## Integrate the FB5 distribution

Hi,

I have a problem in integrating the FB5 distribution. The FB5's probability density function is given by $f(x)=\frac{1}{c(\kappa, \beta)}\exp[\kappa \gamma_1 x +\beta[(\gamma_2 x)^2 - (\gamma_3 x)^2]]$. We know that $\gamma_1$ represents the mean direction of the distribution and that $\gamma_2$ and $\gamma_3$ represent the major and minor axes respectively. The three columns $\gamma_1$, $\gamma_2$ and $\gamma_3$ form an orthogonal (rotation) 3x3 matrix.

I would like to integrate this PDF and therefore I need to express its exponent in polar coordinates. I know that $\gamma_1 x$ can be expressed as: $\gamma_1 x= \cos(\theta) \cos(\theta_o) + \sin(\theta) \sin(\theta_o) \cos(\phi - \phi_o)$ since it is the vector dot product of two 3d vectors. In the above $\gamma_1 = [\sin(\theta_o) \cos(\phi_o) \sin(\theta_o) \sin(\phi_o) \cos(\theta_o) ]$ and $x= [\sin(\theta) \cos(\phi) \sin(\theta) \sin(\phi) \cos(\theta) ]$.

How can we express the $\gamma_2 x$ and $\gamma_3 x$ in a similar way?

Do you think that the following form would be correct? : $\gamma_2 x= \cos(\theta) \cos(\theta major) + \sin(\theta) \sin(\theta major) \cos(\phi - \phi major)$...same for $\gamma_3 x$ ???? . However, I have read that one needs to define only one angle to specify the major and minor axes. This is very confusing to me. As a result, I am not sure if I can express $\gamma_2 x$ and $\gamma_3 x$ as stated above. Please advise!!!!!!!!

Many Thanks

BR

Alex.