# WANTED: Short proof for challenging identity

I have to prove the following without any use of further mathematical theories except basic calculus and linear algebra:

Let $\ell$ be a positive integer and $A$ a real, symmetric and positive definite $n \times n$-Matrix (square Matrix with $n$ columns and $n$ rows). Show that the following identity holds:

$$\int_{\mathbb{R}^n} x_{i_1} \cdots x_{i_{2 \ell}} e^{-\frac{1}{2} \left< \mathbf{ x}, A \ \mathbf{x} \right>} d^n x = \frac{(2 \pi)^{n/2}}{\ell! \sqrt{\det{A}}} \sum\limits_{ \begin{array} \{ \{ k_1,k_1'\},\ldots,\{k_\ell,k_\ell'\}\in P \\ \cup_{j=1}^\ell \{ k_j, k_j' \} = \{ 1,\ldots,2 \ell \} \end{array} } ( A^{-1} )_{i_{k_1},i_{k_1'}} \cdots ( A^{-1} )_{i_{k_\ell},i_{k_\ell'}}$$ with $P = \left\{ \{k,k'\}, k \not= k' \in \{1,\ldots,2 \ell\} \right\}$ and $\left< \mathbf{x}, \mathbf{y} \right>$ the standard scalar product for $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$.

Frankly speaking I have no idea what to do, where to begin. I'm also afraid that I cannot really explain the notation, because I don't get it 100% myself. In fact we are a group of approximately 15 students and we all don't know a smooth short way. This exercise could be part of a 2 hour exam with 5 other exercises. So we suppose there exists a more or less short solution.

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What is $\left< \mathbf{ x}, A \mathbf{x} \right>$? And what is $i$? –  Matt Gregory Feb 1 '11 at 12:46
The standard scalar product: $\forall \mathbf{x}, \mathbf{y} \in \mathbb{R}^n: \left< \mathbf{x}, \mathbf{y} \right> := \sum\limits_{i=1}^n x_i \cdot y_i$. Concerning $i$: It is not written in the exercise, so I do not inted to edit the question, but I suppose $i \in \mathbb{N}$. –  a1337q Feb 1 '11 at 12:52
But isn't Ax a matrix? Sorry, I have no idea how to do this problem. I'm just curious about it. –  Matt Gregory Feb 1 '11 at 12:55
Well, no, by definition of the matrix multiplication, $A \mathbf{x}$ is again in $\mathbb{R}^n$. –  a1337q Feb 1 '11 at 12:57
$Ax$ is a vector. –  Grumpy Parsnip Feb 1 '11 at 12:58

Hint for the case $\ell = 1$. The rest can be done by induction.

Since $A$ is positive definite, can define the matrix $B = (A)^{-1/2}$. Consider the change of variables $x = By$, so $dx = |B| dy$ using the Jacobian determinant. Then your equation can be re-written as

$$\int_{\mathbb{R}^n} x_i x_j e^{-\frac12 xAx} dx = |B| \int_{\mathbb{R}^n} B_{ik}B_{jl}y_ky_l e^{-\frac12 y\cdot y} dy$$

(Notice that $|B| = |A|^{-1/2}$ as desired.). Now, using the linearity of the action by the matrix $B$, you can integrate by parts in $y_l$ using the derivative operator $\partial_l$. This will pick out a term of the form (using that $\partial_ly_k = \delta_{lk}$)

$$B_{il}B_{jl} \int_{\mathbb{R}^n} e^{-\frac12 y\cdot y} dy$$

the integral is the Gaussian integral giving you the factor of $2\pi$. And $B_{il}B_{jl} = B^2 = A^{-1}_{ij}$.

The factorial factor comes from when you integrate by parts, when $\ell > 1$, by counting the terms you actually have.

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Well, the exam went by and this one wasn't part of it. Whee. Thank you. –  a1337q Feb 6 '11 at 20:15