TITLE: Similar to Cauchy inegral formula
QUESTION [10 upvotes]: Let $f=u+iv$ be an analytic function in disk $\mathbb{D}$ and $0<r<1$.
Can you help me to prove that 
$$\pi{r}f'(0)=\int_{0}^{2\pi}\frac{u(re^{i\theta})}{e^{i\theta}}d\theta\;\;\;?$$
I tried with the Cauchy integral formula, but unsuccessfully.

REPLY [2 votes]: Here is a slightly more real-analytic proof: by the Cauchy-Riemann equations
$$ f'=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}$$
and since $f'$ is harmonic (i.e. its value at any point equals the mean of its values on any ball centered at that point) we have
$$f'(0)=\frac{1}{\pi r^2}\int_{\mathbb{D}_r}f'(x+iy)\,dx\,dy=\frac{1}{\pi r^2}\int_{\mathbb{D_r}}\left(\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}\right)=\frac{1}{\pi r^2}\int_{\partial\mathbb{D}_r}u(\nu_x-i\nu_y).$$
Here $\mathbb{D}_r=\{|z|<r\}$ and $\nu=(\nu_x,\nu_y)$ is the outward unit normal. In the last equality we used the divergence theorem.
But if we parametrize $\partial\mathbb{D}_r$ by $\theta\mapsto re^{i\theta}$ (as usual) we notice that $\nu=e^{i\theta}$,
so that $\nu_x-i\nu_y=\overline{\nu}=e^{-i\theta}$ and we finally get
$$f'(0)=\frac{1}{\pi r}\int_0^{2\pi}u(re^{i\theta})e^{-i\theta}\,d\theta.$$