I have a function defined like this:
k=0.0000861733;
Func = 3*(T/D)^3 *
Integrate[x^3/(Exp[x] - 1), {x, 0, D/T}, Assumptions -> T > 0,
Assumptions -> D > 0];
F = (k*T)*(((9*D)/(8*T)) + (3*Log[1 - Exp[-D/T]]) - (Func/T));
I now simply want to have a derivative of F w.r.t D. So I use D[F, {D, 1}]
Which gives me following output:
ConditionalExpression[
0.0000861733 T (9/(8 T) + (3 E^(-(D/T)))/((1 - E^(-(D/T))) T) - (
3 (-(D^3/T^4) - (D^3 E^(D/T))/((1 - E^(D/T)) T^4)) T^2)/D^3 + (
9 T^2 (-(\[Pi]^4/15) - D^4/(4 T^4) + (D^3 Log[1 - E^(D/T)])/
T^3 + (3 D^2 PolyLog[2, E^(D/T)])/T^2 - (
6 D PolyLog[3, E^(D/T)])/T + 6 PolyLog[4, E^(D/T)]))/D^4),
E^(D/T) <= 1]
I have two questions hereafter. 1) First of all how to get rid of the ConditionalExpression? and then 2) How can I plot the derivative of that F as a function of T (say 1 < T < 1000) for a certain value of D let's say D=200? I would really appreciate if someone help me out a bit.
Dis a builtin and should not be used as a variable. – Szabolcs Aug 22 '16 at 12:32Func = 3*(T/d)^3* Integrate[x^3/(Exp[x] - 1), {x, 0, d/T}, Assumptions -> {d > 0, T > 0}]– Feyre Aug 22 '16 at 12:33T>0,d>0ANDE^(d/T) <= 1For that, eitherdorTwould have to be negative. – Feyre Aug 22 '16 at 12:35dandTis +ve. – baban Aug 22 '16 at 12:36Dbyd:( and what about the plot? – baban Aug 22 '16 at 12:39Assumptionsas I suggested? – Feyre Aug 22 '16 at 12:45Plot[F /. d -> 200, {T, 0, 100}]Works fine. – Feyre Aug 22 '16 at 12:46plotplots the functionFitself? But I think I need to plot the derivative i.e.,F'– baban Aug 22 '16 at 13:15