Schwarz integral formula

Remarks

Call the integral $S(z)$. The kernel $(e^{it} + z)/(e^{it} - z)$ is holomorphic in $z$, so $S$ is holomorphic on $\mathbb{D}$, and its real part is the Poisson integral of $u$ with the Poisson kernel. Since $u$ is harmonic in $\mathbb{D}$ and continuous on $\overline{\mathbb{D}}$, the Poisson integral of $u$ equals $u$. So $f - S$ is holomorphic with real part $0$, and hence it is a purely imaginary constant. At $z = 0$ the integral is the mean of $u$ over the circle, which is the real number $u(0)$, so the constant is $i \operatorname{Im} f(0)$. Example: for $f(z) = z$ one has $u(e^{it}) = \cos t$, and the formula gives $\frac{1}{2\pi}\int_0^{2\pi} \frac{e^{it} + z}{e^{it} - z}\cos t \, dt = z$.

Going deeper

For a student

The formula is the holomorphic counterpart of the Poisson integral: taking real parts of both sides gives the Poisson representation of $u$. It solves the problem of finding a holomorphic function with prescribed real part on the boundary, which is a simple Riemann–Hilbert type problem. It is used to prove the Herglotz representation theorem and the Borel–Carathéodory theorem, and it has versions for the upper half-plane.