$X \sim N_n(0, I_n), Y \sim N_n(\mu, \Sigma)$ where $\Sigma^{-1}$ exists, and $g$ is the pdf of $Y$. Find the expression of $E_X[g(X)]$.
This is what I'm trying to solve. (Let $f$ be the pdf of $X$) Then,
$$f(x) = (2\pi)^{-n/2}\text{exp}\{-\frac{1}{2}x^Tx\} \\ g(y) = (2\pi)^{-n/2}|\Sigma|^{-1/2}\text{exp}\{-\frac{1}{2}(y-\mu)^T\Sigma^{-1}(y-\mu)\}.$$
Since $E[g(X)] = \int g(x)f(x)dx$, $$f(x)g(x) = (2\pi)^{-n}|\Sigma|^{-1/2}\text{exp}\{-\frac{1}{2}(x^Tx +(x-\mu)^T\Sigma^{-1}(x-\mu) )\}$$
My initial goal was to make some quadratic form in the exponential term;
like $\text{exp}\{(\tilde x - \tilde \mu)^T\Sigma^{-1}(\tilde x - \tilde \mu)\}$ so that I can calculate the expectation readily.
I tried to use $\Sigma := QQ^T$ where $Q$ is invertible, but it seemed useless.
Is there some ways to obtain $E_X[g(X)]$ without messy calculations?
Any help w.r.t. this problem would be grateful.
We can indeed complete the square. Set $Q^{-1} = I + \Sigma^{-1}$ and $$ \gamma = (I + \Sigma^{-1})^{-1}\Sigma^{-1}\mu $$ then you can check that, $$ (x - \gamma)^T Q^{-1}(x - \gamma) = x^T x + x^T \Sigma^{-1}x - 2\mu^T \Sigma^{-1}x + \text{constant} $$ where the constant above does not depend on $x$. The rest is purely algebra, and I'll leave that to you.