% Date:     Thu, 12 Feb 98 18:02:12 GMT
% From: J P Fitch <jpff@maths.bath.ac.uk>

load_package defint;
lik := ((1 + ( mu - 1 )^2 / ( 3 sigma^2 ) ) *( 1 + ( mu + 1 )^2 / ( 3 sigma^2 )
))^( -2 ) / sigma^2;
% load_package gnuplot;
% plot lik;

% This shows that the function is positive with a bump near the origin.
% so now we integrate...

int(lik,sigma,0,infinity);

% and the answer is zero.  It is suggested that this is wrong.
% Maple gives a silly answer as well, but it would be nice to get it right.

% Date:    Fri, 13 Feb 1998 16:15:14 -0100
% From:    neun@zib.de (Winfried Neun)

% Whatever it is not mine, it is Stan's. I do know that this info
% does not help too much, but the integration is correct if the
% second variable (mu) is substitutes by (3 say).
% This type of integrals is the only one which Stans code is able to
% handle with infinity as upper bound (poly/poly)

end;
