}.$

## 2 Moments

For $a>0$, define for $Z\in\mathbb{C}$,

 $Z(J)=a^{1/2}(2\pi)^{-1/2}\int_{\mathbb{R}}e^{-ax^{2}/2}e^{iJx}dx.$

By Theorem 1,

 $Z(J)=e^{-a^{-1}J^{2}/2}.$ (1)

By the dominated convergence theorem,

 $Z^{(n)}(J)=a^{1/2}(2\pi)^{-1/2}i^{n}\int_{\mathbb{R}}x^{n}e^{-ax^{2}/2}e^{iJx}dx,$

and so

 $a^{1/2}(2\pi)^{-1/2}\int_{\mathbb{R}}x^{n}e^{-ax^{2}/2}dx=i^{-n}\frac{dZ}{dJ}(% 0).$

From (1) we calculate

 $Z^{\prime}(J)=-a^{-1}JZ(J),\quad Z^{\prime\prime}(J)=-a^{-1}Z(J)+a^{-2}J^{2}Z(% J),$

so $Z^{\prime\prime}(0)=-a^{-1}Z(0)=-a^{-1}$, and thus for $t>0$ and $a=t^{-1}$,

 $(2\pi t)^{-1/2}\int_{\mathbb{R}}x^{2}e^{-t^{-2}x^{2}/2}dx=t,$

i.e.

 $\int_{\mathbb{R}}x^{2}p_{t}(x)dx=t.$

## 3 N-dimensional Gaussian integrals

Let $S(x)=\frac{\left\langle x,x\right\rangle}{2}$ for $x\in\mathbb{R}^{N}$. For $\chi\in\mathscr{D}(\mathbb{R}^{N})$ and $t>0$, Laplace’s method tells us that

 $\int_{\mathbb{R}^{N}}e^{-tS(x)}\chi(x)dx=(2\pi t^{-1})^{N/2}(\det\mathrm{Hess}% \,S(0))^{-1/2}e^{-tS(0)}\chi(0)(1+O(t^{-1}))$

as $t\to\infty$. Here, $\mathrm{Hess}\,S(x)=I$ for all $x$ and $S(0)=0$, so

 $\int_{\mathbb{R}^{N}}e^{-t\frac{\left\langle x,x\right\rangle}{2}}\chi(x)dx=(2% \pi t^{-1})^{N/2}\chi(0)(1+O(t^{-1}))$

as $t\to\infty$.

For $A$ an $N\times N$ matrix, we write $A>0$ if $A$ is symmetric and has positive eigenvalues. It is proved that

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

for all $\xi\in\mathbb{R}^{N}$, and

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

for all $b\in\mathbb{R}^{N}$. Let

 $Z_{A}=\int_{\mathbb{R}^{N}}e^{-\frac{1}{2}\left\langle Ax,x\right\rangle}dx=(% \det A)^{-1/2}(2\pi)^{N/2}.$

Let $\lambda_{N}$ be Lebesgue measure on $\mathbb{R}^{N}$ and let $\mu_{A}$ be the following Borel probability measure on $\mathbb{R}^{N}$:

 $d\mu_{A}(x)=\frac{1}{Z_{A}}e^{-\frac{1}{2}\left\langle Ax,x\right\rangle}d% \lambda_{N}(x)=(\det A)^{1/2}(2\pi)^{-N/2}e^{-\frac{1}{2}\left\langle Ax,x% \right\rangle}d\lambda_{N}(x).$

For $\xi\in\mathbb{R}^{N}$,

 $\displaystyle\int_{\mathbb{R}^{N}}e^{-i\left\langle\xi,x\right\rangle}d\mu_{A}% (x)$ $\displaystyle=(\det A)^{1/2}(2\pi)^{-N/2}\int_{\mathbb{R}^{N}}e^{-\frac{1}{2}% \left\langle Ax,x\right\rangle}e^{-i\left\langle\xi,x\right\rangle}d\lambda_{N% }(x)$ $\displaystyle=(\det A)^{1/2}(2\pi)^{-N/2}\cdot(\det A)^{-1/2}(2\pi)^{N/2}e^{-% \frac{1}{2}\left\langle A^{-1}\xi,\xi\right\rangle}$ $\displaystyle=e^{-\frac{1}{2}\left\langle A^{-1}\xi,\xi\right\rangle},$

