Maxwell-boltzman distrubution

[E92M3]

If I have two particles that follows the maxwell velocity distribution:
$$\phi(v_i)dv_i=4 \pi v_i^2 \left ( \frac{m_iv_i}{2\pi kT} \right ) ^{3/2}e^{\frac{-m_iv_i^2}{2kT}}dv_i$$
Why is their combined distribution:
$$\phi(v)dv=4 \pi v^2 \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2}e^{\frac{-\mu v^2}{2kT}}dv$$
where mu is the reduced mass and v=v2-v1
I have these questions because I don't quite follow these derivations.
http://dissertations.ub.rug.nl/FILES/faculties/science/2007/a.matic/c2.pdf [Broken]
http://www.astro.psu.edu/users/rbc/a534/lec11.pdf
Namely, I not sure why the following holds:
$$\int_0^\infty \int_0^\infty \phi(v_1) \phi(v_2) v_1 v_2 \sigma dv_1 dv_2 = 4\pi \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2} \int_0^\infty v^3 \sigma e^{\frac{-\mu v^2}{2kT}}dv$$

 

[mathman]

You have v=v1-v2. Let u=v1+v2. The new integration is straightforward as long as the integral limits are -∞ to ∞. You then need |v1v2| instead of v1v2 in the integrand. The u integration should leave you with what you want.