Isotropic tensor integral

A rotationally invariant rank-two tensor is a scalar multiple of ${\delta }_{ij}$; taking its trace determines the scalar.

For a radial weight $F$ with the required absolutely integrable moments, rotational symmetry makes the second and fourth moments isotropic tensors. The second moment is $A{\delta }_{ij}$; its trace gives $3A=4\pi {\int }_0^{\infty }{r}^{4}F(r)dr$. Complete symmetry of the fourth moment forces equal coefficients of the three Kronecker delta pairings. Its double tensor contraction gives $15B=4\pi {\int }_0^{\infty }{r}^{6}F(r)dr$, proving the displayed expression. Odd moments vanish under $x\mapsto -x$. Choosing $F$ to be the indicator function of a spherical shell gives all the corresponding shell moments without separate angular integrations.

Away from the origin the Hessian matrix of $r^{-1}$ is $(3x_k{x}_l-r^2{\delta }_{kl})/r^5$. Multiplication by $x_i{x}_j$ makes the singularity locally integrable. Rotational invariance and the symmetries in each index pair reduce the integral to $A{\delta }_{ij}{\delta }_{kl}+B({\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})$. The Laplacian trace gives $3A+2B=0$, while contracting the two pairs as $k=i,l=j$ gives $3A+12B={\int }_{r<a}2/rdV=4\pi a^2$. Solving these two equations yields the displayed expression.

Assume the radial moment integral exists. Rotation through $\varphi$ about the third axis multiplies $x_1+i{x}_2$ by $e^{i\varphi }$ and preserves the region, volume element and radial weight. The integral equals $e^{in\varphi }$ times itself for every angle, so it vanishes. For degree two and four this also follows by contracting the isotropic sphere moments with the complex null vector $(1,i,0)$, whose bilinear square is zero. Odd moments vanish under inversion.

Spherical symmetry makes moments over a ball isotropic tensors. Odd moments vanish under $x\mapsto -x$, while the second and fourth moments areFor the second moment, a tensor contraction gives $3\alpha =\int r^{2}dV=4\pi a^5/5$. For the fourth, complete index symmetry makes the three coefficients equal and a double tensor contraction gives $15\beta =\int r^{4}dV=4\pi a^7/7$. These identities turn polynomial volume integrals into scalar radial integrals and are useful for inertia and rotational averages.