I have the following multiple integral I would wish to evaluate

$$\int_0^{2\pi} dx_1 dx_2 \cdots dx_n D(x_1,x_2)\cdots D(x_n,x_1)$$

where the $D$ functions are given by

$$ D(x,x') = \text{Sn}^2\bigg(\theta_3^2(0,q) \frac{x}{2}\bigg)\text{Sn}\bigg(\theta_3^2(0,q)\frac{(x+x')}{2}\bigg)\text{Sn}\bigg(\theta_3^2(0,q)\frac{(x-x')}{2}\bigg) $$

and the $Sn$ functions are Jacobi Elliptic functions with modulus $k=\theta_2^2(0,q)/\theta_3^2(0,q)$.

I have tried using the Fourier representation of the elliptic functions, namely

$$\text{Sn}\bigg(\theta_3^2(0,q) \frac{\alpha}{2}\bigg) = -\frac{4}{\theta_2^2(0,q)}\sum_{n=0}^\infty \frac{\text{Sin}((n+\frac{1}{2})\alpha)}{q^{n+\frac{1}{2}}-q^{-(n+\frac{1}{2})}}$$

However this doesn't seem to simplify the calculation at all as after one integral is computed you are left with a large number of extra terms, which compounds after each integration via this method without any real simplification.

Any help would be greatly appreciated.


I've managed to find a convenient Fourier representation of the Jacobi Elliptic function, namely

$$ \text{sn}\bigg(\theta_3^2 \frac{\alpha}{2}\bigg) = \frac{2i}{\theta_2^2}\sum_{-\infty}^{+\infty} \frac{e^{i(n-\frac{1}{2})\alpha}}{q^{n-\frac{1}{2}}-q^{-(n-\frac{1}{2})}} $$

Using this we can write down the integral over coincident $D(x,x')$ functions:

$$\begin{multline*} \int_0^{2\pi} dx_i \rho(x_{i-1},x_i)\rho(x_i,x_{i+1}) = \\ \sum_{\{\alpha_{i-1},\alpha_i\}} \text{(Lots of q-factors)} (2\pi)\delta_{a_i + b_i + c_i + d_i,d_{i-1}-c_{i-1}+2} e^{i(a_{i-1}+b_{i-1}+c_{i-1}+d_{i-1}-2)x_{i-1}}e^{(c_i-d_i)x_{i+1}} \end{multline*}$$

where ${\alpha_i}=\{a_i, b_i, c_i, d_i\}$. We now have $4N$ integrals and $N$ coupled delta functions and it is unclear to me how one can resolve the computation. For example, ignoring the q-factors for now, if we perform the summation on the kronecker delta functions coming from $x1$ and $x2$, we have

$$\sum_{d_1} \delta_{a_1 + b_1 + c _1 + d_1,d_l - c_l +2} \delta_{a_2 + b_2 + c_2 + d_2 ,d_1 -c_1 +2} = \delta_{a_1 + b_1 +a_2 + b_2 + 2c_1 + c_2 + d_2 , d_l -c_l +4}$$

and repeating this $l-1$ times we get

$$\delta_{\sum_{i=1}^l a_i + b_i + 2c_i - c_l +d_l,d_l-c_l+ 2(l-1)} = \delta_{\sum_{i=1}^l a_i + b_i + 2c_i, 2(l-1)}$$

  • $\begingroup$ A multidimensional integral with such a complex integrand, my first approach would be Monte Carlo. You probably won't get much analytical insight, however. $\endgroup$ – skbmoore Aug 22 '18 at 21:47

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