Cfrgtkky

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_1^2int_{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_1^2int_{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

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

$$mathbf{1}_{{v_1 geqslant v_2}} = begin{cases}1, & v_1 geqslant v_2 \ 0 , & v_1 < v_2end{cases}$$

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 $v_1 iff v_2$.