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i have this integral it must result in all constant values and the value of the integral its self which is Zeta[2/3] it can be done.

ClearAll["Global`*"]

e0 = (3*h^2)/(8*m*L^2)
ee = e0*n^2
ef = n^3/e0;

be[e_] := 1/(Exp[e/(k*t) - u/(k*t)] - 1); 
g[e_] := Defer[(2/Sqrt[Pi])*((2*Pi*m)/h^2)^(3/2)*v*Sqrt[e]]; 

sub1 = {e/(k*t) -> x}
sub2 = {e -> x*k*t}

be[e] /. sub1; 
g[e] /. sub2; 

f[x_] := Refine[(g[e] /. sub2)*(be[e] /. sub1), Assumptions -> {u == 0}]

Integrate[f[x], {x, 0, Infinity}]
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closed as off-topic by Michael E2, MarcoB, m_goldberg, eyorble, Daniel Lichtblau Apr 13 at 20:37

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Because of Defer in your g[e] hold x in substitution so not evaluated

remove Defer in g[e] and Do substitution like this then you find your answer

ClearAll["Global`*"]

e0 = (3*h^2)/(8*m*L^2)
ee = e0*n^2
ef = n^3/e0;

be[e_] := 1/(Exp[e/(k*t) - u/(k*t)] - 1); 
g[e_] := (2/Sqrt[Pi])*((2*Pi*m)/h^2)^(3/2)*v*Sqrt[e]; 


sub1 = {(e/(k t) ) -> x}
sub2 = {(e)^(1/2) -> (x k t )^(1/2)}

f[x_] := Refine[((g[e] /. sub2)*(be[e] /. sub1)), 
  Assumptions -> {u == 0}]


Integrate[f[x], {x, 0, Infinity}]
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