I want to find solution to below integral:
$$J=\int \left(\delta_{ij}-\frac{3 r_i r_j}{r^2}\right)\frac{1}{r^3}\,dx \, dy \,dz $$
In which $\vec{r}=(x_0-x, \: y_0-y, \:z_0-z)$.
I did as below:
\begin{align*} J&=\int \left(\delta_{i \:j}-\frac{3 r_i r_j}{r^2}\right)\frac{1}{r^3} \,r^2\sin(\theta) \,dr\,d\theta \,d \phi\\&= \int \left(1-3 \sin^2(\theta) \cos^2 (\phi)\right)\delta_{ij}\frac{1}{r^3}\,r^2 \,dr \, \sin(\theta) \, d\theta \,d\phi.\end{align*}
After taking angular integration, above integral equals to zero. However I know that, the solution to the integral is zero. Could any one tell me what is the problem with the method I used and how can I find the solution to the integral?
The integral is on infinite space.
I also tried to write it as a divergence of a third rank tensor as below:
$$\frac{1}{2}\int \nabla_k \left(\frac{r_k \delta_{i j}}{2 r^3}-\frac{r_i r_j r_k}{r^5}\right) dx\, dy\, dz,$$ but I don't know how to use Gauss theorem in above integral.