# Proving a Gaussian integral identity

Provide mathematical reasons(no calculation required) why the two-dimensional Gaussian integral is $\int_{\mathbb R^2} x$exp[-($\mathbf x$,A$\mathbf x$)] = $\frac{\pi}{\sqrt {detA}}$

here, $\mathbf x$ = $( x,y)$ $\in$ $\mathbb R^2$, and $A$ is a positive definite 2 x 2 symmetric matrix.

so using the formula for n-dimensional Gaussian integral, the integrand is the following $$\sqrt\frac{ {(2\pi)^n}}{det A}$$ but it seems like the question wants me to solve it without using the formula nor calculation any guide or help will be appreciated.

• Perhaps the intended notation was $\mathbf x=(x,y)$, since in the integral you appear to be using $\mathbf x$ where there should be a vector and $x$ vere there should be a scalar? – user228113 Oct 16 '16 at 23:33
• Is it given that $A$ is symmetric? – robjohn Oct 16 '16 at 23:47
• The integral $\int_{\mathbb{R}^n}\color{#C00000}{x}\,e^{-x\cdot Ax}\,\mathrm{d}x$ seems to be a vector. Is the $\color{#C00000}{x}$ supposed to be there? – robjohn Oct 17 '16 at 0:01
• Apologies, I forgot to write that the matrix is symmetric. um for also x, I have edited.. sorry for any inconvenience! – Michael.K Oct 17 '16 at 1:11
• also for the integral, seems like x is there – Michael.K Oct 17 '16 at 1:12

If $A$ is positive definite and symmetric, then there is a $B$ so that $B^TB=A$, then since $\det(A)=\det(B)^2$, a standard change of variables says that \begin{align} \int_{\mathbb{R}^n}e^{-x\cdot Ax}\,\mathrm{d}x &=\int_{\mathbb{R}^n}e^{-|Bx|^2}\,\mathrm{d}x\\ &=\frac1{\det(B)}\int_{\mathbb{R}^n}\,e^{-|x|^2}\,\mathrm{d}x\\ &=\sqrt{\frac{\pi^n}{\det(A)}} \end{align} However, by symmetry, the vector valued integral $$\int_{\mathbb{R}^n}x\,e^{-x\cdot Ax}\,\mathrm{d}x=0$$