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?

  • $\begingroup$ You must have; it's no longer dimensionally consistent. $\endgroup$
    – J.G.
    Nov 1, 2021 at 17:54
  • $\begingroup$ 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. $\endgroup$
    – user65081
    Nov 1, 2021 at 18:03
  • $\begingroup$ @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 $\sqrt{\frac{2kT \pi}{m}}$ $\endgroup$
    – physBa
    Nov 1, 2021 at 18:14
  • $\begingroup$ You should refresh your knowledge on how to integrate $\endgroup$
    – user65081
    Nov 1, 2021 at 18:17
  • $\begingroup$ @Wolphramjonny Could you please be a bit more specific? Is my formula for the Gaussian integral incorrect? The way I have applied it? $\endgroup$
    – physBa
    Nov 1, 2021 at 18:23

2 Answers 2


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.


Note that:

$$d^3v = v^2dvd\Omega $$

is the differential in the integral, so you are evaluating $\langle 1 \rangle$, not $\langle v \rangle$.

Also: $0\le v\lt\infty$.


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