Showing that the multivariate normal density integrates to 1 This is NOT the same as How to show the normal density integrates to 1?.
Let $\mathbf{x} \in \mathbb{R}^d$ be a multivariate normal random vector, with $\mathbb{E}[\mathbf{x}] = \boldsymbol\mu$ and positive-definite $\text{Var}[\mathbf{x}] = \boldsymbol\Sigma$. Then 
$$\int_{\mathbb{R}^d}\dfrac{1}{(2\pi)^{n/2}|\boldsymbol\Sigma|^{1/2}}\exp\left[-\dfrac{1}{2}(\mathbf{x}-\boldsymbol\mu)^{T}\boldsymbol\Sigma^{-1}(\mathbf{x}-\boldsymbol\mu)\right]\text{ d}\mathbf{x}$$
should equal $1$. How do I show this? I am allowed to use the one-dimensional case as a fact, so I thought, perhaps I should use induction on $d$. One-dimensional case is true, great. Now suppose it's true for $k$ dimensions. At the $k+1$th dimension, the difficulty is working with the new variance-covariance matrix and the Mahalanobis distance term 
 $$(\mathbf{x}-\boldsymbol\mu)^{T}\boldsymbol\Sigma^{-1}(\mathbf{x}-\boldsymbol\mu)\text{.}$$
Does anyone have any suggestions? I don't need a complete solution, but I would like a starting point.
 A: Write $(x-\mu)^T \Sigma^{-1} (x-\mu)$ as $\| \Sigma^{-1/2}(x-\mu) \|^2$ where $\Sigma^{1/2}$ is a matrix square root (well defined because $\Sigma$ is positive definite). Orthogonally diagonalize $\Sigma^{-1/2}$ (again possible because $\Sigma$ was positive definite). Then observe (using the orthogonality) that you have a product of $d$ 1D Gaussian integrals. The variances are different, but you have exactly the factor of $|\Sigma|^{1/2}$ in the denominator that you need to cancel that out.
A: Change variables. First translate everything considering $\mathbf{y} = \mathbf{x}-\boldsymbol\mu$. Second, you can orthogonally diagonalize $\boldsymbol\Sigma^{-1}$ using the fact that it is real symmetric and then you can express this $-\dfrac{1}{2}\mathbf{y}^{T}\boldsymbol\Sigma^{-1}\mathbf{y}$ as a sum of squares. Because the matrix used is orthogonal, that means that the abosulute value of the jacobian will be $1$. After that, factor the exponential as a product of exponentials of one variable, express the integral as a product of equal integrals and so as an $n$-th power of an integral in one variable. Using the case $n=1$ you are done.
