 Solution

First suppose the integral of $g(\xi )/(\xi -z)$ converges, for example if $g\in L^1(\mathbb{R})$. Substitute the complexified coordinates $x=(z+\bar{a})/2$, $y=(z-\bar{a})/(2i)$ into the Poisson integral. The factorization of its denominator givesThe given Schwarz integral formula therefore writes $f(z)$ as $(i\pi {)}^{-1}\int g(\xi )/(\xi -z)d\xi$ plus a constant. For $z=x+iy$, the real part of this integral is exactly the Poisson integral, sinceThe remaining constant has zero real part, giving $\boxed{f(z)=(i\pi {)}^{-1}{\int }_{\mathbb{R}}\frac{g(\xi )}{\xi -z}d\xi +ic}$ with $c\in \mathbb{R}$. The complex substitution is initially understood where the real-analytic harmonic function has its complexification; the holomorphic integral then continues the identity throughout the upper half-plane.

Decay of $u$ alone does not guarantee convergence of the printed unregularized integral. For example, choose a continuous $g$ which vanishes on $(-\infty ,e]$ and equals $1/\log \xi$ for $\xi \ge e^2$, with continuous interpolation. It is bounded and tends to zero at both ends. Its Poisson integral tends to zero as $|z|\to \infty$ in the closed upper half-plane: split the boundary function into a small uniform tail and a compactly supported part. Nevertheless ${\int }^{\infty }g(\xi )/(\xi -z)d\xi$ diverges like $\log \log \xi$, even as a symmetric principal value. With only the stated decay assumptions, the always convergent version isThe subtracted kernel is real on the integration line, so it changes only the imaginary constant and leaves the harmonic function $u$ unchanged. The kernel difference is $O({\xi }^{-2})$, which ensures convergence for bounded $g$.