# Gaussian integration and dimension argument

I made a mistake recently regarding the Gaussian density, by putting the determinant of the variance to the power $\frac{d}{2}$. Would the following argumentation be valid to highlight it should be to the power $\frac{1}{2}$? :

The argument of the exponential must be dimensionless, so whatever unit is $x$ in, the variance has dimension $x^2$.

$$dimension( ~{ x^t \Sigma^{-1} x}) = 0$$

in the density formula, dividing by the determinant yields therefore a quantity with dimension $x^{-2*d}$.

$$dimension~ ( \frac{\alpha}{\det A} ) ~=~ ~x^{-2*d}$$

But since you want a density, that means if you multiply that by a volume (of dimension $x^d$) you get a dimensionless number.

$$dimension ~( \frac{\alpha}{\det A} . dx ^d ~) ~=~ 0$$

So one should raise the determinant to the power $\frac{1}{2}$

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## 1 Answer

I) Well, Gaussian integrals

$$\tag{1} \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^t A x} ~=~ \sqrt{\frac{(2\pi)^n}{\det A}}$$

are easy to calculate exactly, where the matrix ${\rm Re}(A)$ is positive definite.

II) But if OP just wants to confirm that the power $p$ of the determinant $\det A$ on the rhs. of eq. (1) is $p=-1/2$ (as opposed to some other power $p$), then indeed one may use dimensional analysis. If the integration variables $x^i$ have dimension of length $[x^i]=L$, then the matrix elements $A_{ij}$ have dimension $[A_{ij}]=L^{-2}$ to keep the argument of the exponential dimensionless. Therefore $\det A$ has dimension $[\det A]=L^{-2n}$. Moreover both sides of eq. (1) must have dimension $L^n$. Hence the power $p=-1/2$ of the determinant $\det (A)$.

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Thanks Qmechanic. trying to get back my engineering/math reflex ! –  nicolas Dec 6 '12 at 10:07
btw, I was reflecting recently on how sloppy the derivative notation df(x)/dx is (x specifies both positional argument for derivative and the point at which said derivative is evaluated) and was remembering a similar 'notation variation' for integral, which is the one you use, that is to put the integrated variable close to the integral operator. it seems much cleaner, but not sure why. does this notation has a name ? –  nicolas Dec 6 '12 at 10:12
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