Distributions of Quadratic form of a normal random variable

I am quite sure @kjetil b halvorsen's answer at What is the moment generating function of the generalized (multivariate) chi-square distribution? reduces to the expression of MGF in this post on simplification.

A direct proof is not difficult either when $Sigma$ is assumed to be positive definite.

The proof simply relies on the fact that for a symmetric positive definite matrix $B$, we have from the multivariate normal density

$$int_{mathbb R^p}expleft[-frac12(x-mu)' B^{-1}(x-mu)right] dx=(2pi)^{p/2}(det B)^{1/2}$$

Or,

$$int_{mathbb R^p}expleft[-frac12 x' B^{-1}x+mu' B^{-1}x-frac12 mu' B^{-1}muright] dx=(2pi)^{p/2}(det B)^{1/2}$$

Taking $b'=mu' B^{-1}$, this is same as

$$int_{mathbb R^p}expleft[-frac12x' B^{-1}x+b' xright]dx=(2pi)^{p/2}(det B)^{1/2}expleft(frac12 b' B bright) tag{*}$$

For symmetric $A$,

begin{align} M_{X' A X}(t)&=Eleft[e^{tX' AX}right] \&=frac1{(2pi)^{p/2}(det Sigma)^{1/2}}int_{mathbb R^p} exp(tx' Ax)cdot expleft[-frac12(x-mu)' Sigma^{-1}(x-mu)right] dx \\&=frac{exp(-frac12 mu'Sigma^{-1}mu)}{(2pi)^{p/2}(det Sigma)^{1/2}}int_{mathbb R^p} expleft[-frac12 x'(I-2tASigma)Sigma^{-1}x+mu'Sigma^{-1}xright] dx end{align}

We have $(I-2tASigma)Sigma^{-1}=Sigma^{-1}-2tA$, which is assumed positive definite (it is already symmetric) for sufficiently small $|t|$. The MGF is now precisely of the form $(*)$.

Taking $B=(Sigma^{-1}-2tA)^{-1}=Sigma(I-2tASigma)^{-1}$ and $b'=mu'Sigma^{-1}$ finally gives

$$M_{X'AX}(t)=(det(I-2tASigma))^{-1/2}expleft{-frac12 mu'[I-(I-2tASigma)^{-1}]Sigma^{-1}mu right},,$$

whenever the MGF exists. This also generalizes @whuber's answer here.