Gaussian integral in multiple dimensions

Question

I would like to solve $f_k=\int_{-\infty}^{\infty}{x_k}e^{-\frac{1}{2}\sum_{i}^{n}\sum_{j}^{n}A_{ij}(x_{i}-C)(x_{j}-C)}d^{n}x$ for real constant C. I feel like this should be obvious but I'm unable to come up with a solution or find the answer anywhere. Thanks in advance.

Answer 1

First, let us recall that: $$\int_{-\infty}^{\infty}dx e^{-\frac{1}{2}\alpha x^{2}} = \sqrt{\frac{2\pi}{\alpha}}$$ where $\alpha > 0$. This is the usual Gaussian integral in one dimension. Now, let me write: $$\langle x, A x\rangle = x^{T}Ax = \sum_{i,j=1}^{n}x_{i}A_{ij}x_{j}$$ for simplicity. In order to the $n$-dimensional Gaussian integral to be well-defined, we must demand $A$ to be positive definite. In this case, there exists a unitary matrix $V$ such that: $$V^{-1}AV = D = \mbox{diag}\{\alpha_{1},...,\alpha_{n}\}$$ where $\alpha_{1},...,\alpha_{n}$ are the eigenvalues of $A$, which are all strictly positive since $A$ is positive definite. Now, set $y := Vx$. Then, we have: $$\langle x, Ax \rangle = \langle y, D y \rangle = \sum_{i=1}^{n}\alpha_{i}y_{i}^{2}$$ and, also $dx = |\det V| dy = dy$, since $|\det V| = 1$ because $V$ is unitary. Thus: $$\int_{\mathbb{R}^{n}}dx e^{-\frac{1}{2}\langle x, Ax \rangle} = \int dy e^{-\frac{1}{2}\sum_{i=1}^{n}\alpha_{i}y_{i}^{2}} = \prod_{i=1}^{n}\int_{-\infty}^{\infty}dy_{i} e^{-\frac{1}{2}\alpha_{i}y_{i}^{2}} = \prod_{i=1}^{n}\sqrt{\frac{2\pi}{\alpha_{i}}} =\bigg{(}\sqrt{\frac{2\pi}{\alpha_{i}}}\bigg{)}^{n} = \frac{(2\pi)^{\frac{n}{2}}}{{\det A}^{\frac{1}{2}}}$$ where I used the one-dimensional Gaussian integral result stated above and the fact that $\det A = \prod_{i=1}^{n}\alpha_{i}$.

We can proceed in a similar way to evaluate the Laplace transform: $$g(p) := \int_{\mathbb{R}^{n}}dx e^{-\frac{1}{2}\langle x, Ax \rangle}e^{\langle p, x\rangle}$$ with $p \in \mathbb{R}^{n}$. Using the same transformation as before and noticing that: $$\langle p, x \rangle = \langle p, V^{-1}y\rangle = \langle Vp, y\rangle = \langle \zeta, y\rangle$$ with $\zeta := Vp$, we can obtain: $$g(p) = \prod_{i=1}^{n}\int_{-\infty}^{\infty} dy_{i} e^{-\frac{1}{2}\alpha_{i}y_{i}^{2}}e^{\zeta_{i}y_{i}} = \frac{(2\pi)^{\frac{n}{2}}}{(\det A)^{\frac{1}{2}}}e^{\frac{1}{2}\langle p, A^{-1}p\rangle}$$

Finally, if you want to evaluate the object: $$f_{k} = \langle x_{k}\rangle := \int_{\mathbb{R}^{n}}x_{k}e^{-\frac{1}{2}\langle x, Ax\rangle}$$ then the usual trick is to use: $$f_{k} = \frac{\partial g(p)}{\partial p_{k}}\bigg{|}_{p=0} = \bigg{[}\frac{\partial}{\partial p_{k}}\bigg{(}\frac{(2\pi)^{\frac{n}{2}}}{(\det A)^{\frac{1}{2}}}e^{\frac{1}{2}\langle p, A^{-1}p\rangle}\bigg{)}\bigg{]}\bigg{|}_{p=0}$$

I'll leave the details to you!