and for $b\in\mathbb{R}^{N}$,

 $\displaystyle\int_{\mathbb{R}^{N}}e^{\left\langle b,x\right\rangle}d\mu_{A}(x)$ $\displaystyle=(\det A)^{1/2}(2\pi)^{-N/2}\int_{\mathbb{R}^{N}}e^{\left\langle b% ,x\right\rangle}e^{-\frac{1}{2}\left\langle Ax,x\right\rangle}d\lambda_{N}(x)$ $\displaystyle=(\det A)^{1/2}(2\pi)^{-N/2}\cdot(\det A)^{-1/2}(2\pi)^{N/2}e^{% \frac{1}{2}\left\langle A^{-1}b,b\right\rangle}$ $\displaystyle=e^{\frac{1}{2}\left\langle A^{-1}b,b\right\rangle}.$
###### Theorem 2.

For $\xi\in\mathbb{R}^{N}$,

 $\int_{\mathbb{R}^{N}}e^{-i\left\langle\xi,x\right\rangle}d\mu_{A}(x)=e^{-\frac% {1}{2}\left\langle A^{-1}\xi,\xi\right\rangle},$

and for $b\in\mathbb{R}^{N}$,

 $\int_{\mathbb{R}^{N}}e^{\left\langle b,x\right\rangle}d\mu_{A}(x)=e^{\frac{1}{% 2}\left\langle A^{-1}b,b\right\rangle}.$

Let3

 $L=L^{A}=\sum_{j,k=1}^{N}A_{j,k}^{-1}\partial_{j}\partial_{k}.$

We work out the semigroup whose infinitesimal generator is $L/2$.

###### Theorem 3.

For $f\in C^{1}(\mathbb{R}^{N})$ that is $\mu_{A}$-integrable and for $t>0$,

 $(e^{tL/2}f)(x)=\int_{\mathbb{R}^{N}}f(x-t^{1/2}y)d\mu_{A}(y),\qquad x\in% \mathbb{R}^{N}.$
###### Proof.

For $\xi\in\mathbb{R}^{N}$ define $f(x)=e^{\left\langle\xi,x\right\rangle}=e^{\xi_{1}x_{1}+\cdots+\xi_{N}x_{N}}$. On the one hand,

 $Lf=\sum_{j,k=1}^{n}A_{j,k}^{-1}\xi_{j}\xi_{k}f=\left\langle A^{-1}\xi,\xi% \right\rangle f.$

Then

 $\exp(tL/2)f=\exp\left(\frac{1}{2}t\left\langle A^{-1}\xi,\xi\right\rangle% \right)f.$

On the other hand, for $x\in\mathbb{R}^{N}$, applying Theorem 2,

 $\displaystyle\int_{\mathbb{R}^{N}}f(x-t^{1/2}y)d\mu_{A}(y)$ $\displaystyle=\int_{\mathbb{R}^{N}}e^{\left\langle\xi,x-t^{1/2}y\right\rangle}% d\mu_{A}(y)$ $\displaystyle=e^{\left\langle\lambda,x\right\rangle}\int_{\mathbb{R}^{N}}e^{% \left\langle-t^{1/2}\xi,y\right\rangle}d\mu_{A}(y)$ $\displaystyle=e^{\left\langle\lambda,x\right\rangle}e^{\frac{1}{2}\left\langle A% ^{-1}(-t^{1/2}\xi),(-t^{1/2}\xi)\right\rangle}$ $\displaystyle=e^{\frac{1}{2}t\left\langle A^{-1}\xi,\xi\right\rangle}f.$

Therefore

 $\int_{\mathbb{R}^{N}}f(x-t^{1/2}y)d\mu_{A}(y)=e^{tL/2}f.$

## 4 Concentration of measure

Let $\gamma_{N}$ be the Borel probability measure on $\mathbb{R}^{N}$ defined by

 $d\gamma_{N}(x)=(2\pi)^{-N/2}e^{-\frac{1}{2}\left\langle x,x\right\rangle}d% \lambda_{N}(x).$

We estimate the mass $\gamma_{N}$ assigns to a spherical shell about the sphere of radius $N^{1/2}$.44 4 Alexander Barvinok, Measure Concentration, http://www.math.lsa.umich.edu/~barvinok/total710.pdf, p. 5, Proposition 2.2.

