Recently I met an integral as follow: $$\int_0^{2\pi}\cdots\int_0^{2\pi}\left(\prod\limits_{1\leq i<j\leq9}\sin\frac{\theta_i-\theta_j}{2}\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+1}))\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+2}))\right)d\theta_1\cdots d\theta_9,$$ where $\theta_{10}=\theta_1,\theta_{11}=\theta_2$.

I want to know whether the above integral is $0$ or not. I do not know how to connect this integral with some known integral. So I ask it here for some ideas.

Well, my trick was seen through by Fedor Petrov.

Originally I wanted to prove the antisymmetrization of $$f(x_1,x_2,\cdots,x_9)=\prod\limits_{i=1}^9(x_i+x_{i+1})^2\prod\limits_{i=1}^9(x_i+x_{i+2})^2,\ \text{where the indices are module}\ 9,$$ is nonzero, just like what I asked before, see my previous question. Then I did some transformations as follow: $$\text{the antisymmetrization of}\ \prod\limits_{i=1}^9(x_i+x_{i+1})^2\prod\limits_{i=1}^9(x_i+x_{i+2})^2\ \text{is nonzero}$$ $$\Longleftrightarrow$$ $$\text{the constant term of this Laurent polynomial}\\\prod\limits_{1\leq i<j\leq 9}(\frac{1}{z_i}-\frac{1}{z_j})\prod\limits_{i=1}^9(z_i+z_{i+1})^2\prod\limits_{i=1}^9(z_i+z_{i+2})^2\ \text{is nonzero}$$ $$\Longleftrightarrow$$ $$\int_0^{2\pi}\cdots\int_0^{2\pi}\prod\limits_{1\leq i<j\leq9}(e^{-i\theta_i}-e^{-i\theta_j})\prod\limits_{i=1}^9(e^{i\theta_i}+e^{i\theta_{i+1}})^2\prod\limits_{i=1}^9(e^{i\theta_i}+e^{i\theta_{i+2}})^2d\theta_1\cdots d\theta_9\neq0$$ $$\Longleftrightarrow$$ $$\int_0^{2\pi}\cdots\int_0^{2\pi}\left(\prod\limits_{1\leq i<j\leq9}\sin\frac{\theta_i-\theta_j}{2}\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+1}))\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+2}))\right)d\theta_1\cdots d\theta_9\neq 0,$$ where the indices above are all module $9$. That is how my above question came from.

I thought maybe it is easy to turn into an integral, but now it seems that it is not easy to see whether the above integral is zero or not. So come back to my original question:

Is the antisymmetrization of $$f(x_1,x_2,\cdots,x_9)=\prod\limits_{i=1}^9(x_i+x_{i+1})^2\prod\limits_{i=1}^9(x_i+x_{i+2})^2$$ nonzero?

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    $\begingroup$ With all the products being finite, I would guess you can compute this directly using Sage or Mathematica or something. After all, the integrand is just some big polynomial in $e^{i \theta_k}$. Have you tried? $\endgroup$ – Nate Eldredge Apr 12 '16 at 13:53
  • $\begingroup$ the product of sines is the Vandermonde determinant of the matrix $M_{nm}=e^{im\phi_n}$, perhaps that helps. $\endgroup$ – Carlo Beenakker Apr 12 '16 at 13:53
  • $\begingroup$ Incidentally, unbalanced parentheses around the cosines. $\endgroup$ – Nate Eldredge Apr 12 '16 at 14:08

Denote $z_k=e^{i\theta_k}$, where $i$ is imaginary unit. Up to some non-zero factor like power of 2, this is the integral of $\prod_j z_j^{-8}\cdot \prod_{j<k}(z_j-z_k)\cdot \prod_j (z_j+z_{j+1})^2 (z_j+z_{j+2})^2$, indices are modulo 9, integral with respect to $\prod d\theta_j$. Only constant term survives after integration, that is, your question is whether constant term of this Laurent polynomial, or, equivalently, coefficient of $\prod_j z_j^8$ in $\prod_{j<k}(z_j-z_k)\cdot\prod_j (z_j+z_{j+1})^2 (z_j+z_{j+2})^2$, equals 0 or not. At first, you may calculate it on the computer (this is much easier task for them than integrating). At second, this looks to be less or more the same question as your previous question.


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