---
tag: 61WT6CE08W
kind: statement
role: theorem
name: Schwarz integral formula
msc: ["30E20", 31A10]
uses: [0NAHXR1FRH, W19567AVYG, 0DEB0JMN0W]
about: [0NAHXR1FRH, TCYSWCTYYM]
sources:
  - url: https://en.wikipedia.org/wiki/Schwarz_integral_formula
same_as:
  wikidata: Q4491936
  wikipedia: Schwarz integral formula
---

## Statement

Let $\mathbb{D} = \set{z \in \C : \abs{z} < 1}$.
Let $f$ be a [[holomorphic function|0NAHXR1FRH]] on $\mathbb{D}$ that extends continuously to the closed disc $\overline{\mathbb{D}}$, and let $u = \operatorname{Re} f$.
Then for every $z \in \mathbb{D}$,
$$f(z) = \frac{1}{2\pi}\int_0^{2\pi} \frac{e^{it} + z}{e^{it} - z}\, u(e^{it})\, dt + i \operatorname{Im} f(0).$$

## 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|W19567AVYG]].
Since $u$ is [[harmonic|0DEB0JMN0W]] 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 $\frac{1}{2\pi}\int_0^{2\pi} \frac{e^{it} + z}{e^{it} - z}\cos t \, dt = z$.

## Explanations

### L1 human

A holomorphic function on the disc is determined, up to an imaginary constant, by its real part on the boundary circle.
The Schwarz formula writes the function explicitly as an integral of that real part.
So the real part on the boundary carries all the information about the function inside.

### L2 human

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|9ZBAX4ZPX0]] and the Borel–Carathéodory theorem, and it has versions for the upper half-plane.