How can I simplify this triple integral with unknown constants in Mathematica?

[Petar Mali]

I need to calculate integral

\int^{\frac{2\pi}{a}}_{0}\int^{\frac{2\pi}{a}}_{0}\int^{\frac{2\pi}{a}}_{0}\frac{1}{\sqrt{1-\frac{(cosxa+cosya+cosza)^2}{9}}}ctgh(\frac{6SJ\sqrt{1-\frac{(cosxa+cosya+cosza)^2}{9}}}{2T})dxdydz

a,S,J are constants different then 0.

I try to use Mathematica

N[\int^{\frac{2\pi}{a}}_{0}\int^{\frac{2\pi}{a}}_{0}\int^{\frac{2\pi}{a}}_{0}\frac{Coth[\frac{6SJ\sqrt{1-\frac{(Cos[x a]+Cos[y a]+Cos[z a])^2}{9}}}{2T}]}{\sqrt{1-\frac{(Cos[x a]+Cos[y a]+Cos[z a])^2}{9}}}dxdydz]

I try also to define a>0 but Mathematica don't give a result. Where is a problem? How can I dodefine this? Thanks for your answer.

 

[Bill Simpson]

Since it seems that Mathematica cannot do

Integrate[Coth[Sqrt[1 - Cos[x]^2/9]]/Sqrt[1 - Cos[x]^2/9], {x, 0, 2*Pi}]

it seems very unlikely it will be able to do your full problem

 

[Simon_Tyler]

N[] won't work because you have unknown constants in the integral.

All of the a's can be scaled out of the integral to give a factor of a^{-3}.
If you differentiate the integral wrt S then you remove the outer square root.
You're left with an integrand like Csch[S Sqrt[1-()^2]]^2

Then you might need to start being smart -- try using the symmetries and changes of variables to simplify it down. Sqrt[1-()^2] makes you think of cos^2+sin^2=1...

In terms of aiding mathematica -- I've already removed the constant 'a' from the integral. All of the other constants can be combined into one. Too many free parameters (even if they can be combined) often make it more difficult for the integration routine.

Since there is only one parameter left, you can try (numerically) integrating for various values of the parameter -- eg low integers. This might give you some clues and you might be able to guess what the final form is. Also maybe a series expansion...

Also try doing the 1 and 2 dimensional problems first - see if that gives you any ideas.

Post any more questions and results here for us to look at.

Good luck