Is it correct to compute Bhattacharyya distance for Cauchy like, Bell shaped function? I have the algorithm (MF (Membership function) ARTMAP Neural network). Output from this algorithm are clusters in n-dimensional feature space. Over each cluster (in n+1 dimension) there is some membership function (Cauchy like - Bell shaped function), that says the membership value of some point to some cluster.    
Now I want to compute overlap between two clusters. I found the Jeffries-Matusita distance. But JM use Bhattacharyya distance (I am using bhattacharyya.dist() function in R) and the reference to this R function say: Computes Bhattacharyya distance between two multivariate Gaussian distributions.
Is it correct to use this Bhattacharyya distance with Cauchy like - Bell shaped function?
 A: If you have two Cauchy distributions, the definition of the Bhattacharyya distance (and coefficient BC) applies.  But, you cannot expect to apply the formulas you can find for multinormal distributions. Those formulas depend on the mean and covariance matrix, which do not exist in the Cauchy case!
But, we can find expression for the Bhattacharyya coefficient for the case of two Cauchy distribution. Here I will only look at the scalar case.  The cauchy density function (generalized to form a location-scale family) is
$$
   f(x; \mu, \sigma) = \frac1{\pi \sigma (1+ (\frac{x-\mu}{\sigma})^2)}
$$
The Bhattacharyya coefficient
$$
   BC(f,g) = \int \sqrt{f(x) g(x)} \; dx
$$
can be found, maybe with numerical methods, or we can try to find a formula. I will try to use maple:
z := 1/sqrt( Pi^2 * sigma1 * sigma2
        *(1+((x-mu)/sigma1)^2) * (1+((x+mu)/sigma2)^2) );
                                  1                           
  z := -------------------------------------------------------
                                                         (1/2)
          /              /            2\ /            2\\     
          |              |    (x - mu) | |    (x + mu) ||     
       Pi |sigma1 sigma2 |1 + ---------| |1 + ---------||     
          |              |           2 | |           2 ||     
          \              \     sigma1  / \     sigma2  //     
int(z, x=-infinity..infinity) assuming sigma1 > 0, sigma2 > 0, mu > 0;
/   (1/2)          / (1/2) //     4       2       2
\2 2      EllipticK\2      \\16 mu  + 8 mu  sigma1 

         2       2         4           2       2         4\//   
   + 8 mu  sigma2  + sigma1  - 2 sigma1  sigma2  + sigma2 / \16 

    4       2       2       2       2         4
  mu  + 8 mu  sigma1  + 8 mu  sigma2  + sigma1 

             2       2         4     /     4       2       2
   - 2 sigma1  sigma2  + sigma2  + 4 \16 mu  + 8 mu  sigma1 

         2       2         4           2       2         4\       
   + 8 mu  sigma2  + sigma1  - 2 sigma1  sigma2  + sigma2 /^(1/2) 

    2   /     4       2       2       2       2         4
  mu  + \16 mu  + 8 mu  sigma1  + 8 mu  sigma2  + sigma1 

             2       2         4\             2   /     4
   - 2 sigma1  sigma2  + sigma2 /^(1/2) sigma1  + \16 mu 

         2       2       2       2         4           2       2
   + 8 mu  sigma1  + 8 mu  sigma2  + sigma1  - 2 sigma1  sigma2 

           4\             2\\      \\//   /      1       /    2
   + sigma2 /^(1/2) sigma2 //^(1/2)// |Pi |------------- \4 mu 
                                      \   \sigma2 sigma1       

           2         2   /     4       2       2       2       2
   + sigma1  + sigma2  + \16 mu  + 8 mu  sigma1  + 8 mu  sigma2 

           4           2       2         4\      \\      \
   + sigma1  - 2 sigma1  sigma2  + sigma2 /^(1/2)/|^(1/2)|
                                                  /      /

In this expression $\mu$ is the midpoint of $\mu_1, \mu_2$: $\mu=\frac{\mu_1+\mu_2}{2}$ rewriting the integral using $\mu$ before invoking maple results in a simpler and more symmetric expression. 
Maple gives the result in terms of the complete elliptic integral of the first kind, $K$, available in R in the CRAN package gsl.  See https://en.wikipedia.org/wiki/Elliptic_integral
A: Here is a more readable version of the answer by Kjetil and also here. Let the two Cauchy distributions be
$$f_\pm(x) = \frac{b_\pm}{\pi} \frac{1}{(x\mp 1)^2 + b_\pm^2}$$
where I normalized the means to unity. The Bhattacharyya coefficient between them is
$$BC = \frac{4}{\pi}\sqrt{\frac{b_+ b_-}{A_+}} K\left(\frac{A_-}{A_+}\right).$$
Here, I have introduced:
$$A_\pm = 4 + (b_+ \pm b_-)^2,$$
and $K$ is the complete elliptic integral of the first kind as defined by Mathematica's EllipticK[].
It also equals the Chernoff coefficient in this case. See this newly published reference and, less directly, this reference.
