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Let's say I have an integral of the form $$\oint\oint \frac{dz_1}{2\pi i} \frac{dz_2}{2\pi i} \frac{1}{(z_1 + n_1)(z_2 + n_2)(z_1 + z_2 + n)}$$ where $n_1,n_2,n\in\mathbb{Z}$ and both the contours are of the form which go from $c-i\infty$ to $c+i\infty$ in both cases for $c\in(0,1)$ and the contour is closed to the left at infinity.

How would one evaluate the integral? What bothers me is that when $n=n_1+n_2$, we end up getting a double pole, while in other cases there seems to be a single pole. How do we take care of this?

metamorphy
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Bharath Radhakrishnan
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  • You're trying to integrate in $\mathbb{R}^4 \simeq \mathbb{C}^2$, but the classic residue theorem is formulated for $\mathbb{C}$. The singularity at $z_2 + j$ for example is not a singular point, but an entire two-dimensional subspace (since $z_1$ is unconstrained). See this question. – MrArsGravis Sep 05 '20 at 16:49
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    @MrArsGravis. It's an iterated integral with two integrals in $\mathbb{C}$. The residue theorem does apply here. $$ \oint \frac{dz_1}{2\pi i}\frac{1}{z_1+i} \left( \oint \frac{dz_2}{2\pi i} \frac{1}{(z_2 + j)(z_1+z_2+k)} \right) $$ – md2perpe Sep 05 '20 at 16:54
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    The choice of $i$ certainly wasn't the best choice of parameter name here. – md2perpe Sep 05 '20 at 16:58
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    @md2perpe what has got me bothered is that the order of integration seems to matter. If not in this case, then at least upon introducing $f(z_1)g(z_2)$ where both $f$ and $g$ are holomorphic, the order seems to matter – Bharath Radhakrishnan Sep 05 '20 at 16:58
  • I've renamed the parameters, following the remark by @md2perpe. – metamorphy Sep 08 '20 at 12:08
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    Do you have an example where the order seems to matter? – md2perpe Sep 14 '20 at 17:55

1 Answers1

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I suppose (perhaps a tiny bit more generally) that the contours go from $c_{1,2}-i\infty$ to $c_{1,2}+i\infty$, for $z_{1,2}$ respectively, where $c_1,c_2\in(0,1)$. The term "closed to the left at infinity" is not a property of such a contour, but rather a limiting process, when the path from $c-i\infty$ to $c+i\infty$ is replaced by the segment $[c-iR,c+iR]$ and, say, the halfcircle $|z-c|=R$ lying in $\Re z\leqslant c$, and the limit $R\to{\raise 1pt\tiny{(+)}}\infty$ is taken. [I'm using this term myself below.]

Assuming that $z_1+z_2+n$ doesn't vanish (that is, $c_1+c_2+n\neq 0$), the given double integral is absolutely convergent. Thus, Fubini's theorem guarantees that the order of integration does not matter. This is also confirmed by the result below (which is symmetric w.r.t. simultaneous $n_1\leftrightarrow n_2$ and $c_1\leftrightarrow c_2$).

The obvious substitutions show that our integral is $I(n_1+c_1,n_2+c_2,n+c_1+c_2)$, where $$I(a,b,c)=\int_{-i\infty}^{i\infty}\int_{-i\infty}^{i\infty}\frac{1}{(z_1+a)(z_2+b)(z_1+z_2+c)}\frac{dz_1}{2\pi i}\frac{dz_2}{2\pi i}.\qquad(a,b,c\in\mathbb{R}_{\neq 0})$$ The computation of $I(a,b,c)$ is merely a bunch of case distinctions.

To ease it, we use the 1D analogue: for $z_1,z_2\in\mathbb{C}\setminus i\mathbb{R}$, we have $$\int_{-i\infty}^{i\infty}\frac{1}{(z+z_1)(z+z_2)}\frac{dz}{2\pi i}=\begin{cases}1/(z_2-z_1),&\Re z_1>0\wedge\Re z_2<0\\1/(z_1-z_2),&\Re z_1<0\wedge\Re z_2>0\\\hfill 0,\hfill&\text{otherwise}\end{cases}$$ (the proof is easy: if $\Re z_1<0$ and $\Re z_2<0$, we "close to the left" as above, getting the poles out of the contour, so the integral is $0$; note that this works for $z_1=z_2$ too, so that the double pole is not an issue; likewise, if $\Re z_1>0$ and $\Re z_2>0$, we "close to the right", and the integral is $0$ again; in the remaining cases, we go either way, computing the residue at a single simple pole).

Now we just proceed with iterated integration, in any order. The final result is $$I(a,b,c)=\begin{cases}1/(c-a-b)&\text{if }a,b,-c\text{ are of the same sign}\\\hfill 0\hfill&\text{otherwise}\end{cases}$$

metamorphy
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  • Is it correct to first assume the contour to follow one half-circle, then throw out the half-circle part and later choose the other half-circle? – MrArsGravis Sep 09 '20 at 18:33
  • @MrArsGravis: No, it is not, and the logic of the above is different. – metamorphy Sep 09 '20 at 20:39
  • We have $\int_{-i\infty}^{i\infty}=\lim_{R\to+\infty}\int_{-iR}^{iR}$. Now we add a half of $|z|=R$ (either to the left or to the right of the imaginary axis), and obtain a closed contour, giving us the ability to evaluate integrals along it using the residue theorem. And finally we observe that integrals along the halfcircles vanish after $R\to+\infty$. – metamorphy Sep 09 '20 at 20:46
  • Thanks! I agree that if we're interested in the straight line only, this seems OK. I thought OP wanted the closed contour, but maybe I misunderstood. – MrArsGravis Sep 10 '20 at 06:39
  • @MrArsGravis: This is exactly what I'm talking about in the first paragraph of the answer. The words "... and the contour is closed to the left at infinity" (in the question) make no sense; a contour "going from $c-i\infty$ to $c+i\infty$" simply can't be closed, for any reasonable meaning of "$c\pm i\infty$". But if we understand this as a limiting process, then - yes - we can throw a zero-in-the-limit thing out for another one. (I know this should go to the OP rather than you...) – metamorphy Sep 10 '20 at 06:54