What is the integral of this gaussian I want to know what is the following integral
$\int e^{-(y-\mu)^T \Lambda(y-\mu) } dy$
I am trying to see the properties of gaussian integral but I couldn't find anything for this one. Any help guys?
I want to know how given

where Z(x) is the partition function

This is the paper
If the integration that you guys have given is correct there should be $|\Lambda|^{1/2}$ at the end instead of just $|\Lambda|$ in $\frac{1}{Z(x)}$. I doubt that there is a type in the paper. So I must be missing something
Can anyone explain why that guy didn't have $|\Lambda|^{1/2}$ but  $|\Lambda|$ instead???

For the integral to exist, $\Lambda$ has to be positive definite. If $\Lambda$ is positive definite, then we can decompose $\Lambda$ as $\Lambda = R^TR$ (for instance, Cholesky decomposition).
We then have $$I = \int_{\mathbb{R}^n} \exp \left(-(y-\mu)^T R^T R (y- \mu)\right)dy$$
Let $x = R(y -\mu)$. We then have $dx = \det(R) dy$. Hence,
\begin{align}
I & = \int_{\mathbb{R}^n} \exp \left(-x^T x\right)\dfrac{dx}{\det(R)} = \dfrac1{\det(R)} \prod_{k=1}^n \int_{-\infty}^{\infty}\exp(-x^2)dx\\
& = \dfrac1{\det(R)} \prod_{k=1}^n \sqrt{\pi} = \dfrac{\pi^{n/2}}{\det(R)} = \sqrt{\dfrac{\pi^n}{\det(\Lambda)}}
\end{align}
