# Trouble integrating Maxwell-Boltzmann velocity distribution for RMS speed

I am trying to integrate over the Maxwell-Boltzmann velocity distribution to find the $$v_{\text{rms}}$$.

The Maxwell-Boltzmann velocity distribution is given by:

$$P(v) =\sqrt{\frac{m}{2\pi kT}} e^{-mV^2 / 2kT}$$

To find the average square velocity I multiply each possible value for $$v^2$$ by the fraction of molecules with velocity $$v$$ and sum by integration like so:

$$v_{rms} = \int^{\infty}_{-\infty} V^2 P(v) = \sqrt{\frac{m}{2 \pi kT}} \int^{\infty}_{-\infty} V^2 e^{-mV^2 / 2kT}$$

I have attempted to integrate by parts with

$$u = V^2$$, $$dv = e^{-mV^2 / 2kT}$$, $$du= 2V$$, and $$v = \sqrt{\frac{2kT \pi}{m}}$$ ($$dv$$ is a Gaussian integral)

Then using the integration by parts formula:

\begin{align} v_{rms} &= \sqrt{\frac{m}{2 \pi kT}} \left[ uv - \int v du\right] \\ &= \sqrt{\frac{m}{2 \pi kT}} \left[ V^2 \sqrt{\frac{2 \pi kT}{m}} - \int{\infty}_{-\infty} \sqrt{\frac{2 \pi kT}{m}} 2V\right] \\ &= \sqrt{\frac{m}{2 \pi kT}} \sqrt{\frac{2 \pi kT}{m}} [V^2 - \int^{\infty}_{-\infty} 2 V] \\ &= [V^2 - \frac{2}{3} V^3]^{\infty}_{-\infty} \end{align}

which is undefined. The expected answer was $$v_{rms} = \frac{kT}{m}$$ (which is valid for an ideal monatomic gas). Did I make a mistake in my integration?

• You must have; it's no longer dimensionally consistent.
– J.G.
Commented Nov 1, 2021 at 17:54
• when you integrate by parts, you defined dv, but the v you wrote does not come from that dv. You are also missing differentials on many integrals.
– user65081
Commented Nov 1, 2021 at 18:03
• @Wolphramjonny I used the formula for a Gaussian integral: $\int e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}$. Setting $a = \frac{m}{2kT}$ and $x = v$ gives $\sqrt{\frac{2kT \pi}{m}}$ Commented Nov 1, 2021 at 18:14
• You should refresh your knowledge on how to integrate
– user65081
Commented Nov 1, 2021 at 18:17
• @Wolphramjonny Could you please be a bit more specific? Is my formula for the Gaussian integral incorrect? The way I have applied it? Commented Nov 1, 2021 at 18:23

Mistakes include integrating from $$-\infty$$, conflating the PDF of $$\vec{V}$$ in $$\Bbb R^3$$ with that of $$V:=|\vec{V}|$$ in $$\Bbb R$$, and replacing $$v$$ in IBP with $$\int_0^\infty vdV$$, which doesn't even have the same dimension. Let $$\alpha:=m/(2kT)$$. Differentiating $$\int_0^\infty e^{-\alpha V^2}dV=\frac{\sqrt{\pi}}{2}\alpha^{-1/2}$$ twice with respect to $$-\alpha$$ gives $$\int_0^\infty V^2e^{-\alpha V^2}dV=\frac{\sqrt{\pi}}{4}\alpha^{-3/2},\,\int_0^\infty V^4e^{-\alpha V^2}dV=\frac{3\sqrt{\pi}}{8}\alpha^{-5/2}$$, so$$v_\text{rms}^2=\frac{\int_0^\infty V^4e^{-\alpha V^2}dV}{\int_0^\infty V^2e^{-\alpha V^2}dV}=\frac{3}{2\alpha}=\frac{3kT}{m}.$$Unfortunately, each integral is hard to evaluate by IBP because the power of $$V$$ is even.
$$d^3v = v^2dvd\Omega$$
is the differential in the integral, so you are evaluating $$\langle 1 \rangle$$, not $$\langle v \rangle$$.
Also: $$0\le v\lt\infty$$.