Fun With Maxwell Distribution

Its clear that,
We, somehow, have to find $P$ .

We know (I don't know how to prove it, just know it),
$\displaystyle dx$ $dy$ $dz = r^2\sin\theta$ $d\theta$ $d\phi$ $dr$

$\Rightarrow dv_x$ $dv_y$ $dv_z = v^2 \sin\theta$ $d\theta$ $d\phi$ $dv$
where, $\displaystyle \theta,\phi$ are the path angle and course angle respectively

Thus, the expression becomes,
$\displaystyle dP = \left(\frac{m}{2\pi kT}\right)^{3/2} e^{-\frac{mv^2}{2kT}} v^2 \sin\theta d\theta d\phi dv$

If we integrate this probability with,
$\theta$ going from $0$ to (\pi) and,
$\phi$ going from $0$ to $2\pi$ , then,

The integral turns out to be,
$P(v) = 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} v^2e^{-\frac{mv^2}{2kT}}$

Now, the average value of this function is,
< $\displaystyle v$ > $\displaystyle = \int\limits_0^{\infty} v P(v) dv$

Substituting $P(v)$ , and integrating,
we get,
< $\displaystyle v$ > $\displaystyle = \sqrt{\frac{8kT}{\pi m}}$ as expected.

Putting < $v$ > $= 300$ ,
$\displaystyle \frac{m}{kT} = 2.829\times 10^{-5}$

Back to the average value function, substitute $\displaystyle v$ as $\displaystyle \frac{1}{v}$ ,
< $\displaystyle \frac{1}{v}$ > $\displaystyle = \int\limits_0^{\infty} \frac{1}{v} P(\frac{1}{v}) dv$

Integrating this expression, and putting the value of $\displaystyle\frac{m}{kT}$ , we get,
< $\displaystyle \frac{1}{v}$ > $\displaystyle = \boxed{4.24\times 10^{-3}}$