As noted in A FULL BAYESIAN APPROACH FOR INVERSE PROBLEMS, let $ y = Ax + n$, where $y$ is a $m$ dimensional signal and $n$ is white Gaussian noise with precision $\beta$, so we have: $$ y|x, \beta \sim N(Ax, \beta^{-1}I) \Rightarrow p(y|x, \beta)=\frac{1}{Z(\beta)}exp(-\frac{1}{2}\beta\left\lVert y-Ax \right\rVert^2) $$
The partition function is $Z(\beta) = (\frac{2\beta}{\pi})^{m/2}$ as claimed in the paper and my question is why this is the case?
I know $ \int_{-\infty}^\infty p(y|x,\beta)dy=1 $, but I reach to different result: $Z(\beta) = (\frac{2\pi}{\beta})^{m/2}$.
Thanks.
First, I think you have to integrate over $\mathbb{R}^n$, because $y$ and $x$ are multi-dimensional right?
Then $Ax$ is just a translation, so by a change of variables you can see that $Z(\beta)$ does not depend on $Ax$, so you can set $Ax=0$.
So you have to solve $$\int_{\mathbb{R}^n}e^{-\frac{1}{2} \beta ||y||^2}dy$$ You can use Fubini and the well-known fact that $$\int_{\mathbb{R}}e^{-\frac{1}{2} y^2}dy=\sqrt{2\pi}$$