Integration by parts — applying limits

Question

Is the following integration by parts done correctly?

Given that $$\int_{\mathbb{R}^3}d^3x\,\,\,\,\,f(\vec x)=0\\\int_{\mathbb{R}^3}d^3x\,\,\,\,\,\vec xf(\vec x)=0$$ I am trying to evaluate the integral $$\int_{\mathbb{R}^3}d^3x\,\,\,\,\,\vec x\cdot \vec x \,\,f(x)$$

So I take $$u=\vec x\cdot \vec x\implies u'=2\vec x\\ v'=f(\vec x)\implies v=\int d^3x\,\,\,f(\vec x)$$

Am I allowed to apply the "limits" to the integral at this stage or must I leave it as indefinite? In other words, can I take $v=0$ thus the whole integral $=[uv]_{limits} -\int\,d^3x\,\,\,u'v=0$?

Thank you.

Answer 1

Your $v$ does not appear to obey $\nabla v = f$.

This is the product fundamental theorem of calculus and the product rule in 3d:

$$\int_S uv \, dS = \int_V u \nabla v + v \nabla u \, dV$$

To construct $v$ such that $\nabla v = f$, you need to use the Green's function for $\nabla$--$G = x/4\pi|x|^3$. That is,

$$\nabla \int_V G(x-x') f(x') \, dV' = \int_V \delta(x-x') f(x') \, dV' = f(x)$$

This no longer makes it obvious (to me) that $\int_V v \nabla u \, dV$ is zero.