###### Theorem 4.

For $\delta\geq 0$,

 $\gamma_{N}\{x\in\mathbb{R}^{N}:\left\|x\right\|^{2}\geq N+\delta\}\leq\left(% \frac{N}{N+\delta}\right)^{-N/2}e^{-\delta/2},$

and for $0<\delta\leq N$,

 $\gamma_{N}\{x\in\mathbb{R}^{N}:\left\|x\right\|^{2}\leq N-\delta\}\leq\left(% \frac{N}{N-\delta}\right)^{-N/2}e^{\delta/2}.$
###### Proof.

For $0<\lambda<1$, if $\left\|x\right\|^{2}\geq N+\delta$ then $\lambda\left\|x\right\|^{2}/2\geq\lambda(N+\delta)/2$ and then $e^{\lambda\left\|x\right\|^{2}/2}\geq e^{\lambda(N+\delta)/2}$. Hence

 $\displaystyle\gamma_{N}\{x\in\mathbb{R}^{N}:\left\|x\right\|^{2}\geq N+\delta\}$ $\displaystyle=e^{-\lambda(N+\delta)/2}\int_{\left\|x\right\|^{2}\geq N+\delta}% e^{\lambda(N+\delta)/2}d\gamma_{N}(x)$ $\displaystyle\leq e^{-\lambda(N+\delta)/2}\int_{\left\|x\right\|^{2}\geq N+% \delta}e^{\lambda\left\|x\right\|^{2}/2}d\gamma_{N}(x)$ $\displaystyle\leq e^{-\lambda(N+\delta)/2}\int_{\mathbb{R}^{N}}e^{\lambda\left% \|x\right\|^{2}/2}d\gamma_{N}(x)$ $\displaystyle=e^{-\lambda(N+\delta)/2}\cdot(2\pi)^{-N/2}\int_{\mathbb{R}^{N}}e% ^{\lambda\left\|x\right\|^{2}/2}e^{-\frac{1}{2}\left\|x\right\|^{2}}d\lambda_{% N}(x)$ $\displaystyle=e^{-\lambda(N+\delta)/2}\cdot\prod_{k=1}^{N}(2\pi)^{-1/2}\int_{% \mathbb{R}}e^{(\lambda-1)u^{2}/2}du.$

For $a=-\lambda+1>0$, we have by Theorem 1

 $(2\pi)^{-1/2}\int_{\mathbb{R}}e^{-au^{2}/2}du=a^{-1/2},$

so

 $\gamma_{N}\{x\in\mathbb{R}^{N}:\left\|x\right\|^{2}\geq N+\delta\}\leq e^{-% \lambda(N+\delta)/2}a^{-N/2}=e^{-\lambda(N+\delta)/2}(1-\lambda)^{-N/2}.$

For $\lambda=\frac{\delta}{N+\delta}$ this is

 $\gamma_{N}\{x\in\mathbb{R}^{N}:\left\|x\right\|^{2}\geq N+\delta\}\leq e^{-% \delta/2}\left(\frac{N}{N+\delta}\right)^{-N/2}.$

Let $\Sigma_{N}=\{x\in\mathbb{R}^{N}:\left\|x\right\|=N^{1/2}\}$, and let $\mu_{N}$ be the unique $SO(N)$-invariant Borel probability measure on $S^{N-1}$ (any Borel probability measure on a metric space is regular so we need not explicitly demand this to ensure uniqueness). Let $\pi_{N}:\Sigma_{N}\to\mathbb{R}$ be the projection

 $\pi_{N}(x)=\pi_{N}(x_{1},\ldots,x_{N})=x_{1},$

and let $\nu_{N}=(\pi_{N})_{*}\mu_{N}$, the pushforward measure which is itself a Borel probability measure on $\mathbb{R}$. The following theorem states that the measures $\nu_{N}$ converges strongly to the standard Gaussian measure $\gamma_{1}$.55 5 Alexander Barvinok, Measure Concentration, http://www.math.lsa.umich.edu/~barvinok/total710.pdf, p. 54, Theorem 13.2.

###### Theorem 5.

For $A$ a Borel set in $\mathbb{R}$,

 $\nu_{N}(A)\to\gamma_{1}(A)$

as $N\to\infty$.

## 5 Zeta functions

Let $A>0$, with eigenvalues $\lambda_{1},\ldots,\lambda_{N}$, counted according to multiplicity. For $s\in\mathbb{C}$, define66 6 Eberhard Zeidler, Quantum Field Theory I: Basics in Mathematics and Physics, p. 434, §7.23.3.

