Gaussian integrals

1 One dimensional Gaussian integrals

For $p\in ℂ$, let

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

Then we check that

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

Integrating by parts yields

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

Since $h^{\prime }(p)=-ph(p)$,

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

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

$h{(0)}^2$ $={\displaystyle {\int }_{ℝ}}\left({\displaystyle {\int }_{ℝ}}{e}^{-x^2/2}𝑑x\right)e^{-y^2/2}𝑑y$

$={\displaystyle {\int }_{{ℝ}^2}}{e}^{-(x^2+y^2)/2}𝑑x𝑑y$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}\left({\displaystyle {\int }_{S^1}}{e}^{-r^2/2}{e}^{-r^2/2}𝑑\sigma (\theta )\right)r𝑑r$

$=2\pi {\displaystyle {\int }_0^{\mathrm{\infty }}}r{e}^{-r^2/2}𝑑r$

$=2\pi ,$

so

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

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

${(2\pi )}^{-1/2}{\displaystyle {\int }_{ℝ}}{e}^{-a{x}^2/2}{e}^{-ipx}𝑑x$ $={(2\pi )}^{-1/2}{a}^{-1/2}{\displaystyle {\int }_{ℝ}}{e}^{-y^2/2}{e}^{-ip{a}^{-1/2}y}𝑑y$

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

$=a^{-1/2}{e}^{-a^{-1}{p}^2/2}.$

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

${(2\pi t^2)}^{-1/2}{\displaystyle {\int }_{ℝ}}\exp \left(-{\displaystyle \frac{{(x-m)}^2}{2t^2}}\right)𝑑x$ $={(2\pi )}^{-1/2}{t}^{-1}{\displaystyle {\int }_{ℝ}}{e}^{-a{y}^2/2}𝑑x$

$=t^{-1}\cdot a^{-1/2}$

$=1.$

Theorem 1.

For $a>0$ and $p\in ℂ$,

$${(2\pi )}^{-1/2}{\int }_{ℝ}{e}^{-a{x}^2/2}{e}^{-ipx}𝑑x=a^{-1/2}{e}^{-a^{-1}{p}^2/2}.$$

For $t>0$ and $m\in ℝ$,

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

For $t>0$ and $x\in ℝ$, let

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

For $\varphi \in 𝒮(ℝ)$, doing the change of variable $x=ty$,

$${\int }_{ℝ}\varphi (x)p_t(x)𝑑x={(2\pi )}^{-1/2}{\int }_{ℝ}\varphi (ty)e^{-y^2/2}𝑑y={\int }_{ℝ}\varphi (tx)p_1(x)𝑑x.$$

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

$${\int }_{ℝ}\varphi (x)p_t(x)𝑑x\to {\int }_{ℝ}\varphi (0)p_1(x)𝑑x=\varphi (0).$$

For $\varphi \in L^1({ℝ}^N)$, let

$$\widehat{\varphi }(\xi )={(2\pi )}^{-N/2}{\int }_{{ℝ}^N}{e}^{-i⟨\xi ,x⟩}\varphi (x)𝑑x,\xi \in {ℝ}^N.$$

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

${\widehat{p}}_t(\xi )$ $={(2\pi t^2)}^{-1/2}\cdot {(2\pi )}^{-1/2}{\displaystyle {\int }_{ℝ}}{e}^{-a{x}^2/2}{e}^{-i\xi x}𝑑x$

$={(2\pi t^2)}^{-1/2}{a}^{-1/2}{e}^{-a^{-1}{\xi }^2/2}$

$={(2\pi )}^{-1/2}{e}^{-t^2{\xi }^2/2}.$

2 Moments

For $a>0$, define for $Z\in ℂ$,

$$Z(J)=a^{1/2}{(2\pi )}^{-1/2}{\int }_{ℝ}{e}^{-a{x}^2/2}{e}^{iJx}𝑑x.$$

By Theorem 1,

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

By the dominated convergence theorem,

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

and so

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

From (1) we calculate

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

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

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

i.e.

$${\int }_{ℝ}{x}^2{p}_t(x)𝑑x=t.$$

3 N-dimensional Gaussian integrals

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

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

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

$${\int }_{{ℝ}^N}{e}^{-t\frac{⟨x,x⟩}{2}}\chi (x)𝑑x={(2\pi t^{-1})}^{N/2}\chi (0)(1+O(t^{-1}))$$

as $t\to \mathrm{\infty }$.

