$r'=\sqrt {(x'-x)^2+(y'-y)^2+(z'-z)^2}$
$\nabla = \frac {\partial}{\partial x'}+\frac {\partial}{\partial y'}+\frac {\partial}{\partial z'}$
I was studying Poisson's equation in 3D see this link and I stumbled upon the following identity:
$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {\partial f(x',y',z')}{\partial x'}\frac {x'-x}{4\pi r'^3}dx'dy'dz'+\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {\partial f(x',y',z')}{\partial y'}\frac {y'-y}{4\pi r'^3}dx'dy'dz'+$ $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac {\partial f(x',y',z')}{\partial z'}\frac {z'-z}{4\pi r'^3}dx'dy'dz'=-f(x,y,z)$ $\quad$ if $f(x,y,z)$ satisfies the conditions stated in the link.
Infact the identity is equivalent to $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \nabla .( f\frac {x'-x}{4\pi r'^3}+f\frac {y'-y}{4\pi r'^3}+f\frac {x'-x}{4\pi r'^3})dx'dy'dz'=-f(x,y,z)$ because $\nabla .(\frac {x'-x}{4\pi r'^3}+\frac {y'-y}{4\pi r'^3}+\frac {x'-x}{4\pi r'^3})=0$
It is not difficult to show that if $f(x',y'z')=f(x'^2+y'^2+z'^2)$, and the integral above exist, then the identity is true. I did this by simply converting the divergence to polar coordinates and did the integration in polar coordinates.Ofcourse It has already been shown in the link $f(x'^2+y'^2+z'^2,\theta, \varphi)$ can still depend on $\theta$, $\varphi$ and the identity is still valid under the stated conditions. The conditions stated in the link are too restrictive I think.
My question is under which conditions is the identity always true?