Gaussian integration and dimension argument

Question

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 $\frac{1}{2}$

Answer 1

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 $\det (A)$.