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When I tried to prove the Schwarz integral formula I found a sketch here, where I confused by the formula

$$u(z)=\text{Re} \int_0^{2\pi} u(e^{i\theta })\frac{e^{i\theta}+z}{e^{i\theta}-z} \frac{\mathrm{d}\theta}{2\pi}$$

I tried to prove this using the mean value property of harmonic function

$$u(z_0)=\frac{1}{2\pi}\int_0^{2\pi} u(z_0+re^{i\theta})\mathrm{d}\theta $$

But I can't see where to go from this. Thanks for any help.

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Yiyi Rao
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  • I'd say for $|z| = r$ : $u(z) = \lim_{a \to 1^-}Re\int_0^{2\pi} u(re^{i\theta })\frac{e^{i\theta}+az}{e^{i\theta}-az} \frac{\mathrm{d}\theta}{2\pi}$ is the Fourier series theorem, and it stays true for $|z| < r$ by "harmonic continuation" – reuns Mar 29 '17 at 15:10

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