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A very well-known formula in complex analysis is

$ \lim_{\epsilon\to0^+}\int_{-\infty}^\infty\frac{f(x)}{x-x_0\pm i\epsilon}dx = P\int_{-\infty}^\infty \frac{f(x)}{x-x_0}dx \mp i\pi f(x_0), $

known as Sokhotski–Plemelj theorem. I am really puzzled by the meaning of the integral on the left hand side. I understand that this integral comes from an integral over an infinite arc of radius $R$ on the upper-half plane (where the integral goes to zero for $R\to\infty$) plus the real line (after taking the limit $\epsilon\to 0$). Assuming the integrand goes to zero faster than $x^{-1}$ and considering the residue theorem, we get the result:

$ \lim_{\epsilon\to0^+}\int_{-\infty}^\infty\frac{f(x)}{x-x_0\pm i\epsilon}dx=2\pi i \text{Res}[f;x_0]. $

Now, if we take the minus sign for $i\epsilon$ and send $\epsilon$ to 0 (without calculating the integral) the above equation yields $2\pi if(x_0)$, while the plus sign yields 0. What is the problem here? Is this related to the fact I can't commute the limit with the integral sign in this case, i.e.

$ \lim_{\epsilon\to0^+}\int_{-\infty}^\infty\frac{f(x)}{x-x_0\pm i\epsilon}dx\neq\int_{-\infty}^\infty\frac{f(x)}{x-x_0}dx? $

1 Answers1

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Taking $x_0=0$:

"I understand that this integral comes from an integral over an infinite arc on the upper-half plane plus the real line": No, the integral on the left is the integral over a horizontal line. For the one with the plus sign, that's the same as the integral over $(-\infty,-r)$ plus the integral over a small semicircle with center at the origin and radius $r$, plus the integral over $(r,\infty)$. The limit of the integrals over the two intervals is exactly the principal-value integral on the right, by definition. And the integral over that small semicircle is asymptotic to half the integral over a full circle of radius $r$, which would be $2\pi if(0)$.

David C. Ullrich
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    You haven't answered my question. Forget about the right hand side of the first equation. I just wanna know why I get different results when using different signs for $i\epsilon$, since this seems to not matter when I take the limit to 0. For the positive sign the pole will be outside the contour so that the integral vanishes. For the negative sign the pole will be inside the contour and then equal to $i\pi f(x_0)$. By the way, I meant the integral is over the real line after taking the limit. –  May 24 '16 at 18:40
  • I have no idea what contour you're talking about - part of my answer was trying to explain that. And what do you mean, "integral over the real line after taking the limit"? The limit of the integral over the line $y=\epsilon$ is not the integral over the real line. – David C. Ullrich May 24 '16 at 18:43
  • Oh. Looking at the end of your post more carefully: Yes, the reason that there's a theorem to be proved here is precisely that $\lim_{\epsilon\to0^+}\int_{-\infty}^\infty\frac{f(x)}{x-x_0\pm i\epsilon}dx\neq\int_{-\infty}^\infty\frac{f(x)}{x-x_0}dx$. (On a first pass I saw that inequation but didn't realize it was something you were asking about - of course that limit is not that integral, so I assumed you undestood that, sorry.) – David C. Ullrich May 24 '16 at 18:46
  • No worries, now I get it. But is there any specific hypothesis to be satisfied so that I can take the limit under the integral sign? A quick google search told me some on the lines of uniform continuity and I was wondering if this wouldnt be the case here. I just have the impression that the physics community have been ignoring this fact and still getting consistent results. –  May 24 '16 at 18:58
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    The theorem itself shows that here you cannot take the limit under the integral sign! The limit is what's on the right hand side, and the right hand side is precisely "what you get by taking the limit under the integral sign" plus another term. – David C. Ullrich May 24 '16 at 19:02
  • Wonderful! Thanks! –  May 24 '16 at 19:04