The Schwarz integral formula reconstructs holomorphic functions via boundary values of its real part with decay conditions $f(z)=O(z^{-\alpha})$ for $\alpha>0$:$$
f(z)=\frac{1}{\pi i}\int_{-\infty}^{\infty}\frac{\operatorname{Re}\{f(\zeta)\}}{\zeta-z}d\zeta
$$
I know one way to do is that one can call for Poisson integral formula and use conformal mapping that maps unit disk to upper half plane to deriving the Schwarz integral formula. But is there any way of deriving this directly from the Cauchy integral formula or what else way? Thank you.
Best Answer
Let $\gamma_R$ denote the positively oriented contour $[-R,R]$ followed by $Re^{it},0\le t\le \pi.$ If $z$ is in the upper half plane, then for $R>|z|$ we can apply Cauchy's formula to get $$f(z) = \frac{1}{2\pi i}\int_{\gamma _R} \frac{f(w)}{w-z}\,dw$$ Now let $R\to \infty$ and use the assumed decay rate of $f$ to see $$f(z) = \frac{1}{2\pi i}\int_\infty^\infty \frac{f(t)}{t-z}\,dt.$$ Claim: If $f$ is replaced by $\bar f$ in the last integral, the integral equals $0.$ Assuming the claim, we have $$f(z) = \frac{1}{2\pi i}\int_\infty^\infty \frac{f(t)+\bar f (t)}{t-z}\,dt=\frac{1}{2\pi i}\int_\infty^\infty \frac{2\text {Re}f(t)}{t-z}\,dt=\frac{1}{\pi i}\int_\infty^\infty \frac{\text {Re}f(t)}{t-z}\,dt.$$ That's the desired result. I'll leave the proof of the claim for you to ponder for now.