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Suppose $\phi(z)$ is an analytic function in the upper complex plane, so it satisfies the Kramers-Kronig relations, i.e. \begin{equation} \Re\phi(w) = \frac{1}{\pi}\int \frac{\Im\phi(x)}{x-w} dx \end{equation} where $w$ and $x$ are real variables. My question is if I define another function \begin{equation} F(w) = \frac{1}{1+\phi(w)} \end{equation} whether the real part and imaginary part of $F(w)$ still satisfies the Kramers–Kronig relations?

From my understanding, since I can rewrite the equation as \begin{equation} F(w) = \frac{1}{1+\phi(w)} = 1 - \phi(w) + \phi^2(w) - \phi^3(w) + \phi^4(w) + \cdots \end{equation} $F(w)$ is still analytic in the upper complex plane, which means Kramers–Kronig relations still holds for $F(w)$. However, I checked it numerically, (just took an concrete example of $\phi$ the electron-hole bubble) the relations breaks down in this situation. What's the reason of this?

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    $\begingroup$ $F$ has a pole whenever $\phi$ becomes $-1$ at some point, so it is definitely not analytic in general. $\endgroup$ – NDewolf Mar 11 at 10:52
  • $\begingroup$ Do you mean that $1/(1+\phi)$ and its geometric expansion are not equivalent in this case? $\endgroup$ – jwyan1126 Mar 11 at 11:31
  • $\begingroup$ The expansion you wrote down clearly does not converge for $\phi(w)=-1$. So you cannot use this to prove analiticity of the LHS even though every term is analytic on its own (but as I said, the problem was already obvious without a series expansion). $\endgroup$ – NDewolf Mar 11 at 12:56

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