Let $f(z)$ be a Nevanlinna (or Herglotz, or Pick) function, i.e. an analytic mapping of $\mathbb{C}^+$ into itself. It is known that such functions admit the following representation formula: there exist $a\in\mathbb{R}$, $b\geq0$ and a Borel measure $\mu$ satisfying $$ \int_\mathbb{R}\frac{1}{1+x^2}\,\mathrm{d}x<\infty $$ such that $$ f(z)=a+bz+\int_{\mathbb{R}}\left(\frac{1}{x-z}-\frac{x}{1+x^2}\right)\mathrm{d}x. $$ In particular, $b=\lim_{y\to+\infty}f(iy)/iy=\lim_{y\to+\infty}\Im f(iy)/y$.
This result is usually shown by invoking a similar result for analytic functions is the unit circle satisfying $\Re f(z)>0$, and then exploiting the conformal mapping between the half-plane and the unit circle to infer the desired representation.
However, I was searching for a direct proof, not involving the conformal mapping, which I can't seem to find in the literature. My idea was the following: by Cauchy's formula, we can write $f(z)$, for any $z\in\mathbb{C}^+$, via its contour integral on a semicircle with radius $R$: $$ f(z)=\frac{1}{2\pi i}\int_{i\delta-R}^{i\delta+R}\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta+\frac{1}{2\pi i}\int_{0}^\pi\frac{f(Re^{i\theta})}{Re^{i\theta}-z}\,iR\mathrm{d}\theta; $$ "playing around" with this formula, one should be able to show that, by letting $R\to\infty$ and $\delta\downarrow 0$, the first term does converge to the integral transform of a measure $\nu$, and the second term should, hopefully, yield the linear term $bz$ in the representation.
However, it looks like I cannot do anything unless I make some hypotheses about the growth of $f(z)$ as $|z|\to\infty$ (both integrals may diverge), which quite puzzles me since the definition of Nevanlinna functions does not contain any explicit growth condition!
I would like to know whether it is possible to prove the representation formula without additional hypotheses and without resorting to the conformal mapping with the unit circle.