Schwarz integral formula

In complex analysis, a branch of mathematics, the Schwarz integral formula, named after Hermann Schwarz, allows one to recover a holomorphic function, up to an imaginary constant, from the boundary values of its real part.

Unit disc

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

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

for all |z| < 1.

Upper half-plane

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

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

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.

Corollary of Poisson integral formula

The formula follows from Poisson integral formula applied to u:

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

This is equivalent to

${\frac {1}{2\pi }}\int u(e^{i\psi }){\frac {cos(2\psi )}{cos(2\psi )-sin(2\psi )+\Re ^{2}(z)+2\Re (z)\Im (z)-\Im ^{2}(z)}}-{\frac {\Re ^{2}(z)-\Im ^{2}(z)}{\cos(2\psi )-\sin(2\psi )+\Re ^{2}(z)+2\Re (z)\Im (z)-\Im ^{2}(z)}}d\psi$

$={\frac {1}{2\pi }}\int u(e^{i\psi }){\frac {cos(2\psi )}{cos(2\psi )-sin(2\psi )+\Re ^{2}(z)+2\Re (z)\Im (z)-\Im ^{2}(z)}}d\psi -[\arctan({\frac {-\tan(x)+\Re ^{2}(z)\tan(x)+2\Re (z)\Im (z)\tan(x)-1}{\sqrt {\Im ^{4}(z)-4\Re (z)\Im ^{3}(z)+2\Re ^{2}(z)\Im ^{2}(z)+4\Re ^{3}(z)+\Re ^{4}(z)-2}}})({\frac {1}{2\pi }})+\pi \operatorname {sgn}(2\Re ^{2}(z)+2\Im ^{2}(z)+4\Re (z)\Im (z)-2)\lfloor {\frac {1}{2}}+{\frac {x}{\pi }}\rfloor {\frac {1}{2\pi }}]{\frac {\Re ^{2}(z)-\Im ^{2}(z)}{\sqrt {\Im ^{4}(z)-4\Re (z)\Im ^{3}(z)+2\Re ^{2}(z)\Im ^{2}(z)+4\Re ^{3}(z)+\Re ^{4}(z)-2}}}$

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

Notes and references

• Ahlfors, Lars V. (1979), Complex Analysis, Third Edition, McGraw-Hill, ISBN 0-07-085008-9
• Remmert, Reinhold (1990), Theory of Complex Functions, Second Edition, Springer, ISBN 0-387-97195-5
• Saff, E. B., and A. D. Snider (1993), Fundamentals of Complex Analysis for Mathematics, Science, and Engineering, Second Edition, Prentice Hall, ISBN 0-13-327461-6