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Carlo Beenakker
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indeed, integration by parts it is:

$$\int d\mathbf{x}\; \vec{x} \,f(\vec{x})\exp\left(-\tfrac{1}{2}\vec{x}\cdot \mathbf{C}^{-1}\cdot\vec{x}\right)=-\int d\mathbf{x}\; \,f(\vec{x})\,\mathbf{C}\cdot\frac{\partial}{\partial \vec{x}}\exp\left(-\tfrac{1}{2}\vec{x}\cdot \mathbf{C}^{-1}\cdot\vec{x}\right)$$ $$=\int d\mathbf{x}\; \,\exp\left(-\tfrac{1}{2}\vec{x}\cdot \mathbf{C}^{-1}\cdot\vec{x}\right)\mathbf{C}\cdot\frac{\partial}{\partial \vec{x}}\, f(\vec{x})$$

where I have used that $\mathbf{C}$ is a symmetric matrix

Carlo Beenakker
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