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This is a question on contour integration. The particular problem has a (simple) pole on the contour which prohibits a direct application of Cauchy's Residue Theorem.

Daniel Fischer commented as follows

Not really. [...] if the contour is smooth at the pole, it's as if half of the pole lies inside the contour and half outside. If the contour has a corner at the pole, with (inner) angle $\alpha$, the fraction is $\frac{\alpha}{2\pi}$, so you get $\alpha i$ times the residue of the pole instead of $2\pi i$ times as for singularities properly enclosed by the contour.

The same result is mentioned in this question.

Unfortunately, Daniel didn't know a reference for this (generalised) result. Can anyone point me to a book/paper/recourse which covers this result? I'd like to see a proof and some maths underlying this intuition.

Thank you very much!

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    $\begingroup$ This is referred as Fractional Residues (pg. 209) in Gamelin's Complex Analysis. It is also referred as the Indentation Lemma( tartarus.org/gareth/maths/Complex_Methods/rjs/indentation.pdf). I don't think there is a standardized name for this result. $\endgroup$ Commented Jan 6, 2020 at 22:37
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    $\begingroup$ Great first question! Welcome to MSE! $\endgroup$ Commented Jan 6, 2020 at 23:01
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    $\begingroup$ @daruma Thank you for the quick reply! I just have a brief follow up question. Take the simplest case: Suppose we integrate over a rectangular where the simple pole is at the origin. We then have an inner angle of $\frac{\pi}{2}$ and would only count $\frac{1}{4}$ of the residue (times $2\pi i$) according to Daniel Fischer's comment. But as I read and understand the Lemma you pointed to, I would obtain $2\pi-\frac{\pi}{2}=\frac{3}{2}\pi$ times the residue? $\endgroup$ Commented Jan 6, 2020 at 23:06
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    $\begingroup$ If you are integrating along a rectangle like the one in the question, then you would have $-\pi i\cdot \textrm{Res}[f,z=0]$. $\endgroup$ Commented Jan 6, 2020 at 23:20
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    $\begingroup$ @daruma I'm sorry but where does the $-\pi i$ come from? I thought $\frac{\pi}{2}i\mathrm{Res}[f,z=0]$? $\endgroup$ Commented Jan 6, 2020 at 23:23

2 Answers 2

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Very reasonable question! I wondered about this for decades, myself! :)

The integral does not literally converge. It does converge in a "Cauchy principal value" sense, but this requires that we make a convention, or do something. It is not in any way automatic, any more than $\int_{\mathbb R} f(x)/x\;dx$ "automatically" takes the "Cauchy principal value" value.

The more bare, real fact is that that integral "through" the pole is not well-defined, since, after all, as a literal integral (as opposed to something with conventions imposed) it does not converge at all.

This explains why there's no "proof" that a contour integral "through" a pole picks up half the residue. Because the assertion is not literally true, as stated. Sure, we can say something about the related principal value integral, but that is a very different thing.

(And the possibilities of other "angles" of contour through poles likewise need principal value interpretations, otherwise are not well-defined. And, NB, there is no mandate to take the PV interpretation, so, in particular, the literal integrals do not magically/automatically take those values.)

EDIT: also, in case people might too glibly assume that there's not real issue about "regularizing" such integrals, please do consider the precise assertion of the Sokhotski-Plemelj theorem (eminently google-able). That is, it turns out that it is easy to imagine false things in terms of regularization.

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  • $\begingroup$ Thank you for your answer! Great to know that I'm not the only one wondering about it! I see the convergence issue, but we may be able to prove the stated result for the Cauchy PV? And if we can show that $\lim_{z\to z_0} f(z) < \infty$, then we may be able to even talk about the actual integral? $\endgroup$ Commented Jan 6, 2020 at 23:11
  • $\begingroup$ Yes, @Alex, if there is no actual pole (or essential singularity), then the literal integral converges... But is a much more mundane thing, and probably not at issue. It's exactly like the convergence of $\int f(x)/x\,dx$ when $f$ is smooth and vanishes at $0$. $\endgroup$ Commented Jan 6, 2020 at 23:12
  • $\begingroup$ But then I still wonder where I could find a reference/proof for the result being true for (at least) the Cauchy PV? $\endgroup$ Commented Jan 6, 2020 at 23:15
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    $\begingroup$ @Alex, I think if you grant yourself the usual Cauchy integral formula, and the definition of the PV integral, you can obtain the result. $\endgroup$ Commented Jan 6, 2020 at 23:17
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In both cited questions there is not truly a “pole on the contour”, I agree with paul garrett that this is a confusing phrase which requires a careful definition. Instead the principal value is considered, whose purpose is precisely to avoid a pole on the contour. Regarding a (poor man’s) proof of the statement, consider integrating a function $f$ along a contour $C$ consisting of two pieces, one that goes straight from a point $a$ on the negative real axis to the origin, then in a straight line from the origin to a point $b$ in the upper half plane (we do not lose any generality by doing this). Let’s call the (smallest) angle made by the broken contour $\alpha$. (We’ll also assume $f$ is analytic in an open region containing $C$.) The quantity we’d like to consider is of the form $$\lim_{\epsilon \rightarrow 0} \int_C dz \frac{f(z)}{z+i\epsilon}.$$ For any positive $\epsilon$ there is no “pole on the contour”, but it’s also not unforgivable to say “in the limit there is a pole on the contour”. Words aside, for any $0 < \epsilon$ (and small enough compared to $|a|$ and $|b|$) we can draw a circle around the point $-i\epsilon$ of radius, say, $2 \epsilon$. Consider then the points of intersection of this circle with $C$, and define a new contour $C’$ which goes from $a$ to the intersection on the negative real axis, then along a piece of the circle until the second intersection point on the segment $[0,b]$, and finally straight along that segment until $b$. Now, we can deform $C$ onto $C’$ while keeping the value of the integral invariant because $f$ is analytic in the region enclosed by the two curves. Now we use the definition of contour integration along $C’$. The contributions from the straight pieces, in the limit $\epsilon \rightarrow 0$, give the principal value $PV \int_C dz f(z)/z$, by definition. You’ll see that the integral along the arc is equal to $-i \alpha f(0)$ (in the limit $\epsilon \rightarrow 0$).

This should give you some techniques to handle problems in similar contexts.

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