Kinetic Energy in Maxwell-Boltzmann distribution So I was trying to arrange the expression of a Maxwell-Boltzmann velocity vector distribution into a kinetic energy distribution. Since we know that $E= \frac{1}{2} mv^2$ we can rearrange this expression to give $v^2=\frac{2E}{m}$. If we insert this expression in a Maxwell-Boltzmann distribution:
$$f(v)=\left(\frac{m}{2\pi K_BT}\right)^\frac{3}{2} \exp\left(\frac{-m(v_x^2 + v_y^2 + v_z^2)}{2K_BT}\right)$$
we get:
$$f(E)=\left(\frac{m}{2\pi K_BT}\right)^\frac{3}{2} \exp\left(\frac{-E}{K_BT}\right)$$
But Wikipedia says that the kinetic energy distribution is:
$$f(E)=2\left(\frac{E}{\pi}\right)^\frac{1}{2}\left(\frac{1}{K_BT}\right)^\frac{3}{2} \exp\left(\frac{-E}{K_BT}\right)$$
What am I missing here? 
 A: Kinetic energy distribution $f_E(E)$ is a function that gives probability per unit energy interval.
Since kinetic energy $E$ is a function of speed, this distribution is related to speed distribution $f_v(v)$, but its value at any energy $E$ is not merely proportional to value of $f_v(v(E))$. The derivation of $f_E(E)$ from $f_v(v)$ relies on the substitution theorem from the integral calculus.
For positive speeds $v_1,v_2$ and corresponding energies $E_1=\frac{1}{2}mv_1^2, E_2=\frac{1}{2}mv_2^2$, the probability that particle has kinetic energy in the interval $(E_1,E_2)$ is the same as the probability that it has speed in the interval $(v_1,v_2)$. We transform the expression of the second probability in the following way:
$$
P = \int_{v_1}^{v_2}f_v(v)dv => (subst. theorem) => \int_{E_1}^{E_2}f_v(v(E))v'(E) dE
$$
where
$$
v(E) = (2E/m)^{1/2}
$$
and $v'(E)$ is derivative of this function at value $E$.
The expression of the first probability is
$$
P = \int_{E_1}^{E_2}f_E(E)dE.
$$
Since we can choose $v_1,v_2$ arbitrarily close to each other, it must be
$$
f_E(E) = f_v(v(E))\frac{dv(E)}{dE}.
$$
Substituting
$$
f_v(v) = \left(\frac{m}{2\pi k_BT}\right)^{3/2} 4\pi v^2  e^{-\frac{mv^2}{2k_B T}}
$$
we obtain
$$
f_E(E) = \left(\frac{m}{2\pi k_BT}\right)^{3/2} 4\pi 2E/m \cdot e^{-\frac{E}{k_B T}} \cdot (2/m)^{1/2}\frac{1}{2}E^{-1/2}
$$
and after simplification
$$
f_E(E) = \left(\frac{1}{\pi k_BT}\right)^{3/2} 2\pi \cdot E^{1/2} e^{-\frac{E}{k_B T}}.
$$
A: You should think in terms of changing your integration variable.
I think you should procede this way: first of all write the integral in polar coordinates and integrate the angular variables. Then change your variable of integration from $v$ to $E=\frac12mv^2$. I actually haven't done it but it should work.
That's basically because you want to preserve the probability and to set $p(v)dv=p(E)dE$.
