# Mathematical interlude

Posted in math on Tuesday, April 08 2014

Last time I poked around with calculating the mean velocity of the particles in a gas from the Maxwell distribution of velocities. While doing so I bashed my way through some integrals that, on reflection, are much easier when I remembered that they are in the back of some of my textbooks as "useful formulae", especially when doing probability. Namely:

$$\int \limits_0^\infty x^{2n} \cdot \exp \left( -x^2 \right) dx = { {1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)} \over {2^{n+1}} } \cdot \sqrt{\pi}$$

$$\int \limits_0^\infty x^{2n+1} \cdot \exp \left( -x^2 \right) dx = \frac{n!}{2}$$

Using those identities it is pretty straight forward to figure out the mean kinetic energy of the particles in the gas, from the Maxwell distribution:

$$\frac{1}{2} m \bar{u^2}= {{ \int \limits_0^\infty \frac{1}{2}m u^2 \cdot f(u) du } \over { \int \limits_0^\infty f(u) du }}$$

Then in my chicken scratch:

Which ends us off with $$\frac{1}{2} m \bar{u^2} = \frac{3}{2} \kappa T$$