# Question: Unable to plot an integral function containing a piecewise function

November 09 2012 by
false
Maple

0

Here are my codes:

restart;
Delta := proc (beta) options operator, arrow; sqrt(1-1/beta^2) end proc;
T := proc (&epsilon;, Z) options operator, arrow; piecewise(&epsilon; < Delta(beta) and Delta(beta) < &epsilon;, 2*Delta(beta)^2/(&epsilon;^2+(Delta(beta)^2-&epsilon;^2)*(2*Z^2+1)^2), &epsilon; < -Delta(beta) or Delta(beta) < &epsilon;, 2*abs(&epsilon;)/(abs(&epsilon;)+sqrt(&epsilon;^2-Delta(beta)^2)*(2*Z^2+1))) end proc;
g := proc (V, Z, beta) options operator, arrow; 1/(Int(T(&epsilon;, Z)*beta*exp(beta*(&epsilon;-e*V))/(exp(beta*(&epsilon;-e*V))+1)^2, &epsilon; = -infinity .. infinity, method = _NCrule)) end proc;
e := 1;
plot('g'(V, .1, 10), V = -4 .. 4);

what I wanna do is to plot self-defined intrgral function, however the integrand in this integral is also defined by preceding functions. I know that usually in this case if I don't delay evaluation for g(V, .1, 10) in plot, I'm not able to get a plot correctly. However here even I have put the single quotes on g, nothing changes, except for the error prompt. I think the problem results from the piecewise function T(&epsilon;, Z), since if I disregard it in the definition of g(V, Z, beta), I can get the results quite well.

Does anybody have some suggestions? Thank you.