# Integral of a function with compact support

I am reading a book with the following statement

Since $$g$$ has compact support $$\sum_{i=1}^d \int_{\mathbb{R}^d} \frac{\partial (fF_i g)}{x_i} dx = 0$$

where

1. $$\frac{dx}{dt} = F(x)$$ with $$x\in \mathbb{R}^d$$
2. $$g: \mathbb{R}^d \rightarrow \mathbb{R}$$ is continuously differentiable with compact support
3. $$f: \mathbb{R}^d \rightarrow \mathbb{R}$$

The compact support means that if it is zero outside of a compact set.

How can we say the sum of integral is zero?

• This is the volume integral of divergence of the vector field $fgF$. When it is written as a surface integral over a sufficiently large surface, which is beyond the support of $g$, it vanishes. – Amey Joshi Jan 1 '19 at 7:31

The key of the reasoning is that, defined $$V(x)=\left( \begin{matrix} fF_1 g\\ fF_2 g\\ \vdots\\ fF_{d-1} g\\ fF_d g \end{matrix} \right) \implies V\text{ has compact support in }\mathbb{R}^d$$ Since $$\sum_{i=1}^d \int\limits_{\mathbb{R}^d} \frac{\partial (fF_i g)}{\partial x_i} \mathrm{d}x = \int\limits_{\mathbb{R}^d} \nabla\cdot V\, \mathrm{d}x,$$ then, by considering a ball $$B(0,r)\in\mathbb{R}^d$$ of radius $$r>0$$, we have that $$\begin{split} \sum_{i=1}^d \int\limits_{\mathbb{R}^d} \frac{\partial (fF_i g)}{\partial x_i} \mathrm{d}x & = \int\limits_{\mathbb{R}^d} \nabla\cdot V\, \mathrm{d}x\\ &\triangleq\lim_{r\to\infty}\int\limits_{B(0,r)} \nabla\cdot V\, \mathrm{d}x\\ &\text{ by Gauss-Green}\\ &=\lim_{r\to\infty}\int\limits_{\partial B(0,r)} V\cdot\nu_x\, \mathrm{d}\sigma_x=0 \end{split}$$ where $$\nu_x$$ is the normal unit vector to $$\partial B(0,r)$$ in the point $$x\in \partial B(0,r)$$, since $$V$$ has compact support $$\iff$$ $$V\cdot\nu_x=0$$ for all $$\nu_x$$ and all $$r$$ larger than a fixed finite value $$r_0>0$$.