why $\int_\Omega S(t) div(w) = \int_\Omega div(w)$ ? where $S(t)$ is a heat semigroup with neumann condition In this work 
https://www.dropbox.com/s/sygzebrr87ma99z/cao_2015_publicado.pdf?dl=0
page 1984, The author claims that 
Let $w \in (C^{\infty}_0(\Omega))^N$ then $\int_{\Omega}S(t)div(w)=\int_{\Omega}div(w)$ See,
where $div$ is the divergence operator, S(t) is a heat semigroup with Neumann condition (Analytic semigroup) and $\Omega$ is a open set of $R^N$
I understand the right side (use divergence theorem and $w=0$ in boundary) and the continuation of the proof is clear for me. I know that if div (w) = constant the statement is true. I appreciate your help.
Thanks and sorry for my English

The divergence term is not important here. What is important is that solutions of the heat equation with homogeneous Neumann boundary conditions preserve total heat, so that
$$\int_\Omega S(t) f \, dx = \int_\Omega f \, dx$$
for all $t$. To show this, simply note that $u(x,t)=S(t)f(x)$ solves the heat equation $u_t = \Delta u$ with $\partial u/\partial \nu=0$ on the boundary and $u(x,0) = f(x)$. Integrating by parts we have
$$\frac{d}{dt} \int_\Omega u \,dx = \int_\Omega u_t \, dx = \int_\Omega \Delta u \, dx = \int_{\partial \Omega} \frac{\partial u}{\partial \nu} \, dS = 0.$$
So
$$\int_\Omega S(t) f \, dx = \int_\Omega u(x,t) \, dx = \int_\Omega u(x,0) \, dx = \int_\Omega f(x) \, dx.$$
