On the integral of matrix functions Why does the following integral hold?
$$\large \int_{-\infty}^{\infty} \frac{1}{( \det 2 \pi A)^{1/2}} e^{-\frac{1}{2} x^T \cdot A^{-1}\cdot x}dx =1 $$
where $A$ is a $3 \times 3$ positive definite symmetric matrix.
And if we want to generalize the formula to any dimensions, for example
$$\large \int_{-\infty}^{\infty}\frac{1}{f(A)} e^{-{A_{ijkl}} X_i X_j X_k X_l}dX =1 $$
How can we find such a function $f$?
 A: This result possesses a probabilistic interpretation: $f(x)=\frac{1}{(2 \pi \det A)^{1/2}} e^{-\frac{1}{2} x^T \cdot A^{-1}\cdot x}$ is the probability density function (pdf) of a certain multivariate normal (or gaussian) distribution ; it is thus "normal" that its integral is equal to 1, $A$ being interpreted as the variance matrix of this distribution. See for example (http://www.real-statistics.com/multivariate-statistics/multivariate-normal-distribution/multivariate-normal-distribution-basic-concepts/). 
If $x=(x_1,x_2,x_3)$, the case of independance of corresponding random variables $X_k$ nicely matches the case where $A=diag(\sigma_1,\sigma_2,\sigma_3)$ ( $\sigma_k$ being the variance of $X_k$), which itself corresponds to the case where the triple integral splits into 3 separable integrals.
A: Your first formula is wrong. The correct version is:
$$ \int \operatorname{e}^{-\frac{1}{2} x^T \cdot A^{-1} x} \operatorname{d}^3 x = (2\pi)^{3/2} \sqrt{\operatorname{det}(A)}.$$
If $A$ is positive definite and symmetric, then you can diagonalize it into the form:
$$A = V^T \Lambda V,$$
where $V$ is the normalized matrix of $A$'s eigenvectors (an othogonal matrix) and $\Lambda$ is the matrix of the eigenvalues. Do a change of variables on the integral $x' = V x$, and the answer falls right out. The generalized $N$-dimensional version is:
$$ \int \operatorname{e}^{-\frac{1}{2} x^T \cdot A^{-1} x} \operatorname{d}^N x = (2\pi)^{N/2} \sqrt{\operatorname{det}(A)}.$$
Edit to add: the generalized integral provided is not of the same form as the simple 3-d case. The one with $A$ a rank 4 tensor and 4 factors of the $x$ vector in the exponent is, as far as I now, an unsolved problem, especially when there is also a term in the exponent with a different rank 2 $A$ and 2 factors the $x$ vector in it.
