In a paper https://projecteuclid.org/journals/brazilian-journal-of-probability-and-statistics/volume-36/issue-2/Limit-theorems-for-quasi-arithmetic-means-of-random-variables-with/10.1214/22-BJPS531.short , the author said that if $X$ has Cauchy distribution, $$\lim_{R\to \infty}\frac{1}{R}\sup_{|z|=R}|f(z)|=0, \quad (1)$$ $$\lim_{\varepsilon \to 0}\varepsilon\sup_{|z-a|=\varepsilon}|f(z)|=0,\quad (2)$$ then $E(f(X))=f(\mu +\sigma i)$.
I tried to show the above statement but I couldn't.
My try: Since $X$ has Cauchy distribution, we have $$E[f(X)]=\int_{\mathbb{R}}f(x)(\frac{\sigma}{\pi}\frac{1}{(x-\mu)^2+\sigma^2})dx$$ $$=\frac{\sigma}{\pi}\int_{\mathbb{R}}\frac{f(x)}{(x-\mu)^2+\sigma^2}dx$$ $$=\frac{\sigma}{\pi}\int_{\mathbb{R}}\frac{1}{2i\sigma}(\frac{f(x)}{x-\mu-i\sigma}-\frac{f(x)}{x-\mu+i\sigma})dx$$ $$=\frac{1}{2\pi i}\int_{\mathbb{R}}\frac{f(x)}{x-\mu-i\sigma}dx -\frac{1}{2\pi i}\int_{\mathbb{R}}\frac{f(x)}{x-\mu+i\sigma}dx=f(\mu+\sigma i).$$
The author says that the result follows from the Cauchy integral formula, due to (1) and (2) But I didn't use (1) and (2). Is there something wrong in my proof? I'd really appreciate if someone could help me about it. Thanks in advance.