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

for all $\xi \in {ℝ}^N$, and

$${\int }_{{ℝ}^N}\exp \left(-\frac{1}{2}⟨Ax,x⟩+⟨b,x⟩\right)𝑑x={(detA)}^{-1/2}{(2\pi )}^{N/2}\exp \left(\frac{1}{2}⟨A^{-1}b,b⟩\right).$$

for all $b\in {ℝ}^N$. Let

$$Z_A={\int }_{{ℝ}^N}{e}^{-\frac{1}{2}⟨Ax,x⟩}𝑑x={(detA)}^{-1/2}{(2\pi )}^{N/2}.$$

Let ${\lambda }_N$ be Lebesgue measure on ${ℝ}^N$ and let ${\mu }_A$ be the following Borel probability measure on ${ℝ}^N$:

$$d{\mu }_A(x)=\frac{1}{Z_A}{e}^{-\frac{1}{2}⟨Ax,x⟩}d{\lambda }_N(x)={(detA)}^{1/2}{(2\pi )}^{-N/2}{e}^{-\frac{1}{2}⟨Ax,x⟩}d{\lambda }_N(x).$$

For $\xi \in {ℝ}^N$,

${\displaystyle {\int }_{{ℝ}^N}}{e}^{-i⟨\xi ,x⟩}𝑑{\mu }_A(x)$ $={(detA)}^{1/2}{(2\pi )}^{-N/2}{\displaystyle {\int }_{{ℝ}^N}}{e}^{-\frac{1}{2}⟨Ax,x⟩}{e}^{-i⟨\xi ,x⟩}𝑑{\lambda }_N(x)$

$={(detA)}^{1/2}{(2\pi )}^{-N/2}\cdot {(detA)}^{-1/2}{(2\pi )}^{N/2}{e}^{-\frac{1}{2}⟨A^{-1}\xi ,\xi ⟩}$

$=e^{-\frac{1}{2}⟨A^{-1}\xi ,\xi ⟩},$

and for $b\in {ℝ}^N$,

${\displaystyle {\int }_{{ℝ}^N}}{e}^{⟨b,x⟩}𝑑{\mu }_A(x)$ $={(detA)}^{1/2}{(2\pi )}^{-N/2}{\displaystyle {\int }_{{ℝ}^N}}{e}^{⟨b,x⟩}{e}^{-\frac{1}{2}⟨Ax,x⟩}𝑑{\lambda }_N(x)$

$={(detA)}^{1/2}{(2\pi )}^{-N/2}\cdot {(detA)}^{-1/2}{(2\pi )}^{N/2}{e}^{\frac{1}{2}⟨A^{-1}b,b⟩}$

$=e^{\frac{1}{2}⟨A^{-1}b,b⟩}.$

Theorem 2.

For $\xi \in {ℝ}^N$,

$${\int }_{{ℝ}^N}{e}^{-i⟨\xi ,x⟩}𝑑{\mu }_A(x)=e^{-\frac{1}{2}⟨A^{-1}\xi ,\xi ⟩},$$

and for $b\in {ℝ}^N$,

$${\int }_{{ℝ}^N}{e}^{⟨b,x⟩}𝑑{\mu }_A(x)=e^{\frac{1}{2}⟨A^{-1}b,b⟩}.$$

Let

$$L=L^A=\sum _{j,k=1}^N{A}_{j,k}^{-1}{\partial }_j{\partial }_k.$$

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

Theorem 3.

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

$$(e^{tL/2}f)(x)={\int }_{{ℝ}^N}f(x-t^{1/2}y)𝑑{\mu }_A(y),x\in {ℝ}^N.$$

Proof.

