I was reading this paper where it had a gaussian distribution model.
I mean gaussian is given by
$P(y) = \frac{e^{-\frac{1}{2}(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n/2}|\Sigma|^{1/2}}$
But is
$\frac{e^{-(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n}|\Sigma|}$ Gaussian as well?
The paper used their model saying the gaussian had the above form. It is valid?
I think the equation we are talking about is
$$ p(y) = \frac{1}{Z}\exp \{ -y^T \Lambda y - 2 x^T\Theta y \} $$
This is a perfectly normalized, multivariate Gaussian and the normalization factor is precisely $Z$, the partition function. Indeed $Z$ is defined by the normalization
$$ \int p(y) dy =1 . $$
The integration can be performed completing the square, i.e. writing
$$ p(y) = \frac{1}{Z}\exp\{ - (y-y_0)^T \Lambda (y-y_0) + y_0 \Lambda y_0 \} $$ by choosing $y_0=\Lambda^{-1}\Theta^T x$ so that the term linear in $y$ matches. You can now change variable $y'=y-y_0$ and the normalization condition reads
$$ \frac{1}{Z} \exp\{x^T \Theta \Lambda^{-1} \Theta^T x \} \int \exp\{- y'^T \Lambda y'\} dy'=1 $$ The integration over $y'$ gives
$$ \int \exp\{- y'^T \Lambda y'\} dy'=\sqrt{\frac{\pi^n}{\det{\Lambda}}} $$
Finally the normalization is $$ \frac{1}{Z} \exp\{x^T \Theta \Lambda^{-1} \Theta^T x \} \sqrt{\frac{\pi^n}{\det{\Lambda}}} = 1 $$
i.e. $$ \frac{1}{Z} = \exp\{-x^T \Theta \Lambda^{-1} \Theta^T x \} \sqrt{\det{\Lambda}} \pi^{-n/2} $$
which should be Eq. (2) of your paper. The, unimportant, numerical constant $c$ in the paper is $\pi^{-n/2}$ but in Eq. (2) they forgot the sqrt term on $\det{\Lambda}$ (probably a typo).
You can compute the Gaussian integral yourself looking at this link
http://en.wikipedia.org/wiki/Gaussian_integral#n-dimensional_with_linear_term