I am trying to to show: \begin{equation} \int_{\mathbb{R}^{n}}{}d^{n}\vec{x}\frac{\partial{}}{\partial{}x^{i}}\left(B(\vec{x})\exp\left(-\frac{1}{2}\vec{x}^{T}A\vec{x}\right)\right)=0 \end{equation} where $B\left(\vec{x}\right)$ is any function and $A$ is a real symmetric matrix.

Furthermore, I have already proved:

\begin{equation} \int_{\mathbb{R}^{n}}{}d^{n}\vec{x}\exp\left(-\frac{1}{2}\vec{x}^{T}A\vec{x}\right)=\sqrt{\frac{\left(2\pi\right)^{n}}{\det\left(A\right)}} \end{equation}


\begin{equation} \int_{\mathbb{R}^{n}}d^{n}\vec{x}\exp\left(-\frac{1}{2}\vec{x}^{T}A\vec{x}+\vec{j}^{T}\vec{x}\right)=\sqrt{\frac{\left(2\pi\right)^{n}}{\det\left(A\right)}}\exp\left(\frac{1}{2}\vec{j}^{T}A^{-1}\vec{j}\right) \end{equation}

My attempt at a solution was the following: I wrote the Taylor expansion of $B\left(\vec{x}\right)$ \begin{equation} B\left(\vec{x}\right)=\sum_{k_{1}=0}^{\infty}\cdots{}\sum_{k_{n}=0}^{\infty}\frac{x_{1}^{k_{1}}\cdot{}\ldots{}x_{n}^{k_{n}}}{k_{1}!\cdot{}\ldots{}\cdot{}k_{n}!}\frac{\partial{}B}{\partial{}x_{1}^{k_{1}}\cdots{}\partial{}x_{n}^{k_{n}}} \end{equation} and plugged it into the expression that I want to compute. Then I used the product rule and thus obtained two integrals. At this point I should be able to solve them explicitely, but I get a pretty messy expression when I try to change coordinates to diagonalize $A$, because $x_{1}^{k_{1}}\cdot{}\ldots{}\cdot{}x_{n}^{k_{n}}$ and $\frac{\partial{}}{\partial{}x^{i}}$ are not invariant under the orthogonal transformation.

Is there any trick to show this equality? Am I on the right track? If so, could you give me some hints as to how I should proceed?


  • $\begingroup$ i guess this somehow related to the fact that u somehow integrating a tensor of rank one ($\partial_i B(\vec{x})e^{...}$) over an 'even' intervall (whole $R^n$) this expressions tend to vanish. $\endgroup$ – tired Dec 8 '15 at 21:14
  • $\begingroup$ I am guessing you have some kind of assumptions on $B$ (for instance differentiability). If $B(\vec x) \exp\left(-\frac{1}{2} \vec{x}^T A \vec{x}\right)$ vanishes at infinity, then the fundamental theorem of calculus should do the trick. Another approach is to use integration by parts. Try these two approaches in the $1$-dimensional case and then generalize to $\mathbb{R}^n$. $\endgroup$ – Ben CW Dec 14 '15 at 3:04

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