Is this Jeffrey Divergence code correct?

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This function calculates a Jeffrey Divergence between 2 images histogram XI and XJ. I have Jeffrey Divergence function but it contains integration since I don't know how to code integration.
please help
function d=jeffrey_divergence(XI,XJ)
% Implementation of the Jeffrey Divergence
% (cf. "The Earth Movers' Distance as a Metric for Image Retrieval",
% Y. Rubner, C. Tomasi, L.J. Guibas, 2000)
%
% @author: B. Schauerte
% @date: 2009
% @url: http://cvhci.anthropomatik.kit.edu/~bschauer/
% Copyright 2009 B. Schauerte. All rights reserved.
%
% Redistribution and use in source and binary forms, with or without
% modification, are permitted provided that the following conditions are
% met:
%
% 1. Redistributions of source code must retain the above copyright
% notice, this list of conditions and the following disclaimer.
%
% 2. Redistributions in binary form must reproduce the above copyright
% notice, this list of conditions and the following disclaimer in
% the documentation and/or other materials provided with the
% distribution.
%
% THIS SOFTWARE IS PROVIDED BY B. SCHAUERTE ''AS IS'' AND ANY EXPRESS OR
% IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED
% WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE
% DISCLAIMED. IN NO EVENT SHALL B. SCHAUERTE OR CONTRIBUTORS BE LIABLE
% FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
% CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
% SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR
% BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY,
% WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR
% OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF
% ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
%
% The views and conclusions contained in the software and documentation
% are those of the authors and should not be interpreted as representing
% official policies, either expressed or implied, of B. Schauerte.
m=size(XJ,1); % number of samples of p
p=size(XI,2); % dimension of samples
assert(p == size(XJ,2)); % equal dimensions
assert(size(XI,1) == 1); % pdist requires XI to be a single sample
d=zeros(m,1); % initialize output array
for i=1:m
for j=1:p
m=(XJ(i,j) + XI(1,j)) / 2;
if m ~= 0 % if m == 0, then xi == xj == 0
d(i,1) = d(i,1) + (XI(1,j) * log(XI(1,j) / m)) + (XJ(i,j) * log(XJ(i,j) / m));
end
end
end
Thanks

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