PeterDonis said:
The function g(E)g(E)g(E) is never "written in terms of Density of States times dEdEdE". It is not a function of dEdEdE. It's a function of EEE. When you integrate it over a range of values of EEE, you have to include dEdEdE in the integrand because that's how integration works. That's really all there is to it.
Apparently I misunderstood the part how ##g(E)## is transformed for the integrand to work. The way ##g(E)## is transformed to include it in the integrand supports the explanation of the replier at StackExchange.
##g(E)## gives the number of quantum states for a
particular energy ##E## and is 1/8th of a sphere surface in n-dimensions:
$$g(E) = \frac{4 \pi mL^2 \cdot E}{h^2} = 4\pi r^2 \cdot \frac{1}{8}$$
From this:
$$r=\sqrt{\frac{8mL^2\cdot E}{h^2}}$$
For the integrand for the continuous approach, one would want to calculate the number of quantum states over a range ##dE##, which is the surface of 1/8th of an n-sphere times a thickness ##dr##
$$g(E\geq E+dE) = \frac{4 \pi mL^2 \cdot E}{h^2} \cdot d(\sqrt{\frac{8mL^2\cdot E}{h^2}})$$
The ##dr## can be wirtten as:
$$dr = d(\sqrt{\frac{8mL^2\cdot E}{h^2}}) = \frac{\sqrt{2}\cdot mL}{h\cdot \sqrt{mE}}\cdot dE$$
Substuting ##dr## with this in ##g(E \geq E + dE)## and simplifying gives:
$$g(E\geq E+dE) = \frac{2^{2.5}\pi \cdot m^{1.5} \cdot V \cdot \sqrt{E}}{h^3}\cdot dE$$
Since the number of quantum states in the energylevels within ##dE## doesn't change much, one can multiply this ##g(E \geq E+dE)## by the number of particles per 1 quantum state at a particular energy ##E##, which is ##\frac{N}{Z} \cdot e^{-\beta E}##, giving the formula in my previous post #150 that the replier at StackExchange explained.