Maxwell-Boltzmann distribution as the most probable [closed]

Question

In Statistical Physics: An Introduction, by Daijiro Yoshioka, there's a neat derivation of the Maxwell-Boltzmann distribution as the most probable distribution of an Ideal Gas. He uses the method of velocity phase space, then goes on to describe the probability function $$W(n_1,n_2,...) \propto \frac{N!}{n_1!n_2!...n_i!}g_1^{n_1}g_2^{n_2}. $$

Given this he then enuntiates the restrictions of the system, namely the total number of molecules $N$ and the total energy $E$

$$ N = \sum_i n_i\\ E = \sum_i \epsilon_in_i = \sum_in_i\left(\frac{mv_i^2}{2}\right) $$

Now he says "The maximum probability W under the constraints of fixed N and E is the most probable velocity distribution (...) To find the maximum of $W$ under the constraints, we apply Lagrange’s method of undetermined multipliers to $\ln W$ ... Introducing $\lambda$ and $\beta$ as undetermined multipliers, we require the following equations to be satisfied for each $n_j, j = 1,2,3,...$ :

$$ \frac{\partial}{\partial n_j} \left( \ln W - \lambda \sum_in_i-\beta\sum_in_i\epsilon_i \right) = 0 "$$

I went through the derivation and used Stirling's approximation: $\ln (N!) \approx N\ln(N)-N$ to compute the partial derivative and then solved for $n_j$

$$ n_j = N\exp{\{-\lambda-1-\beta\epsilon_j\}}g_j $$

Applying this to the restrictions of the system, one gets

$$ N = \sum_i N\exp{\{-\lambda-1-\beta\epsilon_i\}}g_i $$ $$ N = \int_\vec{v} N\exp{\{-\lambda-1-\beta\epsilon_i\}} c dv_xdv_ydv_z$$

Where c is just the proportionality constant between the infinitesimal velocity volume and the probability $g_i$. Doing each of the 3 (independent) integrals is actually easy, since we know

$$ \int_{-\infty}^{+\infty} \exp{\{-\frac{m\beta}{2}v_i^2\}} dv_i = \sqrt{\frac{2\pi}{m\beta}} $$

It's just a gaussian integral. With this result the equation that restricts the number of molecules becomes

$$ N = Nce^{-1-\lambda}\left(\frac{2\pi}{m\beta}\right)^{3/2} \rightarrow ce^{-1-\lambda} = \left(\frac{m\beta}{2\pi}\right)^{3/2} $$

Now with this I can actually already write down the most probable distribution, since we know

$$ n_j = N\exp{\{-\lambda-1-\beta\epsilon_j\}}g_j$$

Which generalizes to

$$ f(\vec{v}) dv_xdv_ydv_z = N \left(\frac{m\beta}{2\pi}\right)^{3/2}\exp{\left(-\frac{m\beta}{2}v_i^2\right)} dv_xdv_ydv_z $$

Which of course is Maxwell's velocity distribution. However I am not sure how he solves for the $\beta$ multiplier, since if you follow a similar procedure for $E$ as for $N$ you get the following integral equation

$$ E = \frac{m}{2}N\left(\frac{m\beta}{2\pi}\right)^{3/2}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}v^2\exp \left( -\frac{m\beta}{2}v^2\right)dv_xdv_ydv_z $$

Which is an integral I am not sure how to solve. The author just says that the RHS of the equation is $\frac{3N}{2\beta}$ and so $\beta = \frac{3N}{2E}$. The reason why I really want to compute this integral is because from the kinetic theory of gases we know $E = \frac{3}{2}NkT $ and with that we get the familiar value $\beta= \frac{1}{kT}$.

My attempt at solving this integral involved thinking that the integrand is an even function, so I can just integrate from $(0,\infty)$ and add a factor of 2 for each integral. Also, I know $v^2 = v_x^2+v_y^2+v_z^2$, so the most I could do was this

$$ E = 4mN\left(\frac{m\beta}{2\pi}\right)^{3/2}\int_0^{+\infty}\int_0^{+\infty}\int_0^{+\infty}(v_x^2+v_y^2+v_z^2)\exp \left[ -\frac{m\beta}{2}(v_x^2+v_y^2+v_z^2)\right]dv_xdv_ydv_z $$

TL; DR

How can I solve the following integral $$ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}v^2\exp \left( -\frac{m\beta}{2}v^2\right)dv_xdv_ydv_z $$

Answer 1

I'll admit I'm just skipping to the end and addressing the integral, at the risk of bypassing something important which luckily other answers will cover. The trick is mainly to go to spherical coordinates in velocity space. Let us rewrite your integral (I'll call it $I$ for short) as $$I = \int v^2 e^{- \frac{m \beta}{2} v^2} v^2 \ \textrm{d}v \ \textrm{d}\Omega,$$ where $\textrm{d}\Omega$ stands for the angular integrals, which amount to a $4\pi$ factor. Hence, $$I = 4\pi \int v^4 e^{- \frac{m \beta}{2} v^2} \ \textrm{d}v.$$

This is now know as a Gaussian integral. Wikipedia has a formula, and its proof, for this sorts of integrals in here. In particular, we're interested in the formula $$\int_0^{+\infty} x^{2n} e^{- a x^2} \ \textrm{d}x = \frac{(2n - 1)!!}{a^n 2^{n+1}} \sqrt{\frac{\pi}{a}},$$ which, as stated and shown in Wikipedia, can be proven by taking derivatives of the "pure" Gaussian integral ($n = 0$) with respect to $a$. Substituting the constants for the ones we're using, we get $$\begin{align} I &= 4 \pi \frac{3!!}{a^2 2^{3}} \sqrt{\frac{\pi}{a}}, \\ &= 4 \pi \frac{3}{2 m^2 \beta^2} \sqrt{\frac{2\pi}{m \beta}}, \\ &= \frac{6 \pi}{m^2 \beta^2} \sqrt{\frac{2\pi}{m \beta}}, \end{align}$$ assuming I didn't make any mistakes.