For $\xi \in {ℝ}^N$ define $f(x)=e^{⟨\xi ,x⟩}=e^{{\xi }_1{x}_1+\mathrm{\cdots }+{\xi }_N{x}_N}$. On the one hand,

$$Lf=\sum _{j,k=1}^n{A}_{j,k}^{-1}{\xi }_j{\xi }_{k}f=⟨A^{-1}\xi ,\xi ⟩f.$$

Then

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

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

${\displaystyle {\int }_{{ℝ}^N}}f(x-t^{1/2}y)𝑑{\mu }_A(y)$ $={\displaystyle {\int }_{{ℝ}^N}}{e}^{⟨\xi ,x-t^{1/2}y⟩}𝑑{\mu }_A(y)$

$=e^{⟨\lambda ,x⟩}{\displaystyle {\int }_{{ℝ}^N}}{e}^{⟨-t^{1/2}\xi ,y⟩}𝑑{\mu }_A(y)$

$=e^{⟨\lambda ,x⟩}{e}^{\frac{1}{2}⟨A^{-1}(-t^{1/2}\xi ),(-t^{1/2}\xi )⟩}$

$=e^{\frac{1}{2}t⟨A^{-1}\xi ,\xi ⟩}f.$

Therefore

$${\int }_{{ℝ}^N}f(x-t^{1/2}y)𝑑{\mu }_A(y)=e^{tL/2}f.$$

∎

4 Concentration of measure

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

$$d{\gamma }_N(x)={(2\pi )}^{-N/2}{e}^{-\frac{1}{2}⟨x,x⟩}d{\lambda }_N(x).$$

We estimate the mass ${\gamma }_N$ assigns to a spherical shell about the sphere of radius $N^{1/2}$.

Theorem 4.

For $\delta \ge 0$,

$${\gamma }_N\{x\in {ℝ}^N:{\parallel x\parallel }^2\ge N+\delta \}\le {\left(\frac{N}{N+\delta }\right)}^{-N/2}{e}^{-\delta /2},$$

and for ,

$${\gamma }_N\{x\in {ℝ}^N:{\parallel x\parallel }^2\le N-\delta \}\le {\left(\frac{N}{N-\delta }\right)}^{-N/2}{e}^{\delta /2}.$$

Proof.

For , if ${\parallel x\parallel }^2\ge N+\delta$ then $\lambda {\parallel x\parallel }^2/2\ge \lambda (N+\delta )/2$ and then $e^{\lambda {\parallel x\parallel }^2/2}\ge e^{\lambda (N+\delta )/2}$. Hence

${\gamma }_N\{x\in {ℝ}^N:{\parallel x\parallel }^2\ge N+\delta \}$ $=e^{-\lambda (N+\delta )/2}{\displaystyle {\int }_{{\parallel x\parallel }^2\ge N+\delta }}{e}^{\lambda (N+\delta )/2}𝑑{\gamma }_N(x)$

$\le e^{-\lambda (N+\delta )/2}{\displaystyle {\int }_{{\parallel x\parallel }^2\ge N+\delta }}{e}^{\lambda {\parallel x\parallel }^2/2}𝑑{\gamma }_N(x)$

$\le e^{-\lambda (N+\delta )/2}{\displaystyle {\int }_{{ℝ}^N}}{e}^{\lambda {\parallel x\parallel }^2/2}𝑑{\gamma }_N(x)$

$=e^{-\lambda (N+\delta )/2}\cdot {(2\pi )}^{-N/2}{\displaystyle {\int }_{{ℝ}^N}}{e}^{\lambda {\parallel x\parallel }^2/2}{e}^{-\frac{1}{2}{\parallel x\parallel }^2}𝑑{\lambda }_N(x)$

$=e^{-\lambda (N+\delta )/2}\cdot {\displaystyle \prod _{k=1}^N}{(2\pi )}^{-1/2}{\displaystyle {\int }_{ℝ}}{e}^{(\lambda -1)u^2/2}𝑑u.$

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

$${(2\pi )}^{-1/2}{\int }_{ℝ}{e}^{-a{u}^2/2}𝑑u=a^{-1/2},$$

so

$${\gamma }_N\{x\in {ℝ}^N:{\parallel x\parallel }^2\ge N+\delta \}\le e^{-\lambda (N+\delta )/2}{a}^{-N/2}=e^{-\lambda (N+\delta )/2}{(1-\lambda )}^{-N/2}.$$

