Normalization Factor
Let us define a generalized Gaussian probability density function (PDF) as:
$$
f_{s}\left( x \right) = A_{o} \ e^{^{\displaystyle - \frac{ (x - x_{o})^{2} }{ 2 \sigma^{2} } }} \tag{0}
$$
where $A_{o}$ is the normalization constant, $x$ is the argument, and $s$ denotes the set of distributions (e.g., particle species), $x_{o}$ is the offset of the peak from $x = 0$, and $\sigma$ is the variance of the distribution.
The normalization factor $A_{o}$ is found by using the following constraint:
$$
\int_{-\infty}^{+\infty} \ dx \ f_{s}\left( x \right) = 1 \tag{1}
$$
The analytical solution to this integral can be found in any standard integral table or using something like Mathematica, where one finds:
$$
A_{o} = \frac{ 1 }{ \sqrt{2 \ \pi \ \sigma^{2}} } \tag{2}
$$
Maxwellian Velocity Distribution
To convert from the 1D Gaussian PDF in Equation 0 above to a Maxwell-Boltzmann distribution, or Maxwellian, we just convert the variables as follows:
- $x \rightarrow v$, where $v$ is the velocity argument of $f_{s}$ ranging from $-\infty$ to $+\infty$
- $x_{o} \rightarrow v_{o}$, where $v_{o}$ is drift velocity or bulk flow velocity
- $2 \ \sigma^{2} \rightarrow V_{Ts}^{2}$, where $V_{Ts}$ is the thermal speed (here the most probable speed)
Then we can see that the 1D Maxwellian is given by:
$$
f_{s}\left( v \right) = \frac{ 1 }{ \sqrt{\pi} \ V_{Ts} } \ e^{^{\displaystyle - \left( \frac{ v - v_{o} }{ V_{Ts} } \right)^{2} }} \tag{3}
$$
The conversion to a full 3D distribution is simple enough so long as each velocity component, $v_{j}$, is not correlated with any other component. Then (dropping the subscript $s$ for brevity) we can define:
$$
\begin{align}
f\left( v_{x}, v_{y}, v_{z} \right) & = f\left( v_{x} \right) \ f\left( v_{y} \right) \ f\left( v_{z} \right) \tag{4a} \\
& = \prod_{ k = x,y,z } \ A_{k} \ e^{^{\displaystyle - \left( \frac{ v_{k} - v_{ok} }{ V_{Tk} } \right)^{2} }} \tag{4b}
\end{align}
$$
where the total normalization factor is given by:
$$
A_{x} \ A_{y} \ A_{z} \ = \frac{ 1 }{ \pi^{3/2} \ V_{Tx} \ V_{Ty} \ V_{Tz} } \tag{5}
$$
Conversion to Energy
To do this properly, one should use a momentum analog of the velocity-space version described above.
In the non-relativistic limit the conversion from momentum to energy is $E = \tfrac{p^{2}}{2 \ m}$ or $p = \sqrt{2 \ m \ E}$. Therefore, we can see that $dp/dE \propto E^{-1/2}$ or $\propto p^{-1}$. Since energy is a scalar, the limits of integration (e.g., when finding the normalization factor) change from $-\infty \leq v_{j} \leq +\infty$ to $0 \leq E \leq +\infty$.
Then the 3D version (e.g., Equation 4b above) becomes:
$$
f\left( E \right) = \frac{ 1 }{ Z } \ e^{^{\displaystyle - \left( \frac{ E }{ k_{B} \ T } \right) }}
$$
where $Z$ is the partition function, $k_{B}$ is Boltzmann's constant, and $T$ is the temperature.
In the relativistic limit, the situation becomes incredibly complicated as discussed at What is the correct relativistic distribution function?.
Transport Coefficients
One can calculate these through use of the moments of the distribution. I wrote a detailed answer for a non-relativistic velocity distribution at https://physics.stackexchange.com/a/218643/59023.
To make the probability distributions shown above consistent with the velocity moments discussed in my other answer, you can redefine $f_{s}$ such that when integrated over all velocity space one gets the number density of set $s$.
You have to go back to see where your equation $(1)$ came from.
For one dimension the probability of particles having a velocity between $\vec v_{\rm x}$ and $\vec v_{\rm x}+ d\vec v_{\rm x}$ is given by
$$f(\vec v_{\rm x})\;d\vec v_{\rm x} =\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,d\vec v_{\rm x}$$
The speed distribution is given by
$$f(v_{\rm x})\;dv_{\rm x} =2\,\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}=\sqrt{\frac{2m}{ \pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}$$
the factor $2$ being there because the speed (magnitude) of $\vec v_{\rm x}$ is the same as that of $-\vec v_{\rm x}$.
This is in agreement with your equation $(3)$.
Equation $(1)$ came from the idea that in three dimensions there is no preferred direction so
$f(\vec v_{\rm x},\vec v_{\rm xy},\vec v_{\rm z})\,d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z} = f(\vec v_{\rm x})d\vec v_{\rm x}\,f(\vec v_{\rm y})d\vec v_{\rm y}\,f(\vec v_{\rm z})d\vec v_{\rm z}$
You now have to count all the speeds which are the same ie the magnitude of the velocity $v$ is the same where $v^2 = v^2_{\rm x}+v^2_{\rm y}+v^2_{\rm z}$.
In this three dimensional case the volume of a shell of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z}= 4 \pi v^2 dv$ is being considered which results in your equation $(1)$.
$$f(v) \,dv = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$
The equivalent distribution for two dimensions with an area of a ring of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}= 2 \pi v\, dv$ and $v^2 = v^2_{\rm x}+v^2_{\rm y}$ is
$$f(v) \,dv = \left(\frac{m}{2 \pi kT}\right)\, 2\pi v\, \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$
Best Answer
You have to take into account the differentials. The actual equation is $$ f_\text{MB}(\mathbf{v})\,\text{d}v_x\text{d}v_y\text{d}v_z = n\left(\frac{m}{2\pi k_BT}\right)^{3/2}e^{-mv^2/2k_BT}\,\text{d}v_x\text{d}v_y\text{d}v_z. $$ Changing to spherical coordinates, we get $$ \text{d}v_x\text{d}v_y\text{d}v_z = v^2\sin\theta\,\text{d}\theta\,\text{d}\varphi\,\text{d}v. $$ Integrating $\theta$ and $\varphi$, this becomes $$ v^2\,\text{d}v\int_0^{2\pi}\text{d}\varphi\int_0^\pi\sin\theta\,\text{d}\theta = 4\pi v^2\,\text{d}v, $$ so we have $$ f_\text{MB}(v)\,\text{d}v = 4\pi n\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2e^{-mv^2/2k_BT}\,\text{d}v. $$ Now you can change $v$ to $E$. Using $$ \text{d}E = mv\,\text{d}v = \sqrt{2mE}\,\text{d}v, $$ you eventually obtain $$ f_\text{MB}(E)\,\text{d}E = 2n\left(\frac{1}{k_BT}\right)^{3/2}\sqrt{\frac{E}{\pi}}e^{-E/k_BT}\,\text{d}E. $$