# Cauchy representation and branch point order

This question is about the nature of branch points which arise in certain Cauchy-integral representations of functions of a single complex argument, $$z$$. The main application is for dispersion relations but we can ignore that connection since this is a straightforwardly mathematical question. (I posted to math.stackexchange but no attention, even with 50 point bounty.)

Suppose we have the following representation: \begin{align} f(z) &= \frac{1}{\pi}\int_0^\infty dx \frac{\Im f(x)}{x-z}, \end{align} where we assume the integral exists for $$z=0$$. This function has branch-point singularities at $$z=0,\infty$$ due to the fact that the (constant) endpoints of integration coincide with the pole of the integrand at those points.

The question that I have regarding the order of the branch point is equivalent to understanding the behavior of $$f(z)$$ as a circuit is traced around the branch point $$z=0$$.

Suppose I define the "first sheet" as the values $$f^{(I)}$$ of $$f(z)$$ where $$z=x_0+i\epsilon$$ ($$\epsilon$$ infinitesimal, as usual). I consider a circuit in the $$z$$-plane as $$z=x_0e^{i\theta}$$ ($$0<\theta<2\pi$$). The function is analytic at all points on this trajectory. Once, however, we consider $$2\pi < \theta$$ we must distort the contour to avoid the singularity in the integrand at $$x=x_0$$. This effects the analytic continuation of $$f(z)$$ onto what we'll call the "second sheet". The value $$f^{(II)}$$ of $$f(x_0)$$ on the second sheet is \begin{align} f^{(II)}(x_0) &= \frac{1}{\pi}\int_0^\infty dx \frac{\Im f^{(I)}(x)}{x-(x_0-i\epsilon)},\\ &= \frac{1}{\pi}\int_0^\infty dx \frac{\Im f^{(I)}(x)}{x-x_0}-i\,\Im f^{(I)}(x_0). \end{align} This demonstrates that $$f(x_0+i\epsilon)-f(x_0-i\epsilon) = 2i\,\Im f(x_0)$$ and shows that the integration contour is the branch cut.

We can continue the above procedure, I believe, an arbitrary number of times. So the order of the branch point at $$z=0$$ appears to be of the logarithmic variety (of infinite order).

My question: Is the branch point that arises from an endpoint singularity always logarithmic? (I suspect there should be way to obtain a square-root type of (or "first order") branch point from this Cauchy representation. But I don't see how to construct it given the above arguments.)