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I am reading the book "Analytic S-matrix" by Eden,Landshoff, Olive and Polkinghorne.

In Sec.2, they discuss the different kinds of singularities of a function $f(z)$ defined as an integral of an analytic function $g(z,\omega)$ in the complex $\omega-$plane along some closed contour $\mathcal{C}$, i.e.

$$ f(z) = \int_\mathcal{C}\, d\omega\, g(z,\omega)\,. $$

It is assumed that we know the singularities of $g(z,\omega)$. They discuss the case of pinch singularities, i.e. those singularities that approach the contour from the opposite sides and eventually coincide. As an example, they consider

$$ f(z) = \int_0^1 \frac{d\omega}{(\omega-z)(\omega-a)} $$ where the contour is not really specified. This is already strange... shouldn't they specify the contour?

Their result is

$$ f(z) = \frac{1}{z-a}\log{\left[\frac{a(1-z)}{(1-a)z}\right]}\,,\qquad(a>1) $$ To perform this integral, I was thinking to take the parameterization $\omega(t) = \sqrt{t} e^{i \arccos{\sqrt{t}}}$ where $t\in [0,1]$. This parameterization actually selects the path of a semi-cicle in the upper-half complex $\omega-$plane, running from $\omega=0$ to $\omega=1$. The integral then can be computed as

$$ f(z) = \int_0^1 dt \frac{d\omega(t)}{dt}\frac{1}{(\omega(t)-z)(\omega(t)-a)} $$

If $z=a>1$ or $z=a<0$, the singularities of the integrand lie outside my integration path, and the integral is perfectly convergent.

Under Eq.(2.1.5) of their book, the authors write

the singularity at $z=a$ is only encountered on encircling one of the logarithmic singularities, so that now $\log{1}$ is $\pm 2\pi i$ instead of zero and no longer cancels the pole.

Can anyone explain to me what they mean by the last sentence? It is not clear to me which kind of contour can give a singularity at $z=a$.

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  • $\begingroup$ I noticed that you deleted a question you had posted regarding solution to the 2-D Poisson Equation. I had finished my solution, which did use complex analysis, but to my dismay the problem was deleted. If you reopen and let me know, I'll be happy to post my soluiton. $\endgroup$
    – Mark Viola
    Apr 14, 2021 at 3:08
  • $\begingroup$ @MarkViola I've restored the question! Thank you $\endgroup$
    – apt45
    Apr 14, 2021 at 3:38

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