computing the expectation of Trigonometric functions, "sin" and "cos", with respect to their variables

Question

How can I compute the expectation of $\cos((s\odot y)^T\alpha)$ with respect to all its variable $s$, $y$ and $\alpha$ where all of them are Gaussian distributions? $$\int \mathcal{N}(y|\mu_y,\Sigma_y)\mathcal{N}(s|\mu_s,\Sigma_s)\mathcal{N}(\alpha|\mu_{\alpha},\Sigma_{\alpha})\cos((s\odot y)^T\alpha)\mathrm{d}y\;\mathrm{d}s\;\mathrm{d}\alpha$$ here $\odot$ denotes element by element multiplication of two vectors.

Answer 1

We could try computing $$ \mathbb{E}[\cos((s\odot y)^T\alpha)]=\mathbb{E}[\mathbb{E}[\mathbb{E}[\cos((s\odot y)^T\alpha)\mid s\odot y]\mid y]] $$ one integral at a time.

The integral over $\alpha$ is easy: $$ \mathbb{E}[\cos(t^T\alpha)]=\Re\varphi_\alpha(t) $$ where $\varphi_\alpha$ is the characteristic function of $\alpha$, which we know: $$ \varphi_\alpha(t)=\exp\left(i\mu_\alpha^Tt-\tfrac12t^T\Sigma_\alpha t\right) $$ so $$ \mathbb{E}[\cos((s\odot y)^T\alpha)\mid s\odot y]=\cos(\mu_\alpha^T(s\odot y))\exp\left(-\tfrac12(s\odot y)^T\Sigma_\alpha (s\odot y)\right) $$ The next one is also double $$ \int(2\pi)^{-k/2}\det(\Sigma_s)^{-1/2}\exp\left[-\frac12(s-\mu_s)^T\Sigma_s(s-\mu_s)\right]\cos(\mu_\alpha^T(s\odot y))\exp\left(-\tfrac12(s\odot y)^T\Sigma_\alpha (s\odot y)\right)\,\mathrm{d}s $$ the point to note is that $$ (s\odot y)^T\Sigma_\alpha(s\odot y)=s^T\operatorname{diag}(y)\Sigma_\alpha\operatorname{diag}(y)s $$ is a symmetric positive-definite quadratic form in $s$, so it is, up to the $\sqrt{\det(\tilde{\Sigma}_{s}(y))/\det(\Sigma_s)}$ and an exponential factor involving $y$ which you can determine (coming from change of mean), the real part of a normal distribution where the mean and variance are $y$-dependent.

But the final integral I don't know how to do, with the $\sqrt{\det(\operatorname{diag}(y)\Sigma_\alpha\operatorname{diag}(y)+\Sigma_s)}$ multiplied by exponential of a rational function of $y$.