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In a probability theory paper I found this rather pleasant result:

Theorem 4.1 Let $n \geq 2$ and $f_1, \dots, f_n : \mathbb{C} \to \mathbb{C}$ be meromorphic with possible poles at $\{ \mathfrak{p}_1, \dots, \mathfrak{p}_n \} $ Then we have identity $$ \sum_{\sigma \in S^{cyc}(n)} \frac{1}{(2\pi \mathbf{i})^n }\oint \dots \oint \frac{f_{\sigma(1)}(u_1)\dots f_{\sigma_n(n)}(u_n) }{(u_2 - u_1)\dots (u_n - u_{n-1})}\;du_1 \dots du_n = \frac{1}{2\pi \mathbf{i} }\oint f_1(u)\dots f_n(u) du $$

It is proven with a lengthy induction process. The authors obtain the result they needed. I wondered if there is more conceptual argument (perhaps using divisors) and could this be generalized beyond cyclic permutations.

Is there a name for when contour integrals simplify under permutation group action in this way?

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    $\begingroup$ Just a comment: I encountered this too, see Lemma 7 of arxiv.org/abs/1510.00342. Though not lengthy, the proof I give there is still by induction. $\endgroup$ Commented Oct 27, 2017 at 14:34
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    $\begingroup$ I would add that $S^{cyc}$ means the cyclic subgroup of $S_n$ generated by a long cycle. $\endgroup$ Commented Oct 27, 2017 at 15:34
  • $\begingroup$ Hope I wrote it correctly now. Usually they just wrote $\Sigma_{cyc}$ as you might see in Olympiad. These integral identities were convenient as they evaluated $n$-point correlations. $\endgroup$ Commented Oct 27, 2017 at 15:47
  • $\begingroup$ There should be $\sigma$ in the denominator too, else we get something wrong for $n=2$ (the difference $u_2-u_1$ does not cancel) $\endgroup$ Commented Oct 27, 2017 at 19:53
  • $\begingroup$ I do not understand what the identity is already for $n=2$. If the integrand is $(f_1(u_1)f_2(u_2)-f_1(u_2)f_2(u_1))/(u_2-u_1)$, the identity does not hold: for pairs $(f_1,f_2)=(f,1)$ and $(f_1,f_2)=(1,f)$ the values of the integral have opposite sign $\endgroup$ Commented Oct 28, 2017 at 8:51

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