The fourier-laplace transform of the gaussian kernel

Question

I have no idea about this intergral

$$ \int \exp(\mathbf y ^{*} \mathbf V ^{*} \mathbf \Sigma^{-1} \mathbf S_X) N(\mathbf y\mid\mathbf 0,\mathbf b^{-1})\,d\mathbf y = \exp\left({-1 \over 2}log|\mathbf b|+{1 \over 2}\mathbf S_X ^{*} \mathbf \Sigma^{-1} \mathbf V \mathbf b^{-1} \mathbf V ^{*} \mathbf \Sigma^{-1} \mathbf S_X\right)$$

$$\mathbf y= R \times 1, \mathbf V = CF \times R, \mathbf \Sigma = CF \times CF $$ $$\mathbf S_X = CF \times 1, \mathbf b = R \times R $$

and $N(\cdot\mid\mathbf 0,\mathbf b^{-1})$ denotes the Gaussian kernel with mean $0$ and covariance matrix $\mathbf b^{-1}$

I just read that from the paper a link!

"By the formula for the Fourier-Laplace transform of the Gaussian kernel this simplifies ... "

Even if I tried to fine the Fourier-Laplace transform of the Guassian kernel, I couldn't to find the understandable things for solving this problem. So I would like to ask you to give me some hint or explanation :)

Thank you very much in advance.

Answer 1

Let's try a simple special case: Our Gaussian kernel will be $$ N(y) = \frac 1 {\sigma\sqrt{2\pi}} e^{-(1/2)(y/\sigma)^2} $$ with mean $0$ and variance $\sigma^2$. \begin{align} & \int_{-\infty}^\infty \exp (ay) N(y) \, dy = \frac 1{\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left( \frac{-1} 2 \left( \frac y \sigma\right)^2 + ay \right) \, \frac{dy} \sigma \\[10pt] = {} & \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left( \frac{-1}{2\sigma^2} (y-a\sigma^2)^2 \right) \exp\left(\frac{a^2\sigma^2} 2 \right) \, \frac{dy}\sigma \\[10pt] = {} & \exp\left(\frac{a^2\sigma^2} 2 \right) \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left( \frac{-1}{2\sigma^2} (y-a\sigma^2)^2 \right) \, \frac{dy} \sigma \end{align} where, on the second line above, we completed the square. Then we have $$ \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left( \frac{-1}{2\sigma^2} (y-a\sigma^2)^2 \right) \, \frac{dy} \sigma = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left( \frac{-1}{2\sigma^2} u^2 \right) \, \frac{du} \sigma = 1. $$

So the bottom line is $\exp\left(\dfrac{a^2\sigma^2} 2 \right)$.