I am reading Maxwell-Boltzmann distribution for describing the velocities of ideal gas molecules. I went through the PSE question Derivation of the Maxwell-Boltzmann speed distribution and the Wikipedia article and several other resources for some clarification. The following is my understanding of the concept, derivation, and the discrepancy I am not able to resolve.
For discrete energy levels, the fraction of molecules $q(\epsilon)$, from statistical mechanics, of a particle having energy $\epsilon$ is proportional to $\exp{\dfrac{-\epsilon}{kT}}$. For continuous energy levels, the expression is similar, except that we have to multiply by a factor of $4\pi v^2$, which I will do in the end. The energy of an ideal gas particle is given by $\dfrac{1}{2}mv^2$ and therefore $q(v) = C\exp{\dfrac{-mv^2}{2kT}}$. I am trying to find $C$.
Since the particles may have speeds in one, $x$, $y$, or $z$, direction ranging from $0$ to $\infty$,
$$ \int_0^\infty C_x\exp{\dfrac{-m(v_x^2)}{2kT}}dv_x = 1\\ \implies C_x=\dfrac{1}{\int_0^\infty \exp{\dfrac{-m(v_x^2)}{2kT}}dv_x} $$
Discrepancy: Several books write that $\int_{-\infty}^\infty \exp{\dfrac{-mv_x^2}{2kT}}dv_x = 1$. I don't think that it is correct, so I am going to carry on with my derivation and give the result using this method in the end.
My Derivation
Since $\exp{\dfrac{-mv_x^2}{2kT}}$ is an even function of $v$, we can evaluate the integral as $\int_0^\infty \exp{\dfrac{-mv_x^2}{2kT}}dv = \dfrac{1}{2}\int_{-\infty}^\infty \exp{\dfrac{-mv_x^2}{2kT}}dv$. Therefore
$$ C_x=\dfrac{2}{\int_{-\infty}^\infty \exp{\dfrac{-m(v_x^2)}{2kT}}dv_x} $$
Using trick I learned from MSE question Integral of $e^{−x2}$ and the error function, which I verified using WolframAlpha, I obtained the following result.
$$ C_x = \dfrac{2\sqrt{\dfrac{m}{2kT}}}{\sqrt{\pi}} = \sqrt{\dfrac{2m}{\pi kT}}\\ q(v_x) = \sqrt{\dfrac{2m}{\pi kT}}\exp{\dfrac{-mv_x^2}{2kT}} $$
I understand $q(v_x)=q(v_y)=q(v_z)$ and $C_x=C_y=C_z$ and $q(v) = q(v_x)q(v_y)q(v_z)$:
$$ q(v) = \left(\dfrac{2m}{\pi kT}\right)^{\frac{3}{2}}\exp{\dfrac{-mv^2}{2kT}} $$
Speed is continuous, and we need to multiply by $4\pi v^2$ (Justification) to obtain the final result.
$$ q(v) = 4\pi\left(\dfrac{2m}{\pi kT}\right)^{\frac{3}{2}}v^2\exp{\dfrac{-mv^2}{2kT}} $$
Most textbooks and ChemLibreTexts follow the discrepancy and derive the following result.
$$ q'(v) = 4\pi\left(\dfrac{m}{2\pi kT}\right)^{\frac{3}{2}}v^2\exp{\dfrac{-mv^2}{2kT}} $$
Problems
This is problematic because my result would give different values of well-established $v_{\text{mean}}$, $v_{\text{average}}$, and $v_{\text{rms}}$, being greater by a factor of 8. Could someone help me out here?
