Janev et al. in their book Elementary Processes in Hydrogen-Helium Plasmas say that the reaction rate $\langle\sigma v\rangle$ of a certain collision with cross-section $\sigma(v_r)$ for a particle of mass $m$ and energy $E=\frac{1}{2}mV^2$ incident on a stationary Maxwellian distribution of particles with mass $M$ and temperature $k_BT\equiv Mu^2/2$ is:
$$\langle\sigma v\rangle(T,V)=\frac{1}{\pi^{1/2}uV}\int_{0}^{\infty}dv_r v_r^2\sigma(v_r)\{\exp[-(v_r-V)^2/u^2]-\exp[-(v_r+V)^2/u^2]\}$$
where $v_r$ is the relative velocity between the incident paticle and the particles from the Maxwellian.
I really have no idea where this expression comes from. I mean, I know that the general expression is
$$\langle\sigma v\rangle=\frac{1}{n_1n_2}\int d^3v_1d^2v_2 f_1f_2\sigma(v_r)v_r$$
and I know the calculation I would have to do if, instead of an incident particle on a Maxwellian, we would have two species in equilibrium (so both $f_1$ and $f_2$ Maxwellians with same $T$).
But I really have no idea on how it should be in this particular case. Maybe what I am missing is on how to write the distibuition function to the incident particles and how that develops into that term with the difference of two exponentials...
