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A Cartesian tensor invariant under all proper rotations fixing a unit vector $n$ is axisymmetric. Without assuming a symmetric tensor, its general second-order form is the displayed combination of the transverse identity, axial projector and axial antisymmetric part. For $n=e_3$ this isIndeed, imposing $RT{R}^T=T$ for a quarter-turn about $e_3$ forces $T_{11}=T_{22}$, $T_{12}=-T_{21}$ and all mixed axial-transverse entries to vanish. The planar block is a linear combination of the identity and the planar rotation generator, so it commutes with every rotation about the axis. Thus quarter-turn invariance already suffices for a second-order tensor. A symmetric tensor has $\omega =0$, recovering the axisymmetric symmetric rank-two tensor form. The antisymmetric term is allowed because only proper rotations about the fixed axis are imposed.

The integrand is symmetric in $i,j$, so $T$ is a symmetric tensor. The sphere and its area element are invariant under rotations, and the integrand transforms covariantly when $\mathbf{x},\mathbf{y}$ are rotated together. For fixed $\mathbf{y}$, rotations about its direction therefore leave $T$ unchanged. In axes with $\mathbf{y}$ along the third axis, the mixed transverse-axial entries vanish and the two transverse eigenvalues are equal. Thus it is an axisymmetric symmetric rank-two tensor,where $\alpha ,\beta$ depend on $a,n$ but not on the direction of $\mathbf{y}$. The restriction $n>-1$ makes the integral convergent near $\mathbf{x}=\mathbf{y}$: the integrand is of size distance${}^{2n}$ and the local area element has one further radial power.

Take the polar axis along $\mathbf{y}$. Because $|\mathbf{x}|=|\mathbf{y}|=a$,The area element is $a^2\sin \theta d\theta d\varphi$. Hence with $s=1-\cos \theta$,

For the second scalar integral, take the trace. It contracts the two chord-vector factors into their squared length:Put $K=\pi (2a{)}^{2n+2}$. The tensor form gives $3\alpha +\beta a^2=K/(n+1)$, while its axial contraction gives $\alpha +\beta a^2=K/(n+2)$. Solving yields the weighted chord tensor of a sphere coefficientsBecause $a>0$, this is an isotropic tensor exactly when $\beta =0$. Therefore $n=-1/2$ is the unique permitted isotropic exponent.

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 $\mathrm{tr}T=K/(n+1)$ and $y^{T}Ty=a^{2}K/(n+2)$. Solving $3\alpha +\beta a^2=\mathrm{tr}T$ and $\alpha +\beta a^2=y^{T}Ty/a^2$ 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 $n=-1/2$.