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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&gt;} d^n x = \frac{(2 \pi)^{n/2}}{\ell! \sqrt{\det{A}}} \sum\limits_{
\begin{array}
\{ \{ k_1,k_1&#39;\},\ldots,\{k_\ell,k_\ell&#39;\}\in P \\
\cup_{j=1}^\ell \{ k_j, k_j&#39; \} = \{ 1,\ldots,2 \ell \}
\end{array} }
( A^{-1} )_{i_{k_1},i_{k_1&#39;}} \cdots ( A^{-1} )_{i_{k_\ell},i_{k_\ell&#39;}}
$$
with $P = \left\{ \{k,k&#39;\}, k \not= k&#39; \in \{1,\ldots,2 \ell\} \right\}$ and $\left&lt; \mathbf{x}, \mathbf{y} \right&gt;$ the standard scalar product for $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$.</p>

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.

a1337q
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1 Answers1

6

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.

Willie Wong
  • 73,139