How do I integrate this distribution? I have a multinomial multivariate normal distribution of the form:
$$\exp\left[-\frac{1}{2\sigma^2}(({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu)\right]$$
I wish to integrate with respect to $\boldsymbol \beta$.
I have found a form of the Gaussian integral from wikipedia to be as following:
$$\int\limits_{-\infty}^\infty\exp\left[-\frac{1}{2}\sum\limits_{i,j=1}^{n}{\bf A}_{ij}x_ix_j\right] d^nx=\sqrt{\frac{(2\pi)^n}{\det A}}  $$
I do not know how to work out this integral or use this 'rule', but have come out with:
$$\int\limits_{-\infty}^\infty\exp\left[-\frac{1}{2\sigma^2}(({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu)\right] d^n\beta = \sqrt{\frac{(2\pi)^n}{\det \Sigma^{-1}}}  $$
This probably is not right? How do I do the integral? How is the working out done?
 A: You wrote
$$\exp\left[-\frac{1}{2\sigma^2}({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu)\right]$$
If you let the new value of $\Sigma$ be $\sigma^2\Sigma$, then you have
$$\exp\left[-\frac{1}{2}({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu)\right].$$
There's no reason to separate out that scalar, and it's not conventionally done.
The finite-dimensional case of the spectral theorem says every real symmetric matrix can be diagonalized by an orthogonal matrix, and you have
$$
\Sigma = G^T \begin{bmatrix} \lambda_1 \\ & \lambda_2 \\ & & \lambda_3 \\ & & & \ddots \end{bmatrix} G.
$$
Since $\Sigma$ is a variance (a "variance-covariance matrix" if you like), all of the $\lambda$s are non-negative, and since $\Sigma$ is nonsingular, all of them are positive.  So let $\Sigma^{1/2}$ denote the matrix
$$
\Sigma^{1/2} = G^T \begin{bmatrix} \sqrt{\lambda_1} \\ & \sqrt{\lambda_2} \\ & & \sqrt{\lambda_3} \\ & & & \ddots \end{bmatrix} G.
$$
and then $\Sigma^{1/2}$ is a positive-definite symmetric matrix, and $(\Sigma^{1/2})^2=\Sigma$, and we let $\Sigma^{-1/2}$ denote the inverse.  And since $\Sigma^{1/2}$ is symmetric, we have $(\Sigma^{1/2})^T\Sigma^{1/2}=\Sigma$.
Then we have
$$
({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu) = \Big( \Sigma^{-1/2}({\boldsymbol\beta}-\mu) \Big)^T \Big( \Sigma^{-1/2}({\boldsymbol\beta}-\mu) \Big) = \gamma^T\gamma,
$$
where $\gamma=\Sigma^{-1/2}({\boldsymbol\beta}-\mu)$.
Then
$$
\begin{align}
\int_{\mathbb{R}^n} \cdots\cdots d\beta = \int_{\mathbb{R}^n} \cdots\cdots |\det\Sigma^{1/2}| \, d\gamma & = |\det\Sigma^{1/2}|\int_{\mathbb{R}^n} \cdots \cdots \\[10pt]
& = |\det\Sigma^{1/2}|\int_{\mathbb{R}^n} \exp\left[ \frac{-1}{2} \gamma^T\gamma \right]\,d\gamma.
\end{align}
$$
This integral becomes
$$
\int_{\mathbb{R}^n} \exp\left(\frac{-1}{2} \gamma_1^2 \right)\cdots\exp\left(\frac{-1}{2} \gamma_n^2 \right) \, d\gamma_1\cdots d\gamma_n.
$$
Then it becomes the $n$th power of
$$
\int_\mathbb{R} \exp\left(\frac{-1}{2}\gamma^2\right)\,d\gamma.
$$
(And it's not hard to show that $\det(\Sigma^{1/2}) = \left(\det\Sigma\right)^{1/2}$.)
