Similar to Cauchy inegral formula 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.
 A: By the Cauchy formula for derivatives:
$$\int_0^{2\pi}\frac{f(re^{i\theta})}{e^{i\theta}}d\theta=\int_0^{2\pi}\frac{f(re^{i\theta})}{r^2e^{2i\theta}}\frac{rd(re^{i\theta})}{i}=2r\pi\frac{1!}{2i\pi}\int_{|z|=r}\frac{u(z)+iv(z)}{z^2}dz=2r\pi f'(0)$$
Now,
$$\int_{|z|=r}\frac{u-iv}{z^2}dz=\int_{|z|=r}\frac{\bar{f}(z)}{z^2}dz=0$$
as is easily seen, e.g. considering the series
$$\bar{f}(z)=\sum_{n}a_n \bar{z}^n$$
and the fact that
$$\int_{|z|=r}z^a\bar{z}^bdz=0$$
unless $a=b$.
NB The Cauchy formula for derivatives is
$$f^{(n)}(w)=\frac{n!}{2i\pi}\int_{|z-w|=r}\frac{f(z)}{(z-w)^{n+1}}dz$$
A: An idea, (but perhaps not the best answer, there is some computations). Put $f(z)=\sum a_k z^k$, and $a_k=u_k+iv_k$ with $u_k, v_k \in \mathbb{R}$. Then
$$u(r\exp(i\theta))=\sum (u_kr^k \cos(k\theta)-v_kr^k\sin(k\theta))$$
And these series of functions are normally convergent.
Note that as the functions are periodic with period $2\pi$:$$\int_0^{2\pi}u(r\exp(i\theta))\exp(-i\theta)d\theta=\int_{-\pi}^{+\pi}u(r\exp(i\theta))\exp(-i\theta)d\theta$$
and that 
$$\int_{-\pi}^{+\pi}\cos(k\theta)\sin(\theta)d\theta=
\int_{-\pi}^{+\pi}\sin(k\theta)\cos(\theta)d\theta=0$$
for all $k$ (odd functions)
and
$$\int_{-\pi}^{+\pi}\cos(k\theta)\cos(\theta)d\theta=
\int_{-\pi}^{+\pi}\sin(k\theta)\sin(\theta)d\theta=0$$
 if $k\not =1$, and their value is $\pi$ if $k=1$. 
This gives
$$\int_{-\pi}^{+\pi}u(r\exp(i\theta))\exp(-i\theta)d\theta=\pi r u_1+i\pi rv_1=\pi r a_1=\pi r f^{\prime}(0)$$ 
A: You can take the power series of $f$ and integrate termwise.
If $f(z)=\sum\limits_{n=0}^\infty a_nz^n$ then 
$$ u(z)=\frac12\left(\sum_{n=0}^\infty a_nz^n+\sum_{n=0}^\infty \overline{a_n}\cdot \overline{z}^n \right) $$
so
$$
\int_0^{2\pi} u(re^{it})e^{-it}dt = 
\int_0^{2\pi} \left(\sum_{n=0}^\infty 
\frac{ a_n r^ne^{nit} + \overline{a_n}r^ne^{-nit}}2 \right) e^{-it}dt = \\
= \sum_{n=0}^\infty \frac{a_nr^n}2 \int_0^{2\pi}e^{(n-1)it}dt
+ \sum_{n=0}^\infty \frac{\overline{a_n}r^n}2 \int_0^{2\pi}e^{(-n-1)it}dt= \\
=  \sum_{n=0}^\infty \frac{a_nr^n}2 
\begin{Bmatrix}
2\pi & \text{if $n-1=0$} \\
0 & \text{if $n-1\ne0$} \\
\end{Bmatrix}
+ \sum_{n=0}^\infty a_n r^n 
\begin{Bmatrix}
2\pi & \text{if $-n-1=0$} \\
0 & \text{if $-n-1\ne0$} \\
\end{Bmatrix}
= \\ = \frac{a_1r}2 \cdot 2\pi = \pi r a_1 = \pi r f'(0).
$$
A: 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
$$f'(0)=\frac{1}{\pi r}\int_0^{2\pi}u(re^{i\theta})e^{-i\theta}\,d\theta.$$
A: Maybe, next perform is correct. Let
$g(z)=\overline{f(\overline{z})}$. Than $g(z)$ is analytic in $\mathbb{D}$. By Cauchy integral formula we have 
$$f'(0)=\frac{1}{2\pi{i}}\int_{\partial D(0,r)}\frac{f(z)}{z^2}dz=\frac{1}{2\pi{i}}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{r^2e^{i2\theta}}re^{i\theta}id\theta=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta$$. 
On the other hand we have 
$$\frac{1}{\pi{r}}\int_{0}^{2\pi}\frac{u(re^{i\theta})}{e^{i\theta}}d\theta=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta+\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\overline{f(re^{i\theta})}}{re^{i\theta}}d\theta=$$
$$=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta+\frac{1}{2\pi}\int_{0}^{2\pi}\frac{g(re^{-i\theta})}{re^{i\theta}}d\theta=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta+\frac{1}{2\pi{r^2}i}\int_{\partial D(0,r)}g(z)dz=$$
$$=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta=f'(0).$$
