If $\Delta u=0, \text{then}\, \exists v \ni \Delta v=0$ with $u$ and $v$ are harmonic functions.

Question

Could you please explain to me how we get the last integral?

Answer 1

From the definition, $v(x,y) = \int_\gamma - \frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy$, where $\gamma$ is any path between $p$ an $(x,y)$. Denoting $\gamma_{(x,y)}$ to emphasize the point, one can get $$ v(x+h,y) - v(x,y) = \left(\int_{\gamma_{(x+h,y)}} - \frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy\right) - \left(\int_{\gamma_{(x,y)}} - \frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy\right) = \\ = \left(\int_{\gamma_{(x+h,y)}} - \frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy\right) + \left(\int_{-\gamma_{(x,y)}} - \frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy\right) = \int_{\gamma_{(x+h,y)}-\gamma_{(x,y)}} - \frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy $$ where the last equality is due to additivity of line integral. As $\gamma_{(x+h,y)}$ is some path from $p$ to $(x+h,y)$ and $-\gamma_{(x,y)}$ is a path from $(x,y)$ to $p$, it follows that $\gamma_{(x+h,y)} - \gamma_{(x,y)}$ is a path from $(x,y)$ to $(x+h,y)$. As all pathes are arbitrary, we the path from $(x,y)$ to $(x+h,y)$ to be $\gamma_h(t) = (x+th, y)$ as it defined in the given theorem.