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    $\begingroup$ "I am fine with either an exact answer or an approximate one." By "approximate one" do you mean a symbolic solution like series solution, or numeric solution? $\endgroup$ Commented Apr 10, 2020 at 4:53
  • $\begingroup$ @xzczd Fine with either, however in both cases I would like to see a plot of the $\rho(\beta)$ with $\beta$ for some values of $\lambda$. $\endgroup$ Commented Apr 10, 2020 at 19:28
  • $\begingroup$ I suspect the question itself is wrong. Assuming $\rho(\beta)$ can be expanded as Fourier sine series on $[-\alpha_c, \alpha_c]$, take $\alpha=\pi/2$, $\alpha_c=\pi$, and compute the first term of the series expansion, Integrate and NIntegrate both complain the integral doesn't converge: Integrate[(Cot[(Pi/2 - b)/2] Sin[b/2 + Pi/2])/Sin[(Pi/2 - b)/2], {b, -Pi, Pi}, PrincipalValue -> True], NIntegrate[(Cot[(Pi/2 - b)/2] Sin[b/2 + Pi/2])/Sin[(Pi/2 - b)/2], {b, -Pi, Pi/2, Pi}, Method -> PrincipalValue]. $\endgroup$ Commented Apr 11, 2020 at 3:08
  • $\begingroup$ @xzczd Why this particular integral? It just shows that Sin[b/2 + Pi/2] isn't the solution, right? I would suggest to cross check any method with Case 1, it might be helpful. I don't think that the question itself is wrong. $\endgroup$ Commented Apr 11, 2020 at 6:02
  • $\begingroup$ As mentioned above, this is the first term of a Fourier sine series expansion: $\rho(\beta)=c_1 \sin(\beta/2 +\pi/2)+c_2 \sin(2(\beta/2+\pi/2))+…$ $\endgroup$ Commented Apr 11, 2020 at 6:17