Schwarz integral formula

Let f be a function holomorphic on the closed unit disc {z ∈ C | |z| ≤ 1}. Then

${\displaystyle f(z)={\frac {1}{2\pi i}}\oint _{|\zeta |=1}{\frac {\zeta +z}{\zeta -z}}\operatorname {Re} (f(\zeta ))\,{\frac {d\zeta }{\zeta }}+i\operatorname {Im} (f(0))}$

for all |z| < 1.

Let f be a function holomorphic on the closed upper half-plane {z ∈ C | Im(z) ≥ 0} such that, for some α > 0, |z^α f(z)| is bounded on the closed upper half-plane. Then

${\displaystyle f(z)={\frac {1}{\pi i}}\int _{-\infty }^{\infty }{\frac {u(\zeta ,0)}{\zeta -z}}\,d\zeta ={\frac {1}{\pi i}}\int _{-\infty }^{\infty }{\frac {\operatorname {Re} (f)(\zeta )}{\zeta -z}}\,d\zeta }$

for all Im(z) > 0.

Note that, as compared to the version on the unit disc, this formula does not have an arbitrary constant added to the integral; this is because the additional decay condition makes the conditions for this formula more stringent.

The formula follows from Poisson integral formula applied to u:^[1][2]

${\displaystyle u(z)={\frac {1}{2\pi }}\int _{0}^{2\pi }u(e^{i\psi })\operatorname {Re} {e^{i\psi }+z \over e^{i\psi }-z}\,d\psi ,\quad |z|<1.}$

By means of conformal maps, the formula can be generalized to any simply connected open set.