# 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 matrix entries have dimension $x^2$.

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

The density formula is some dimensionless $\alpha$ coming from normalization constant and the exponential, divided by the determinant yields therefore a quantity with dimension $x^{-2*d}$

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

But we want this to be a density, meaning that multiplied by the volume element $dx$ (of dimension $x^d$) you get a dimensionless number, the "count" of how many elements are in that box

$$dimension ~( \frac{\alpha}{\det \Sigma}. dx ~) ~=~ 0$$

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

$$\tag{1} \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^t A x} ~=~ \sqrt{\frac{(2\pi)^n}{\det A}}$$
are easy to calculate exactly, where the matrix $${\rm Re}(A)$$ is positive definite, cf. e.g. this math.SE post.
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 $$\det (A)$$.