The integral is zero by using a theoretical result Set $F(x)=|u(x)|^2\cdot x$, where $u\in L^2(\mathbb{R}^N)$, then 
$$
\begin{split}
\mathrm{div} \ F(x)
 &= \nabla |u(x)|^2 \cdot x\ +\  |u(x)|^2 \mathrm{div} (x)\\
 &= \overline{u}(x)\nabla u(x)+  u(x)\nabla \overline{u} (x)+N   |u(x)|^2
\end{split}
$$
I want to assert that:
$$
\int_{\mathbb{R}^N} \mathrm{div}\ F(x)\ dx=0
$$
using a theoretical result that states if $f\in L^1(\mathbb{R}^N)$ and if $\partial f/\partial x_1\in L^1(\mathbb{R}^N)$, then 
$$
\int_{\mathbb{R}^N} \frac{\partial f}{\partial x_1}(x)\ dx=0    
$$
So it seems I must find a function $G\in L^1(\mathbb{R}^N)$ with that $  \partial G/\partial x_1 =\mathrm{div}\ F $. Any idea of how can I find such  function $G$? Thanks in advance.
 A: Not sure what is the "theoretical result" you mention.
This seems like a question about the Divergence Theorem, which states:
$$
\int_{\Omega} \mathrm{div}\ F(x)\ dV = \oint_{\partial\Omega} F.\hat{n}.dS  
$$
Now, let $\Omega=B_r(0)$, the ball centered around zero with radius $r$. Then:
$$
\int_{\mathbb{R}^N} \mathrm{div}\ F(x)\ dx = \lim_{r\to \infty}\int_{B_r(0)} \mathrm{div}\ F(x)\ dV
$$
Now let's try to find:
$$
\oint_{\partial B_r(0)} F.\hat{n}.dS =\oint_{\partial B_r(0)} |u(x)|^2\cdot x\cdot \frac{x}{r} dS = \oint_{\partial B_r(0)} |u(x)|^2 \frac{|x|^2}{r} dS
$$
Now, I'm temped to ask if you have more assumptions for $u(x)$.
A: This puzzled me for a while... Notice that:
For $N=1$ we have:
$$
\int_\mathbb{R} \text{div} F(x)dx = \int_\mathbb{R} \frac{\partial F(x)}{\partial x}dx = \lim_{r\to \infty} F(r)-F(-r) 
$$
Naturally, because $u(x)$ is square-integrable, we have $\lim_{r\to \infty} u(r)=\lim_{r\to \infty} u(-r) =0$. So there is a gap I cannot fill yet to assert that  $\lim_{r\to \infty} F(r)=0$.  This would demonstrate that for $N=0$ the result is indeed zero.
For higher dimensions:
$$
\int_\mathbb{R^N} \text{div} F(x)dx = \int_\mathbb{R^N} \sum_{i=1}^N\left( \frac{\partial F(x)}{\partial x_i} \right) dx_1dx_2 \cdots dx_N
$$
Swapping integration and summation and reordering each integral with the variable ebign derived:
$$
\int_\mathbb{R^N} \text{div} F(x)dx =  \sum_{i=1}^N \int_\mathbb{R^{N-1}}\left( \int_\mathbb{R}\frac{\partial F(x)}{\partial x_i} dx_i\right) dx_1dx_2 \cdots  dx_{i-1}dx_{i+1}  \cdots dx_N
$$
Then each term in parenthesis is zero by the same logic. Hence the whole integral is zero.
