I'm asked to find the mean velocity of gas particles knowing that their velocity can be modeled by the Maxwell distribution whose probability function is given by : $$ \mathrm{f}\left(\, \vec{v}\,\right) = 4\pi\,\left(\, m \over 2\pi\mathrm{k_{B}}T\,\right)^{3/2}v^{2} \exp\left(\,-\,{v^{2}\over 2m\,\mathrm{k_{B}}T}\right) $$ I guess there is some trick to apply, some change of variable to do or some already-known distribution function to link to but I didn't manage to find it on my own so any help would be very appreciated.
Thanks
I think this is the Maxwell distribution for speed, $0$ to $\infty$. Think of it as $av^2e^{-bv^2}$, where $a$ and $b$ are messy constants.
We want $\int_0^\infty av^3 e^{-bv^2}\,dv$. Maybe make the change of variable $v^2=t$. Then $2v dv=dt$, and we want $$\int_0^\infty \frac{a}{2} te^{-bt}\,dt.$$ To finish, one integration by parts will do it.