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Lubos Motl has already provided a correct answer. This answer uses a different approach in the spirit of perturbation theory with sources: $$\int \! d^nx ~f(x)~e^{-\frac{1}{2}x^TAx +j^Tx} ~=~f\left(\frac{\partial}{\partial j}\right) \int \! d^nx ~e^{-\frac{1}{2}x^TAx +j^Tx} ~~\stackrel{\begin{matrix}\text{Gauss.}\\ \text{int.}\end{matrix}}{=}~~C~ ...


The case with the linear term is obtained from the original one by a simple shift, i.e. the substitution $$ x = X + A^{-1} B $$ Substitute it to the exponent in your more general integral: $$ -\frac 12 x^T A x + B^T x = -\frac 12 (X^T+B^T A^{-1}) A(X+A^{-1}B)+B^T (X+A^{-1}B)=\dots $$ I used $A=A^T$. Now, all the terms that are schematically $BX$ i.e. linear ...

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