Facts from calculus and their $d$-dimensional analogues

The basic fact from calculus that powers the whole discussion is:

Proposition 1  

$$\int_{-\infty}^{\infty} dx~~ e^{ -\frac{\scriptstyle a}{2}x^2} = \sqrt{\frac{2\pi}{a}}.$$

The identity with $a=1$ is proved by the familiar trick of calculating the square of the integral in polar coordinates. The general identity follows by change of variable from $x$ to $x{\sqrt a}$ .

This fact generalizes to higher-dimensional integrals. Set ${\bf v} = (v^1, \dots, v^d)$ and $d{\bf v} = dv^1\cdots dv^d$, and let $A$ be a symmetric $d$ by $d$ matrix.

Proposition 2  

$$\int_{{\bf R}^d} d{\bf v} ~~\exp(-{\scriptstyle\frac{1}{2}}{\bf v}^tA~{\bf v}) = (2\pi)^{d/2} (\det A)^{-1/2}.$$

We work the calculation for the case $A$ diagonalizable: in that case there exists an orthogonal matrix $U$ (so $U^t = U^{-1}$) such that $UAU^{-1}$ is the diagonal matrix $B$ whose only nonzero entries are $b_{11}, \dots, b_{dd}$ along the diagonal. Then $A = U^{-1}B~U$ and ${\bf v}^tA~{\bf v} = {\bf v}^tU^{-1}B~U{\bf v} = {\bf v}^tU^tB~U{\bf v} = {\bf w}^tB~{\bf w}$ where ${\bf w} = U{\bf v}$, using $U^t = U^{-1}$ and $(U{\bf v})^t = {\bf v}^tU^t$. Since $U$ is orthogonal $\det U =1$ and the change of variable from ${\bf v}$ to ${\bf w}$ does not change the integral:

$$\int_{{\bf R}^d} d{\bf v} ~~\exp(-{\scriptstyle\frac{1}{2}}{\bf v... ...\int_{{\bf R}^d} d{\bf w} ~~\exp(-{\scriptstyle\frac{1}{2}}{\bf w}^tB~{\bf w})=$$

$$\int_{{\bf R}^d} d{\bf w} ~~\exp(b_{11}(w^1)^2 +\cdots + b_{dd}(w^d)^2) =$$

$$\int_{{\bf R}^d} d{\bf w} ~~\exp(b_{11}(w^1)^2)~ \cdots ~\exp( b_{dd}(w^d)^2) =$$

$$(\int_{-\infty}^{\infty} dw^1~\exp(b_{11}(w^1)^2))~\cdots~(\int_{-\infty}^{\infty} dw^d~\exp(b_{dd}(w^d)^2))=$$

$$(\sqrt{2\pi}/\sqrt{b_{11}})~\cdots~ (\sqrt{2\pi}/\sqrt{b_{dd}})=$$

$$(2\pi)^{d/2} (\det B)^{-1/2} = (2\pi)^{d/2} (\det A)^{-1/2}.$$

The argument in general uses two additional observations: both sides of the equation vary continuously as functions of the entries in $A$, and any matrix $A$ with complex coefficients can be approximated to arbitrary accuracy by diagonalizable matrices.

Proposition 3  

$$\int_{-\infty}^{\infty} dx \, e^{-\frac{a}{2}x^2 + bx} = \sqrt{\frac{2\pi}{a} } e^{b^2/2a}\, .$$

This follows from Proposition 1 by completion of the square in the exponent and a change of variables.

The generalization to $d$ dimensions replaces $a$ with $A$ as before and $b$ with the vector ${\bf b} = (b^1,\dots,b^d)$.

Proposition 4  

$$\int_{{\bf R}^d} d{\bf v} ~~\exp(-{\scriptstyle\frac{ 1}{ 2}}{\bf... ...)^{d/2} (\det A)^{-1/2} \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}).$$

This is proven exactly like Proposition 2. If we write this integral as $Z_{\bf b}$, then the integral of Proposition 2 is $Z_0$ and this proposition can be rewritten as

$$Z_{\bf b} = Z_0 \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}) .$$