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

$${\gamma }_N\{x\in {ℝ}^N:{\parallel x\parallel }^2\ge N+\delta \}\le e^{-\delta /2}{\left(\frac{N}{N+\delta }\right)}^{-N/2}.$$

∎

Let ${\mathrm{\Sigma }}_N=\{x\in {ℝ}^N:\parallel x\parallel =N^{1/2}\}$, 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 ${\pi }_N:{\mathrm{\Sigma }}_N\to ℝ$ be the projection

$${\pi }_N(x)={\pi }_N(x_1,\mathrm{\dots },x_N)=x_1,$$

and let ${\nu }_N={({\pi }_N)}_{*}{\mu }_N$, the pushforward measure which is itself a Borel probability measure on $ℝ$. The following theorem states that the measures ${\nu }_N$ converges strongly to the standard Gaussian measure ${\gamma }_1$.

Theorem 5.

For $A$ a Borel set in $ℝ$,

$${\nu }_N(A)\to {\gamma }_1(A)$$

as $N\to \mathrm{\infty }$.

5 Zeta functions

Let $A>0$, with eigenvalues ${\lambda }_1,\mathrm{\dots },{\lambda }_N$, counted according to multiplicity. For $s\in ℂ$, define

$${\zeta }_A(s)=\sum _{k=1}^N{\lambda }_k^{-s}=\sum _{k=1}^N{e}^{-s\log {\lambda }_k}.$$

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

$$e^{-{\zeta }_A^{\prime }(0)}=\prod _{k=1}^N{\lambda }_k=detA.$$

Theorem 6.

For $\xi \in {ℝ}^N$,

$${(2\pi )}^{-N/2}{\int }_{{ℝ}^N}\exp \left(-\frac{1}{2}⟨Ax,x⟩-i⟨\xi ,x⟩\right)𝑑x=e^{{\zeta }_A^{\prime }(0)/2}\exp \left(-\frac{1}{2}⟨A^{-1}\xi ,\xi ⟩\right).$$

Let ${\lambda }_k>0$, $k\ge 1$, and let $A{e}_k={\lambda }_k{e}_k$, and if it makes sense let

$$detA=\prod _{k=1}^{\mathrm{\infty }}{\lambda }_k.$$

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

$${\zeta }_A(s)=\sum _{k=1}^{\mathrm{\infty }}{\lambda }_k^{-s}=\sum _{k=1}^{\mathrm{\infty }}{e}^{-s\log {\lambda }_k}.$$

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

$${\zeta }_A^{\prime }(0)=-\sum _{k=1}^{\mathrm{\infty }}\log {\lambda }_k,$$

so

$$e^{-{\zeta }_A^{\prime }(0)}=detA.$$

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

$\mathrm{\Gamma }(s){\zeta }_A(s)$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{t}^{s-1}{e}^{-t}𝑑t\cdot {\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\lambda }_k^{-s}$

$={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\lambda }_k^{-s}{\displaystyle {\int }_0^{\mathrm{\infty }}}{t}^{s-1}{e}^{-t}𝑑t$

$={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\lambda }_k^{-s}{\displaystyle {\int }_0^{\mathrm{\infty }}}{({\lambda }_{k}u)}^{s-1}{e}^{-{\lambda }_{k}u}{\lambda }_k𝑑u$

$={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{u}^{s-1}{e}^{-{\lambda }_{k}u}𝑑u.$

Thus

$${\zeta }_A(s)=\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{u}^{s-1}\sum _{k=1}^{\mathrm{\infty }}{e}^{-{\lambda }_{k}u}du.$$

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

${\zeta }_{\gamma A}(s)$ $={\displaystyle \frac{1}{\mathrm{\Gamma }(s)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{u}^{s-1}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{e}^{-\gamma {\lambda }_{k}u}du$

$={\displaystyle \frac{1}{\mathrm{\Gamma }(s)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{\gamma }^{-s}{v}^{s-1}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{e}^{-{\lambda }_{k}v}dv$

$={\gamma }^{-s}{\zeta }_A(s).$

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

$$det(\gamma A)=e^{-{\zeta }_{\gamma A}^{\prime }(0)}=e^{\log \gamma \cdot {\zeta }_A(0)-{\zeta }_A^{\prime }(0)}={\gamma }^{{\zeta }_A(0)}detA.$$