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The Sokhotski-Plemelj theorem states, $$\lim_{\epsilon\rightarrow 0^+}\int_a^b\frac{f(x)dx}{x+i\epsilon} = \mathcal P \int_a^b \frac{f(x)dx}{x} - i\pi f(0). $$

Is there a numerically stable way to take this limit, without explicitly using the above theorem? I find that if I use $f(x)=1$, $a=-b$, I almost converge to $-i\pi$, but not before things become unstable.

interoception
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    If you're trying to evaluate the limit, then can't you just take a finite non-zero value of $\epsilon$ and integrate over the rectangle $(a,b)\times(0,\epsilon)$ without its edge on the real line? You can choose $\epsilon$ here freely up to the lowest singularity in the upper half-plane, and if $a,b=\pm\infty$ then you don't even have the contributions from the vertical rectangle edges. – Kirill Dec 15 '18 at 01:11
  • For a = -b, this is a great idea, and I cannot believe I did not think of it. However, it does not help for the case when a and/or b are finite. – interoception Dec 16 '18 at 00:44
  • Why not? You'd have an integral over three line segments, $a\to a+i\epsilon\to b+i\epsilon\to b$. I'm not sure what you mean. – Kirill Dec 16 '18 at 01:00
  • I misinterpreted your original idea. I thought you meant that we would integrate around some rectangle $(a,b)\times(0,d)$, where d is some distance from the real axis. Since the rectangle contains no poles, then the contributions from the other three segments must be equal to minus the integral along the real line.

    This allows us to avoid integrating $\frac{f(x)}{x+i\epsilon}$ near $x=0$.

     – interoception Dec 16 '18 at 01:43
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    Yes, that's what I meant. Why doesn't it work when $a,b$ are finite? – Kirill Dec 16 '18 at 01:48
  • Actually, it works in the case that $a$ and $b$ are nonzero. Otherwise you would approach infinity as you approach the real axis on one of the vertical lines in the complex plane.

    Another thing to note is that we cannot assume $f(x)$ can be analytically continued to the complex plane, so the general method must include only integrating along the real line.

     – interoception Dec 16 '18 at 16:33
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    Taking an arbitrary sequence that converges as $O(\epsilon)$ and computing its limit as $\epsilon\to0$ (especially if as you say you can't assume the function has a continuation) is a substantially harder problem than evaluating the Cauchy principal value directly using numerical quadrature on the real line. (I.e., explicitly evaluating the r.h.s. and ignoring the l.h.s. entirely.) Can you say anything about the actual $f(x)$ you're trying to integrate to make the question less open-ended? – Kirill Dec 16 '18 at 16:51
  • In my case, I am actually solving an integral equation, that looks somewhat like $f(x,x^\prime) = \int_0^\infty \frac{g(x^{\prime\prime})}{x^{\prime\prime}-x_0+i\epsilon} f(x^{\prime\prime},x^\prime) dx^{\prime\prime}$, so I must integrate along the real line. This becomes a large system of linear equations after discretization.

    I can attempt to solve the Cauchy PV, but that method carries its own complications (namely in the discretization I use), so I wanted to fully explore this route.

     – interoception Dec 16 '18 at 20:41
  • Also, how do you know that the series converges as $\mathcal{O}(\epsilon)$? – interoception Dec 16 '18 at 21:39
  • The relative error between $1/x$ and $1/(x+i\epsilon)$ is $O(\epsilon)$, so I assumed the total relative error is $O(\epsilon)$. – Kirill Dec 16 '18 at 21:40
  • @Kirill If the series does converge as $\mathcal O(\epsilon)$, then I should be able to just do linear extrapolation once I confirm I am in the linear regime. – interoception Dec 21 '18 at 16:47

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