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Why does the following integral hold? $$\large \int_{-\infty}^{\infty} \frac{1}{( \det 2 \pi A)^{1/2}} e^{-\frac{1}{2} x^T \cdot A^{-1}\cdot x}dx =1 $$ where $A$ is a $3 \times 3$ positive definite symmetric matrix.

And if we want to generalize the formula to any dimensions, for example $$\large \int_{-\infty}^{\infty}\frac{1}{f(A)} e^{-{A_{ijkl}} X_i X_j X_k X_l}dX =1 $$ How can we find such a function $f$?

sam
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2 Answers2

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This result possesses a probabilistic interpretation: $f(x)=\frac{1}{(2 \pi \det A)^{1/2}} e^{-\frac{1}{2} x^T \cdot A^{-1}\cdot x}$ is the probability density function (pdf) of a certain multivariate normal (or gaussian) distribution ; it is thus "normal" that its integral is equal to 1, $A$ being interpreted as the variance matrix of this distribution. See for example (http://www.real-statistics.com/multivariate-statistics/multivariate-normal-distribution/multivariate-normal-distribution-basic-concepts/).

If $x=(x_1,x_2,x_3)$, the case of independance of corresponding random variables $X_k$ nicely matches the case where $A=diag(\sigma_1,\sigma_2,\sigma_3)$ ( $\sigma_k$ being the variance of $X_k$), which itself corresponds to the case where the triple integral splits into 3 separable integrals.

Jean Marie
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Your first formula is wrong. The correct version is: $$ \int \operatorname{e}^{-\frac{1}{2} x^T \cdot A^{-1} x} \operatorname{d}^3 x = (2\pi)^{3/2} \sqrt{\operatorname{det}(A)}.$$

If $A$ is positive definite and symmetric, then you can diagonalize it into the form: $$A = V^T \Lambda V,$$ where $V$ is the normalized matrix of $A$'s eigenvectors (an othogonal matrix) and $\Lambda$ is the matrix of the eigenvalues. Do a change of variables on the integral $x' = V x$, and the answer falls right out. The generalized $N$-dimensional version is: $$ \int \operatorname{e}^{-\frac{1}{2} x^T \cdot A^{-1} x} \operatorname{d}^N x = (2\pi)^{N/2} \sqrt{\operatorname{det}(A)}.$$

Edit to add: the generalized integral provided is not of the same form as the simple 3-d case. The one with $A$ a rank 4 tensor and 4 factors of the $x$ vector in the exponent is, as far as I now, an unsolved problem, especially when there is also a term in the exponent with a different rank 2 $A$ and 2 factors the $x$ vector in it.

Sean Lake
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  • Thanks. Could you provide some reference on this topic? Especially when $A$ is a symmetric tensor. – sam Aug 24 '16 at 10:03
  • The proofs involved are very short -- 1 or 2 lines each. Any introductory text on quantum mechanics, for example R. Shankar's Principles of Quantum Mechanics, should suffice. Or look in your linear algebra text for eigenvectors and eigenvalues, and it should have proofs on the orthogonality of eigenvectors. Mathworld has a nice bit on orthogonal matrices. Axler's Linear Algebra Done Right is also good for fundamentals (basis sets, etc). – Sean Lake Aug 24 '16 at 11:05
  • I mean on the second part of my question. – sam Aug 24 '16 at 11:50
  • You mean the one where $A$ is a rank 4 tensor? For that one I have no help to give and would, honestly, be very interested in an answer, myself. If that was a mistake, though, and you meant something like what I typed (i.e. an $N$-dimensional Gaussian integral), then I outlined the proof. After you change variables, the multi-integral factors into $N$ single integrals. You then only need to know that the determinant of a matrix is the product of its eigenvalues. – Sean Lake Aug 24 '16 at 12:45
  • Thanks. I have understand your answer. I want to know the rank 4 case. – sam Aug 24 '16 at 12:58
  • I don't know if this will work, but you could try something like the following: $$\int \operatorname{e}^{-A_{ijkl} x_i x_j x_k x_l} \operatorname{d}^N x = \int \delta^N(x-y) \operatorname{e}^{-A_{ijkl} x_i x_j y_k y_l} \operatorname{d}^N x \operatorname{d}^N y$$, and go from there, but I don't know if it would pan out. ($\delta^n(x)$ is an $N$-dimensional delta function). – Sean Lake Aug 24 '16 at 13:11