Generalization of integration of e^{-x^2}

By | August 14, 2015

Lemma. Let T be a self-adjoint operator on a finite-dimensional inner product space V. Then

  • Every eigenvalue of T is real.
  • Suppose that V is a real inner product space. Then the characteristic polynomial of T splits.

Theorem (Spectral Theorem for self-adjoint operator). Let T be a linear operator on a finite-dimensional inner product space V over \mathbb{R}. Then T is self-adjoint if and only if there exists an orthonormal basis \beta for V consisting of eigenvectors of T.

Definition. We say a symmetric n\times n real matrix M is said to be positive definite if z^{T}Mz is positive for every non-zero column vector z of n real numbers.

It is a quite easy consequence that if A is positive definite symmetric matrix with real coefficients, all its eigenvalues are positive.

Exercise. Let A be a d\times d positive definite symmetric matrix with
real coefficients. Show that

    \[ \int_{\mathbb{R}^{d}}e^{-\pi\left(x,A\left(x\right)\right)}dx=\left(\det\left(A\right)\right)^{-\frac{1}{2}}. \]

This generalizes the fact that \int_{\mathbb{R}^{d}}e^{-\pi\left|x\right|^{2}}dx=1, which corresponds to the case where A is the identity.

Proof. Since A is positive definite symmetric matrix with real coefficients, by spectral theorem, write A=R^{-1}DR where R is a rotation
and D is diagonal with entries \lambda_{1},\dots,\lambda_{d}, where \left\{ \lambda_{i}\right\} are the eigenvalues of A. Then we have

    \begin{align*} & \int_{\mathbb{R}^{d}}e^{-\pi\left(x,A\left(x\right)\right)}dx\\ & =\int_{\mathbb{R}^{d}}e^{-\pi\left(x,RDR^{-1}\left(x\right)\right)}dx\\ & =\int_{\mathbb{R}^{d}}e^{-\pi\left(R^{-1}x,DR^{-1}x\right)}dx\\ & =\int_{\mathbb{R}^{d}}e^{-\pi\left(y,Dy\right)}\left|\det R\right|dy\\ & =\int_{\mathbb{R}^{d}}e^{-\pi\sum_{i=1}^{d}\lambda_{i}y_{i}^{2}}dy\\ & =\prod_{j=1}^{d}\int_{\mathbb{R}}e^{-\pi\lambda_{i}\left|y_{i}\right|^{2}}dy_{i}. \end{align*}

Since \int_{\mathbb{R}}e^{-\pi\lambda y^{2}}dy=\frac{1}{\sqrt{\lambda}} for \lambda>0, we have

    \[ \int_{\mathbb{R}^{d}}e^{-\pi\left(x,A\left(x\right)\right)}dx=\prod_{j=1}^{d}\frac{1}{\sqrt{\lambda_{j}}}=\left(\det A\right)^{\frac{1}{2}}.\qedhere \]

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