# Kramers-Kronig relations for geometric series

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

From my understanding, since I can rewrite the equation as $$$$F(w) = \frac{1}{1+\phi(w)} = 1 - \phi(w) + \phi^2(w) - \phi^3(w) + \phi^4(w) + \cdots$$$$ $$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?

• $F$ has a pole whenever $\phi$ becomes $-1$ at some point, so it is definitely not analytic in general. – NDewolf Mar 11 at 10:52
• Do you mean that $1/(1+\phi)$ and its geometric expansion are not equivalent in this case? – jwyan1126 Mar 11 at 11:31
• 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). – NDewolf Mar 11 at 12:56