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I am trying to solve the following integral:

$\int \exp \left(-\frac{1}{2} \mathbf{x}^T \mathbf{A} \mathbf{x} \right ) d\mathbf{x}$

where $\mathbf{x}$ is a $N \times 1$ column vector and $\mathbf{A}$ is a $N \times N$ matrix.

I could expand the $\mathbf{x}^T \mathbf{A} \mathbf{x}$ term and then integrate component by component, but I am wondering if there is a nice expression for this integral in terms of the original vector and matrices. If it helps everything is real and $\mathbf{A}$ is symmetric.

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I assume the integral is over all of $\mathbb{R}^N$. If $\mathbf{A}$ is real symmetric and positive definite, then it has a square root $\mathbf{S}, \mathbf{A} = \mathbf{S}^T\mathbf{S}$. So by making a change of variable $\mathbf{y} = \mathbf{Sx}$ the integral is \begin{align} \int_{\mathbb{R}^N} \exp(-\mathbf{x}^T\mathbf{S}^T\mathbf{Sx}/2) \,d\mathbf{x} &= \int_{\mathbb{R}^N} \frac{1}{\det \mathbf{S}} \exp(-\mathbf{y}^T\mathbf{y}/2) \,d\mathbf{y} \\ &= \frac{1}{\det \mathbf{S}} \int_{\mathbb{R}^N} \exp\biggl(\sum_{i=1}^n-y_i^2/2\biggr) \,dy_1 \ldots dy_n \\ &= \frac{1}{\sqrt{\det \mathbf{A}}} \prod_{i=1}^N \int_{-\infty}^\infty \exp(-y_i^2/2) \,dy_i \\ &= \sqrt{\frac{(2\pi)^N}{\det\mathbf{A}}}. \end{align}

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Note the similarities between the integral you write and the multivariate normal density function

Since $$\int (2\pi)^{-N/2}\vert A^{-1}\vert^{-1/2} e^{-\frac{1}{2} x^T A x}dx = 1$$ we see your integral is $$\int e^{-\frac{1}{2} x^T A x} dx= \big[(2\pi)^{-N/2}\vert A^{-1}\vert^{-1/2}\big]^{-1}$$

Noting that $\vert \cdot \vert$ denotes the determinant

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  • $\begingroup$ Why the accumulation of minus signs? $\endgroup$
    – Did
    Sep 30 '16 at 22:17

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