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In the textbook Thermal physics, by Garg, Bansal and Ghosh [1], there exist two integrals on page 640, which are claimed to be equivalent (without demonstration) by a change in the order of integration.

\begin{align} I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\\ I_2=&\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2 \end{align}

It is stated that an interchange of the order of integration will show them to be equivalent. To show this, it must be done in such a way that the limits over $v_1$ in $I_1$, change from $0$ to $\infty$ into $v_1$ to $\infty$.

However, when attempting this problem myself I notice two things. The change in limits on $I_1$ doesn't appear to yield an equivalent region to integrate over, and even if it did, the functions to be integrated are not the same. This indicates to me that there is some other trick at play.

So the question follows: how can I prove these two integrals are the same, and what other technique should be employed, besides a change in integration, in order to show it?

To combine the integrals into something I can change the order in, I have been relabeling the first $v_1$ as $x$, then writing something like this:

\begin{align} I_1&=\int_{0}^{v_1}\int_{0}^{\infty}e^{-\frac{mx^2}{2k_BT}}x^2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2dxdv_2\\ \end{align}

I'm asking this question because I promised @Thorondor I'd show them to be equivalent in this question [2], but unfortunately I got stumped.

References:

[1] Garg S.C. Bansal R.M. Ghosh C.K. (2012) Thermal Physics: Kinetic Theory, Thermodynamics and Statistical Mechanics. Tata McGraw Hill Education, New Delhi, 638-641.

[2] https://physics.stackexchange.com/questions/448761/rigorous-derivation-of-the-mean-free-path-in-a-gas

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Write the inner integral of $I_1$ as

$$\int_{0}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2 \mathbf{1}_{\{v_1 \geqslant v_2\}}$$

where

$$\mathbf{1}_{\{v_1 \geqslant v_2\}} = \begin{cases}1, & v_1 \geqslant v_2 \\ 0 , & v_1 < v_2\end{cases}$$

Apply Fubini's theorem to interchange integrals noting that

$$\int_{0}^{\infty}dv_1f(v_1,v_2)\mathbf{1}_{\{v_1 \geqslant v_2\}} = \int_{v_2}^{\infty}dv_1f(v_1,v_2) $$

and then change notation $v_1 \iff v_2$.

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  • $\begingroup$ Would it be correct to say here, that we are integrating over different regions, but the areas of both regions are the same, justifying the change in notation? I'm just trying to verify if my understanding of the answer is correct. $\endgroup$ – user400188 Jan 2 '19 at 5:46
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    $\begingroup$ If you follow my steps you get $I_1 = \int_{0}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^2\int_{v_2}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_1^2$ before changing notation. I think you can see why this works just by looking at a graph of the original region of integration -- an unbounded triangle-like region. Now switching variable names requires no deep mathematical justification for a definite integral. $\endgroup$ – RRL Jan 2 '19 at 5:55
  • $\begingroup$ Thanks @RRL, I drew the graphs earlier and wanted to confirm my suspicion. $\endgroup$ – user400188 Jan 2 '19 at 5:56

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