 $\zeta_{A}(s)=\sum_{k=1}^{N}\lambda_{k}^{-s}=\sum_{k=1}^{N}e^{-s\log\lambda_{k}}.$

The derivative of $\zeta_{A}$ is

 $\zeta_{A}^{\prime}(s)=\sum_{k=1}^{N}-\log\lambda_{k}\cdot\lambda_{k}^{-s},$

so

 $\zeta_{A}^{\prime}(0)=-\sum_{k=1}^{N}\log\lambda_{k},$

hence

 $e^{-\zeta_{A}^{\prime}(0)}=\prod_{k=1}^{N}\lambda_{k}=\det A.$
###### Theorem 6.

For $\xi\in\mathbb{R}^{N}$,

 $(2\pi)^{-N/2}\int_{\mathbb{R}^{N}}\exp\left(-\frac{1}{2}\left\langle Ax,x% \right\rangle-i\left\langle\xi,x\right\rangle\right)dx=e^{\zeta_{A}^{\prime}(0% )/2}\exp\left(-\frac{1}{2}\left\langle A^{-1}\xi,\xi\right\rangle\right).$

Let $\lambda_{k}>0$, $k\geq 1$, and let $Ae_{k}=\lambda_{k}e_{k}$, and if it makes sense let

 $\det A=\prod_{k=1}^{\infty}\lambda_{k}.$

For those complex $s$ for which the expression makes sense, let

 $\zeta_{A}(s)=\sum_{k=1}^{\infty}\lambda_{k}^{-s}=\sum_{k=1}^{\infty}e^{-s\log% \lambda_{k}}.$

Then, if the above makes sense in a neighborhood of $s=0$,

 $\zeta_{A}^{\prime}(0)=-\sum_{k=1}^{\infty}\log\lambda_{k},$

so

 $e^{-\zeta_{A}^{\prime}(0)}=\det A.$

We calculate, doing the change of variables $t=\lambda_{k}u$,

 $\displaystyle\Gamma(s)\zeta_{A}(s)$ $\displaystyle=\int_{0}^{\infty}t^{s-1}e^{-t}dt\cdot\sum_{k=1}^{\infty}\lambda_% {k}^{-s}$ $\displaystyle=\sum_{k=1}^{\infty}\lambda_{k}^{-s}\int_{0}^{\infty}t^{s-1}e^{-t% }dt$ $\displaystyle=\sum_{k=1}^{\infty}\lambda_{k}^{-s}\int_{0}^{\infty}(\lambda_{k}% u)^{s-1}e^{-\lambda_{k}u}\lambda_{k}du$ $\displaystyle=\sum_{k=1}^{\infty}\int_{0}^{\infty}u^{s-1}e^{-\lambda_{k}u}du.$

Thus

 $\zeta_{A}(s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}u^{s-1}\sum_{k=1}^{\infty}e^{% -\lambda_{k}u}du.$

For $\gamma>0$, the eigenvalues of $\gamma A$ are $\gamma\lambda_{k}$, and doing the change of variables $v=\gamma u$,

 $\displaystyle\zeta_{\gamma A}(s)$ $\displaystyle=\frac{1}{\Gamma(s)}\int_{0}^{\infty}u^{s-1}\sum_{k=1}^{\infty}e^% {-\gamma\lambda_{k}u}du$ $\displaystyle=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\gamma^{-s}v^{s-1}\sum_{k=1}% ^{\infty}e^{-\lambda_{k}v}dv$ $\displaystyle=\gamma^{-s}\zeta_{A}(s).$

Taking the derivative,

 $\zeta_{\gamma A}^{\prime}(s)=-\log\gamma\cdot\gamma^{-s}\cdot\zeta_{A}(s)+% \gamma^{-s}\gamma_{A}^{\prime}(s),$

and then

 $\zeta_{\gamma A}^{\prime}(0)=-\log\gamma\cdot\zeta_{A}(0)+\zeta_{A}^{\prime}(0).$

Then

 $\det(\gamma A)=e^{-\zeta_{\gamma A}^{\prime}(0)}=e^{\log\gamma\cdot\zeta_{A}(0% )-\zeta_{A}^{\prime}(0)}=\gamma^{\zeta_{A}(0)}\det A.$