# Integration by parts — applying limits

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.

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Is $f$ isotropic, i.e. depend only on $\vec{x} \cdot \vec{x}$? –  Ron Gordon Jan 8 '13 at 16:24
@rlgordonma : Not necessarily... –  user55502 Jan 8 '13 at 16:30
Think about what you wrote. If you write $v = \int d^3 x f(\vec{x})$, how does the function $v$ depend on its argument? –  Willie Wong Jan 8 '13 at 16:34
@WillieWong : What do you mean? –  user55502 Jan 8 '13 at 16:36
It looks like you are trying to apply the fundamental theorem of calculus, which states that $v'(x) = f(x) \implies v(x) = \int_0^x f(y) \mathrm{d}y + C$. Note that $v$'s dependence on $x$ is captured in the "upper limit" of the integral. This works well in 1 dimension. How do you propose your $v$ (now a function of $\mathbb{R}^3$) depend on its argument? Where does the argument come in in the integral on the right hand side? What is $v'$ for $v$ a function whose domain is $\mathbb{R}^3$? –  Willie Wong Jan 8 '13 at 16:40

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.

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