Let \(I\) be a finite set, \(c\in \mathbb R^I\), \(v\gt 0\), and \(\mu _{c,v}=\bigotimes _{i\in I}N(c_i,v)\). If \(f:\mathbb R^I\to \mathbb R\) is measurable with \(|f|\le 1\) and \(i\in I\), then \(z\mapsto ((z_i-c_i)/v)f(z)\) is \(\mu _{c,v}\)-integrable and \(|\int ((z_i-c_i)/v)f(z)\, d\mu _{c,v}(z)|\le 1+1/v\).

Let \(I\) be a set, \(c,z\in \mathbb R^I\), \(v\gt 0\), \(i,j\in I\), and \(a\in \mathbb R\) with \(|a|\le 1\). Then \(|(((z_i-c_i)(z_j-c_j)/v^2-\delta _{ij}/v)a)|\le ((z_i-c_i)^2+(z_j-c_j)^2)/(2v^2)+1/v\), where \(\delta _{ij}=1\) if \(i=j\) and zero otherwise.

Let \(I\) be finite, \(c\in \mathbb R^I\), \(v\ge 0\), and \(i,j\in I\). The function \(z\mapsto ((z_i-c_i)^2+(z_j-c_j)^2)/(2v^2)+1/v\) is integrable under \(\bigotimes _kN(c_k,v)\), with division by zero interpreted as zero.

Let \(I\) be a finite set, \(c\in \mathbb R^I\), \(v\gt 0\), and \(\mu _{c,v}=\bigotimes _{i\in I}N(c_i,v)\). Put \(B_{ij}(z)=((z_i-c_i)^2+(z_j-c_j)^2)/(2v^2)+1/v\). For all \(i,j\in I\), \(\int B_{ij}\, d\mu _{c,v}=2/v\).

Let \(I\) be a finite set, \(c\in \mathbb R^I\), \(v\gt 0\), and \(\mu _{c,v}=\bigotimes _{i\in I}N(c_i,v)\). For \(i,j\in I\), put \(H_{ij}(z)=(z_i-c_i)(z_j-c_j)/v^2-\delta _{ij}/v\), where \(\delta _{ij}\) is one if \(i=j\) and zero otherwise. If \(f:\mathbb R^I\to \mathbb R\) is measurable and \(|f|\le 1\), then \(H_{ij}f\) is \(\mu _{c,v}\)-integrable for every \(i,j\in I\).

Let \(I\) be a finite set, \(c\in \mathbb R^I\), \(v\gt 0\), and \(\mu _{c,v}=\bigotimes _{i\in I}N(c_i,v)\). For \(i,j\in I\), put \(H_{ij}(z)=(z_i-c_i)(z_j-c_j)/v^2-\delta _{ij}/v\), where \(\delta _{ij}\) is one if \(i=j\) and zero otherwise. For any \(f:\mathbb R^I\to \mathbb R\) with \(|f|\le 1\) and \(i,j\in I\), \(|\int H_{ij}f\, d\mu _{c,v}|\le 2/v\), assigning zero to a nonintegrable integral.