# Substitution problem in boltzmann factor

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?