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The question is about Boltzmann factor.

Under continuous energy, the following equality holds.

$$\int_{E=0}^{\infty} \; \frac{1}{kT} \; e^{-E/kT} \; dE \; = \;1 \tag 1$$ $$ < E > \; = \; \int_{E=0}^{\infty} \; \frac{E}{kT} \; e^{-E/kT} \; dE \,= \, kT \tag 2$$

which implies that the expected energy of molecules is $kT$.

On one dimensional space, since $E = \frac{1}{2} m\,v_{x}^2$, $dE = m\,v_x \, dv_x$. Therefore $(2)$ changes into $(3)$ by substitution.

$$ < E > \; = \; \int_{v_{x}=0}^{\infty} \; \frac{mv_{x}^2}{2kT} \; e^{-mv_{x}^2/2kT} \; mv_{x} \, dv_{x} \,= \, kT \tag 3$$

so I think that the probability distribution function of speed(one dimensional) should be $g(\ v_{x} ) = \; \frac{1}{kT} \; e^{-mv_{x}^2/2kT} \; mv_{x} \, $, not $g(\ v_{x} ) = \; \frac{1}{kT} \; e^{-mv_{x}^2/2kT} \; $.

When applying $dE = mv_x \, dv_x$, $ < E > \; = \; kT$. But the known expected energy is $ < E > \; = \; \frac{1}{2}kT$.

$ < E > \; = \; \frac{1}{2}kT$ occurs when you ignore the substitution $dE = mv_x \, dv_x$, and just put $dE = \, dv_x$ as follows:

$$ < E > \; = \; \int_{v_{x}=0}^{\infty} \; \frac{mv_{x}^2}{2kT} \; e^{-mv_{x}^2/2kT} \, dv_{x} \,= \, \frac{1}{2}kT \tag 4$$

So I realized the intrinsic trouble occured between two possibilities:

(a) probability $= c e^{-mv_{x}^2/2kT}$

(b) probability $= c e^{-mv_{x}^2/2kT} mv_x \, $

why is (a) right and (b) wrong, even though (b) seems more natural under substition integral?

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