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

November 09 2012 Budaoy 56
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.

Please Wait...