# In what sense does analyticity guarantee the following equality?

I was reading a paper$^1$ on particle physics, and at some point it is stated that, provided $f(x)$ is analitic, we have $$f(x)-f(0)=\frac{x}{\pi}\int_0^\infty \frac{\text{Im}\;f(y)}{y(y-x-i\varepsilon)} \;\mathrm dy\tag{1}$$ where the $i\varepsilon$ is supposed to be taken $\varepsilon\to 0^+$ after integrating.

This looks very similar to (what we physicists) call the Kramers-Kronig relations, though I believe in mathematics it is called the Sokhotski-Plemelj theorem: $$\int_a^b\frac{f(x)}{x-i\varepsilon}\mathrm dx=i\pi f(0)+\mathcal P\!\int_a^b\frac{f(x)}{x}\mathrm dx \tag{2}$$ where $\mathcal P$ means Cauchy principal value.

My questions: is the relation $(1)$ true in general? under what circumstances? is it possible to prove $(1)$ from $(2)$? or is $(2)$ irrelevant here?

$^1$ The Muon g-2, by F. Jegerlehner and A. Nyffelerpage, arXiv:0902.3360v1, page 39.

Write the Kramers-Kronig relations as $$\text{Re}\; g(x)=\frac{1}{\pi}\mathcal P\int \frac{\text{Im}\;g(y)}{y-x} \mathrm dy$$
Using the Sokhotski-Plemelj theorem, namely $$\mathcal P\int \frac{\text{Im}\;g(y)}{y-x} \mathrm dy=\int \frac{\text{Im}\;g(y)}{y-x-i\varepsilon} \mathrm dy-i\pi\ \text{Im}\; g(x)$$ we find $$g(x)=\frac{1}{\pi}\int \frac{\text{Im}\;g(y)}{y-x-i\varepsilon} \mathrm dy$$
Finally, by taking $g(x)\equiv (f(x)-f(0))/x$ we get the expression in the OP. The answer to the question "under what assumptions" is: we must have $\text{Im}\;f(0)=0$.