= Weighted chord tensor of a sphere
{title2=$T(y)=\int_{|x|=a}(y-x)(y-x)^T|y-x|^{2n-2}\,dS_x$}
For $|y|=a$ and $n>-1$, this surface integral is finite: near $x=y$ its integrand has size $O(|x-y|^{2n})$ and the two-dimensional area element supplies one further power of distance. Rotational covariance and symmetry make it an <axisymmetric symmetric rank-two tensor>. Put $K=\pi(2a)^{2n+2}$. Polar integration about the $y$ axis gives $\operatorname{tr}T=K/(n+1)$ and $y^TTy=a^2K/(n+2)$. Solving $3\alpha+\beta a^2=\operatorname{tr}T$ and $\alpha+\beta a^2=y^TTy/a^2$ gives $\alpha=K/[2(n+1)(n+2)]$ and $\beta=\alpha(2n+1)/a^2$. Thus the weighted chord tensor becomes an <isotropic tensor> exactly at $n=-1/2$.
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