The domain is an open unit box (if required) and $u$ is harmonic, $v$ is harmonic conjugate defined below. Prove Complex Differentiabilty
Diffirentiate with respect to $x$ and $y$:
$$v(x,y)=-\int_0^x u_y(t,0)dt+\int_0^y u_x (x,t) dt$$
I assumed the answer would be:
$v_x=-u_y$ and $v_y=u_x$ (which would imply complex differentiability using Cauchy Riemann), but can somebody help me do the differentiation explicitly, with explanations?
I tried using $dv/dx=dv/dt \times dt/dx$
If I'm correct $dv/dt=-u_y(x,0)+u_x(x,y)$
Let $v(x,y)$ be given by
$$v(x,y)=-\int_0^x u_y(t,0)\,dt+\int_0^y u_x (x,t)\, dt$$
Then, taking the partial derivative with respect to $x$ yields
$$\begin{align} v_x(x,y)&=-u_y(x,0)+\int_0^y u_{xx} (x,t) \,dt \tag 1\\\\ &=-u_y(x,0)-\int_0^y u_{yy} (x,t) \,dt \tag 2\\\\ &=-u_y(x,0)-\left(u_y(x,y)-u_y(x,0)\right)\\\\ &=-u_y(x,y) \end{align}$$
which is a Cauchy-Riemann equation. Note in going from $(1)$ to $(2)$, we exploited the fact that $u$ is harmonic.
Then, taking the derivative with respect to $y$ yields
$$v_y(x,y)=u_x(x,y)$$
which is the other Cauchy-Riemann equation. And we are done!