The solution of the Dirichlet problem \begin{equation*} \begin{split} \Delta u=&0, ~~~ in~~\Omega\\ u=&f, ~~~ on ~~\partial\Omega \end{split} \end{equation*} interms of Green's function, $ G(\textbf{r'}, \textbf{r}) $, is given by \begin{equation} u(\textbf{r})=-\int_{\partial\Omega}f(\textbf{r'})\dfrac{\partial}{\partial n}G(\textbf{r'}, \textbf{r})d\sigma, ~~~ \textbf{r}\in\Omega. \end{equation} Where $ \Omega $ is a domain in $ \mathbb{R}^3 $.
The question is show that $ u $, in the above equation, is harmonic in $ \Omega $. i.e $ \Delta u=0 $.
I proceed like this: From Green's first identity for $ u $ and $ G $ $$\int_{\partial\Omega}u\dfrac{\partial G}{\partial n}d\sigma=\int_{\Omega}u\Delta G+(\nabla u).(\nabla G)dv$$ But as $ G $ is harmonic, $ \Delta G $ vanishes and as $ u=f $ on $ \partial\Omega $, the integral becomes $$\int_{\partial\Omega}f\dfrac{\partial G}{\partial n}d\sigma=\int_{\Omega}(\nabla u).(\nabla G)dv$$
Hence $$u(\textbf{r})=-\int_{\Omega}(\nabla u).(\nabla G)dv$$ And I need to show $$\nabla^2 u(\textbf{r})=\nabla^2 \int_{\Omega}(\nabla u).(\nabla G)dv=0$$
But how could I do this? Any help?
Advance thanks.