The code is
C1 = 1.1910427E-12
c2=1.43883
T=300
c=c1*(c2^(-4))*T^4
llimit=c2*2380/T
ulimit=c2*2940/T
print,"X=",llimit,ulimit
lin= c*qpint1d('x^3/(EXP(X)-1)', llimit, +inf, /expr)
rin=c*qpint1d('x^3/(EXP(X)-1)', ulimit, +inf, /expr)
print,abs(lin-rin)
end

IDL reports *any* overflow or underflow that occurs.  That's not necessarily bad.

What do you think happens when you try to evaluate EXP(+inf)?  Or EXP(+3000)?  Your user function generates an overflow, that's what happens.  IDL is warning you that it is treating EXP(+3000) equivalent to +infinity.  That is what happens when IDL cannot evaluate the full range of your user function with full precision.  EXP(+3000) simply cannot be represented on a modern computer.  

If you don't want overflow errors, then rewrite your expression to use EXP(-X) instead.  Of course, then you will get underflow errors.  Life sucks, doesn't it?  :-)

If you really don't want overflow errors for this function, then you must avoid using +inf as your upper bound.

T=300.0

Doing this, the value of c changes from 3.48598e-009 (T is integer) to 0.00225099 (T is floating point).

Maybe also change the definitions of llimit and ulimit to:
llimit=c2*2380.0/T
ulimit=c2*2940.0/T

This might help. But I have not tried the integration.