If $X \sim N(\mu, \Sigma)$ is an $n$-dimensional Gaussian, and $A \in R^n$, so that $X^TA \in R$, and $f(x) = x^2$.
I want to know the solution to the following expectation:
$$ \int_R \int_R \dots\int_R f(Ax)N(x\mid\mu, \Sigma)\, dx = E_X[f(AX)] $$
noting that $v := AX \sim N(\mu^TA, A^T\Sigma A)$ is a one dimensional Gaussian, it seems intuitive to write this expectation as the second moment of $v$:
$$ E_{v}[v^2] = \text{Var}[v] + E[v]^2 = A^T\Sigma A + (\mu^TA)^2 $$
but I am unsure if this is correct, if someone knows that it is, or has a proof, that would be great.
That's quite straightforward, and you don't even need to assume a gaussian distribution:
$$ E[f(X)]=E[(X^t A)^2]=E[A^t X X^t A] = A^t E[X X^t] A$$
But $$\Sigma = Cov(X)=E[(X-\mu)(X-\mu)^t]=E[X X^t] - \mu \mu^t$$
Then, putting all together:
$$ E[f(X)] = A^t (\Sigma + \mu \mu^t) A = A^t \Sigma A + (\mu^t A )^2$$