I need to evaluate an integral of the form

$$\int_{-\infty}^\infty ... \int_{-\infty}^\infty \exp[-\mathbf{x}^T\mathbf{M}\mathbf{x} + \mathbf{r}^T\mathbf{x}] dx_1...dx_N$$

I know that when $\mathbf{r} = \mathbf{0}$, the integral is simply $\sqrt{\frac{\pi^N}{\det\mathbf{M}}}$. The matrix, $\mathbf{M}$, is large but its structure is block

$$\mathbf{M} = \begin{bmatrix} \mathbf{A} & \mathbf{G} \\ \mathbf{G}^T & \mathbf{0} \end{bmatrix}$$

where $\mathbf{A}$ is also block diagonal, so that the determinant of $\mathbf{M}$ is not difficult to find in closed form (by decomposing into a product of matrices with determinants that are much easier to calculate). However, when $\mathbf{r} \neq \mathbf{0}$ then the only way I know to evaluate the integral is to diagonalize $\mathbf{M}$ by $\mathbf{P} \mathbf{M} \mathbf{P}^T$, rotate $\mathbf{r}$, and complete the square. This requires that I find the eigenvectors and eigenvalues of $\mathbf{M}$. Is there either (1) a simpler way to find the eigenvalues and eigenvectors of $\mathbf{M}$ that makes use of its structure ($\mathbf{A}$ is $5M \times 5M$ consisting of $5 \times 5$ blocks, $\mathbf{G}$ is $3 \times 5M$ and $N = 5M + 3$) or (2) an alternative way to evaluate the integral that does not require finding the eigenvalues and eigenvectors of $\mathbf{M}$?

  • $\begingroup$ Do the completion of the squares business and compute. At the end in the result, the factorization of $M$ gets back together, and the result can be put in terms of $M$ and $r$ without computing factorizations of $M$. There is a factor, of the result that is the determinant of a square root of $M$, but that is just the square root of of the determinant of $M$, which doesn't require the factorization of $M$ to be computed. $\endgroup$ – Peyton Aug 2 '17 at 3:10

Got some time, so I can detail what I said in the comment.

Let $M=A^TA$, where $A$ might be complex.


\begin{align} -x^TMx+r^Tx&=-(Ax)^T(Ax)+r^Tx-\frac{1}{4}r^TA^{-1}(A^{-1})^{T}r+\frac{1}{4}r^TA^{-1}(A^{-1})^{T}r\\ &=-(Ax-(A^{-1})^Tr)^T(Ax-(A^{-1})^Tr)+\frac{1}{4}r^TA^{-1}(A^{-1})^{T}r\\ &=-(Ax-(A^{-1})^Tr)^T(Ax-(A^{-1})^Tr)+\frac{1}{4}r^TM^{-1}r \end{align}

Therefore, making the change of variable $y=Ax-(A^{-1})^Tr$ in the integral we get that it is equal to

$\det(A^{-1})e^{\frac{1}{4}r^TM^{-1}r}\int e^{-y^Ty}=\frac{1}{\sqrt{\det(M)}}e^{\frac{1}{4}r^TM^{-1}r}\sqrt{\pi^N}$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.