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For a one-dimensional x,

$$\int_{-\infty}^{\infty}x^{2}e^{-x^{2}}dx=\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^{2}}dx$$

This can be shown through integration by parts. There is a good derivation of that here.

I'm wondering, does the same exist for a multivariate Gaussian $$\ p(x|\mu,\Sigma)=\left(2\pi\right)^{-\frac{N}{2}}|\Sigma|^{-\frac{1}{2}}\exp\left(-\frac{1}{2}(x-\mu)^{\textrm{T}}\Sigma^{-1}(x-\mu)\right)$$

Where $\Sigma$ (positive-definite) is the variance and $\mu$ is the mean. Here, I'm using $N$ for the dimension of $x$. That is, does

$$\int\left(2\pi\right)^{-\frac{N}{2}}|\Sigma|^{-\frac{1}{2}}\left[-\frac{1}{2}(x-\mu)^{\textrm{T}}\Sigma^{-1}(x-\mu)\right]\exp\left(-\frac{1}{2}(x-\mu)^{\textrm{T}}\Sigma^{-1}(x-\mu)\right)dx$$
$$=\frac{1}{2}\int\left(2\pi\right)^{-\frac{N}{2}}|\Sigma|^{-\frac{1}{2}}\exp\left(-\frac{1}{2}(x-\mu)^{\textrm{T}}\Sigma^{-1}(x-\mu)\right)dx$$

?

My first thought is that the identity should be identical because after carrying out the inner products "upstairs" in the exponential and "downstairs" in the coefficient, it becomes a positive scalar.

This calculation is involved in determining the differential entropy of a multivariate Gaussian. For my answer to agree with the answer given in theorem 9.4.1 here, there should be an additional factor of N on the right hand side.

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