Assume that $C(\mu ,\sigma )$ is the Cauchy distribution with location $\mu \in \mathbb{R}$ and scale $\sigma >0$ and the density function of $C(\mu ,\sigma)$ is $p(x;(\mu ,\sigma))=\frac{\sigma}{\pi}\frac{1}{(x-\mu)^2 +\sigma^2}$, $x\in \mathbb{R}$. If $\displaystyle\lim_{R\to \infty}\sup_{|z|=R}|f(z)|=0$ and $\displaystyle\lim_{\epsilon \to 0}\epsilon \sup_{|z-a|=\epsilon}|f(z)|=0$, then how can I show using Cauchy integral formula that $\mathbb E[f(X)]=f(\mu +i\sigma)$, when $X$ is a random variable with Cauchy distribution and $f$ is an injective holomorphic function. Thanks in advance.
My try: Cauchy integral formula is: $f(z_0)=\frac{1}{2\pi i}\int_C \frac{f(z)}{z-z_0}dz$. By definition, we have $$\begin{split} \mathbb E[f(X)]&=\int f(x)\Big( \frac{\sigma}{\pi}\frac{1}{(x-\mu)^2 +\sigma^2}\Big)dx\\ &=\frac{\sigma}{\pi}\int \frac{f(x)}{(x-\mu)^2 +\sigma^2}dx\\ &=\frac{\sigma}{\pi}\int \frac{f(x)}{(x-\mu-i\sigma)(x-\mu+i\sigma)}dx\\ &=\frac{1}{2\pi i}\int \frac{f(x)}{(x-\mu-i\sigma)}dx-\frac{1}{2\pi i}\int \frac{f(x)}{(x-\mu+i\sigma)}dx\\ &=f(\mu +i\sigma)-f(\mu -i\sigma) \end{split}$$
The paper
'Which functions preserve Cauchy laws?' Proc. Amer. Math. Soc. Volume 67, 277-286. Letac, G. (1977)
does not contain exactly the answer to your question but it should be of interest.