From the wikipedia article, for a Gaussian integral of an analytic function we have that
This is equivalent to the Wick theorem when f(x) is a polynomial.
Now I'm trying to obtain a similar formula when there is a linear term in the Gaussian (ie the Gaussian has a nonzero mean).
My guess is that $$ \int f(x) \exp \left( - \frac{1}{2} x^T A x + B^T x \right) d^n x = \\ = \sqrt{\frac{(2 \pi)^n}{\det A}} \exp \left[ \frac{1}{2} \left( B^T + \frac{d}{dx_i} \right) A^{-1} \left( B + \frac{d}{dx_j} \right) \right] f(x) \bigg|_{x=0} $$
but I can't prove it. Is this equation correct? How can I prove it?
Answer
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 in $X$ cancel: the coefficient is $-1/2-1/2+1=0$. The remaining terms give $$ - \frac 12 X^T A X + \frac 12 B^T A^{-1} B $$ The coefficient $+1/2$ in the second term came from $-1/2+1$. The second term only gives a simple factor (the exponential of that), and it's a part of your result – except that the last $B^T$ in your result should be simply $B$.
The hard, quadratic/Gaussian part of the expression may be rewritten in the Wick way from your first identity. It could be enough if you were satisfied with the evaluation of the $x$-derivatives not at $X=0$ but at the right original value $x=0$ which means, thanks to my substitution $$ X = -A^{-1}B. $$ However, if you want to use the values of the derivatives at $X=0$, you have to Taylor-expand the shift operator from $X=0$ to $X=-A^{-1}B$. The shift is the operator $$ \exp(B^T A^{-1} \frac d{dx}) $$ which is exactly what you get from the mixed terms in your guessed exponent, up to an overall sign perhaps that you will surely be able to catch correctly.
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