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I was working on a problem of trying to determine synchronization of neurons and wanted to compare an analytical result to some numerical results I achieved. The analytical result depends upon evaluating this triple integral:

$$\int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi} \sqrt{3+2\cos(\theta_1-\theta_2)+2\cos(\theta_2-\theta_3)+2\cos(\theta_3-\theta_1)} d\theta_1 d\theta_2d\theta_3$$

The triple integral comes from trying to evaluate $E\left[\sqrt{|e^{i\theta_1}+e^{i\theta_2}+e^{i\theta_3}|^2} \right]$ where the $\theta$'s are i.i.d uniform random variables on the unit circle.

Wolfram alpha wasn't able to produce a value so I question if it's even possible to evaluate, whether exactly or a decimal approximation.

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    I find that the triple integral can be reduced to the single variable integral, $\mathcal{I}=32\pi\int_{0}^{1}\mathrm{d}u\,\frac{1+2u}{\sqrt{1-u^{2}}}\,E{\left(\frac{\pi}{2}\mid\frac{8u}{\left(1+2u\right)^{2}}\right)}$. Integrals of elliptic integrals like these occasionally have nice closed forms, but more often their solutions require hypergeometric functions and the like. – 2017-02-10

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It shouldn't be too hard to approximate this numerically. I did a quick Monte Carlo integration in R, taking $n = 10^8$ samples:

set.seed(42)
n = 10^8
theta1 = runif(n, 0, 2*pi)
theta2 = runif(n, 0, 2*pi)
theta3 = runif(n, 0, 2*pi)
integrand = sqrt(3+2*cos(theta1-theta2)+2*cos(theta2-theta3)+2*cos(theta3-theta1))
mu = mean(integrand)
sigma = sd(integrand)

gives that the estimated mean of the integrand is $\mu = 1.574591$ and the standard deviation of the integrand is $\sigma = 0.7215917$. This gives a standard error for that estimate of $\sigma/\sqrt{n} = 0.0000722$.

Thus we get an estimated value for the integrand of $$(8\pi^3) \left( \mu \pm {\sigma \over \sqrt{n}} \right)$$ since the region of integration has volume $8\pi^3$. This is $390.578 \pm 0.018$.

As an analytical note, your integrand is unchanged when $\theta_1, \theta_2, \theta_3$ are rotated by the same angle around the unit circle. So you may as well set $\theta_3 = 0$. Let $\theta_3 = 0$ and you get

$$ \int_0^{2\pi} \int_0^{2\pi} \int_0^{2\pi} \sqrt{3 + 2 \cos (\theta_1 - \theta_2) + 2 \cos \theta_2 + 2 \cos \theta_1} \: d\theta_3 d\theta_2 d\theta_1$$

and since $\theta_3$ isn't in the integrand and you're working over a rectangular region, this is

$$ 2\pi \int_0^{2\pi} \int_0^{2\pi} \sqrt{3 + 2 \cos (\theta_1 - \theta_2) + 2 \cos \theta_2 + 2 \cos \theta_1} \: d\theta_2 d\theta_1$$

This probably isn't easier to do analytically but I'd think a double integral would be easier than a triple integral numerically.

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    Hey this is great, thank you! You're right, I likely will have to do this numerically. I appreciate the R code (I personally don't know how to use R, but they keywords you provided let me find python code which works). – 2017-02-10