Relationship between Cauchy's Integral formula and Poisson kernel If we have a function $\tilde{f}\in L^p(T)$ on the unit circle $T$ with $p \geq 1$ we can regain a harmonic function $f$ on the unit disk using the Poisson kernel $P_r$:
$$
f\left(re^{i\theta}\right)=\frac{1}{2\pi} \int_0^{2\pi} P_r(\theta-\phi) \tilde f\left(e^{i\phi}\right) \,\mathrm{d}\phi, \quad r < 1
$$
It looks somewhat similar to the Cauchy's Integral formula. The latter states that a holomorphic function defined on a disk is completely determined by its values on the boundary $\gamma$ of the disk.
$$
f(a)={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{z-a}}\,dz
$$
Questions:
1) Could you please explain the logic behind the integral transformation with the Poisson kernel? Why does the kernel have the form $P_{r}(\theta )=\operatorname {Re} \left({\frac {1+re^{i\theta }}{1-re^{i\theta }}}\right),\ 0\leq r<1$ ?
2) What is the difference between the integral transformation with the Poisson kernel and the Cauchy's integral formula?
Thank you for your help in advance.
 A: The usual Cauchy kernel is a member of a wider family: we can 
add an arbitrary holomorphic function. If $h$ is holomorphic on the closed disk then
$$
f(a)
= \frac1{2\pi i}\oint_{|z|=1} f(z) \left(\frac1{z-a}+h(z)\right) \mathrm{d}z
= \frac1{2\pi}\oint_{|z|=1} f(z) \left(\frac1{1-a\bar{z}}+zh(z)\right) \frac{\mathrm{d}z}{iz}.
$$
$$
= \frac1{2\pi}\oint_{|z|=1} f(z) \left(1+\frac{a\bar{z}}{1-a\bar{z}}+zh(z)\right) \frac{\mathrm{d}z}{iz}.
$$
From this point we can look for a function $zh(z)$ (which must have a root at $0$) such that the modified kernel
$1+\frac{a\bar{z}}{1-a\bar{z}}+zh(z)$ is real along the unit circle. (Notice that $\frac{\mathrm{d}z}{iz} = \mathrm{d}(\arg z)$ is real.) This can be done by choosing
$$
 zh(z) = \overline{\left(\frac{a\bar{z}}{1-a\bar{z}}\right)} =
\frac{\bar{a}z}{1-\bar{a}z},
$$
so
$$
h(z) = \frac{\bar{a}}{1-\bar{a}z},
$$
that is holomorphic in the disk $|z|<\frac1{|a|}$.
Hence, the modified kernel is
$$
1+\frac{a\bar{z}}{1-a\bar{z}} + \overline{\left(\frac{a\bar{z}}{1-a\bar{z}}\right)} 
= \mathrm{Re}\left(1+2\frac{a\bar{z}}{1-a\bar{z}}\right)
= \mathrm{Re}\left(\frac{1+\bar{a}z}{1-\bar{a}z}\right).
$$
Therefore, the difference between the Cauchy and the Poission kernels is a holomorphic function. :-)
