Volume integral over dyadic: Is it true that $\int_{\mathbb{R}^3} d^3 \vec{x}\; x_i x_j f(|\vec{x}|) \ \propto \ \delta_{ij}$?

Question

Let $f(a)$ be some function of $a>0$. Is it true that $$ \int_{\mathbb{R}^3} d^3 \vec{x}\; x_i x_j f(|\vec{x}|) \ \propto \ \delta_{ij} $$ where $\delta_{ij}$ is the Kronecker delta, and where $x_i$ are the cartesian components of $\vec{x}$?

This seems to be true when pick various values for $i$ and $j$. Is there an easy way to prove this?

Answer 1

Change the variables by a reflection: $$I_{12}=\int_{(x_1,x_2,x_3)\in\mathbb R^3}x_1x_2\,f\left(\sqrt{x_1^2+x_2^2+x_3^2}\right)dV$$ $$=\int_{(x_1,-x_2,x_3)\in\mathbb R^3}x_1(-x_2)\,f\left(\sqrt{x_1^2+(-x_2)^2+x_3^2}\right)|{-1}|dV$$ $$=-I_{12}$$ $$\,$$ $$I_{11}=\int_{(x_1,x_2,x_3)\in\mathbb R^3}x_1^2\,f\left(\sqrt{x_1^2+x_2^2+x_3^2}\right)dV$$ $$=\int_{(x_2,x_1,x_3)\in\mathbb R^3}x_2^2\,f\left(\sqrt{x_2^2+x_1^2+x_3^2}\right)|{-1}|dV$$ $$=I_{22}$$