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I am currently studying V.I. Arnold's course, and I am stuck on this exercise:

Evaluate $$ \underbrace{\idotsint}_{n} \exp\left(-\sum_{1\le i\le j\le n}^n x_i x_j\right) \mathrm{d}x_1\cdots\mathrm{d}x_n $$

Can anyone suggest a hint? Thank you in advance!

zytsang
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2 Answers2

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Having read a lot of things by Arnold, I think his point of view was that "every student should know" the formula $$\int\ldots\int_{\mathbb{R}^n}\exp\left\{-\frac12 x^TAx\right\}dx_1\ldots dx_n=\frac{(2\pi)^{n/2}}{\sqrt{\mathrm{det}\,A}},\tag{1}$$ where $x^T=(x_1,\ldots,x_n)$ and $A$ is a real symmetric positive definite $n\times n$ matrix. Actually, every physics student should know it indeed.

What remains is a simple linear algebra exercise.

Added: The gaussian integration formula (1) is obtained by noticing that the matrix $A$ can be brought to diagonal form by orthogonal transformation, characterized by unit jacobian. Then one is left with a product of gaussian integrals $\prod_{k=1}^n\int_{-\infty}^{\infty}\exp\{-\lambda_k x^2/2\}dx$, where $\lambda_{1\ldots n}$ denote the eigenvalues of $A$.

Community
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Start wearing purple
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Hint: Diagonalize the symmetric matrix $$ A=-\frac12\left( \begin{array}{ccccc} 2&1&1&\cdots&1\\ 1&2&1&\cdots&1\\ 1&1&2&\cdots&1\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&\cdots&2 \end{array}\right)? $$ Diagonalization of a quadratic form in the exponential = separation of variables.

Pedro
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Jyrki Lahtonen
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  • I changed what looked like a typo. Let me know. – Pedro Jul 30 '13 at 14:01
  • Then the answer should be like $\displaystyle |A^{\frac{1}{2}}|\left(\int \exp(-x^2)dx\right)^n$ right? – Samrat Mukhopadhyay Jul 30 '13 at 14:03
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    I worked it out. The diagonal matrix is $\left[ \begin{matrix} 1& & & & \ &1& & & \ & &\ddots& & \ & & &1& \ & & & & n+1 \end{matrix}\right]$. So by n-dimensional Gaussian integral the final result is $\frac{(2 \pi )^{n/2}}{\sqrt{n+1}}? $ – zytsang Jul 30 '13 at 16:22
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    @YorkTsang: That's what I got, too (don't forget that $-1/2$). Note that as $A$ is symmetric, it can be diagonalized with an orthogonal matrix, so the Jacobian of this linear change of variables is $=1$ (orthogonality not essential here, but sometimes else it is). – Jyrki Lahtonen Jul 30 '13 at 16:26
  • @JyrkiLahtonen: Thank you for your hint! – zytsang Jul 30 '13 at 16:31
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    OOPS. Orthogonality IS essential. Diagonalization naturally uses the inverse of a matrix, but the quadratic form uses the transpose. Transpose = inverse, when the matrix is orthogonal. Sorry about that. – Jyrki Lahtonen Jul 30 '13 at 21:44