Consider the Riemann-Hilbert problem, taken from this paper on p 49: $$ \phi^+(s)+\phi^-(s)=s^2-1 \,, \quad s\in [-\ell,\ell]\,, \quad \ell>0\,, \phi(\infty)=0\,,$$ where the functions $\phi^\pm(s)$ mean $$\phi^\pm(s)\equiv\lim_{\epsilon\rightarrow0}\phi(s\pm i \epsilon)\,. $$

As mentioned in the text, to look for solutions to the problem, one needs to compute the integral $$\int_{-\ell}^{\ell}\frac{1-s}{\sqrt{(s-\ell)(s+\ell)}^+}\frac{ \mathrm d s}{2\pi i} =-\frac{i}{2}(\ell^2-2)\,,$$ note that the square root is considered from value of $s$ hovering just above the real axis as denoted by the $+$.

My problems:

  1. How do I use in practice the definition of $\phi^+$ to compute this integral. That is what is really $\sqrt{(s-\ell)(s+\ell)}^+$?

  2. How do you get $-i(\ell^2-1)/2$?


I tried to evaluate this integral by picking some branch-cut for the $\sqrt{(s-\ell)(s+\ell)}^+ = \sqrt{\rho_1\rho_2}e^{i(\phi_1+\phi_2)/2}$, where $\sqrt{(s-\ell)}=\sqrt{\rho_1}e^{i\phi_1/2}$ and $\sqrt{(s+\ell)}=\sqrt{\rho_2}e^{i\phi_2/2}$ with $\sqrt{\rho_2}=|s-\ell|$ and $\sqrt{\rho_2}=|s+\ell|$.

The contour is

Complex contour

As no poles are enclosed by the contour this contour integral sums op zo zero. However, I have the feeling that this is wrong as e.g. the contribution of the big and two small circles don't seem to go to zero when to limits to $\epsilon$ and $R$ are taken to 0 and $\infty$, respectively.

  • $\begingroup$ PDF link does not seem to load. $\endgroup$ – Qmechanic Jun 7 '17 at 11:54
  • $\begingroup$ @Qmechanic Thanks, it should be fixed now! $\endgroup$ – Anne O'Nyme Jun 7 '17 at 19:35
  • $\begingroup$ PDF link does still not seem to load. $\endgroup$ – Qmechanic Jun 7 '17 at 23:06
  • $\begingroup$ @Qmechanic Third try, helpfully it works now! $\endgroup$ – Anne O'Nyme Jun 8 '17 at 8:34

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