Feature Column from the AMS

Finite-dimensional Feynman Diagrams

2. Facts from calculus and their d-dimensional analogues

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

Proposition 1

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

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

This fact generalizes to higher-dimensional integrals. Set v = (v¹, ..., v^d)and dv = (dv¹ ... dv^d),and let A be a symmetric dby d matrix.

Proposition 2

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

We use the fact that a symmetric matrix A isdiagonalizable: there exists an orthogonal matrix U(so U^t = U^-1)such that UAU^-1 is the diagonal matrix Bwhose onlynonzero entries are b₁₁, ... , b_ddalong the diagonal. Then A = U^-1BUand v^tAv = v^t U^-1B U v =v^tU^tB U v =w^tB w where w = Uv,using U^t = U^-1 and (Uv)^t =v^tU^t. Since Uis orthogonal detU = 1 and the change of variable from v to wdoes notchange the integral:

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

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

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

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

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

Proposition 3

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

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

The generalization to ddimensions replaces awith Aas before and bwiththe vector b = (b¹, ... , b^d)

Proposition 4

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

This is proven exactly like Proposition 2.If we write this integral as Z_bthen the integral of Proposition 2 is Z₀and this proposition can be rewritten as

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