How Do You Prove RMS Speed Using Maxwell-Boltzmann Distributions?

[ppyadof]

Homework Statement

This is the problem: I need to prove from the Maxwell-Boltzmann distributions the expression for the rms speed.

Homework Equations

$$v_{rms} = \sqrt{\frac{3kT}{m}}$$
$$< v^2 > = 4\pi(\frac{m}{2\pi kT})^{3/2} \int v^4 e^{\frac{-mv^2}{2kT}} dv$$
The limits on the integral are infinity and 0.

From what I know;
$$v_{rms} = sqrt{ <v^2> }$$

The Attempt at a Solution

I did a similar one like this to prove the corresponding expression for <v>. I used a substitution and then used parts on the resulting expression, but if use that method on this one, you get a factor of $u^{3/2}$ times by the exponential (assuming the substitution is $u = v^2$). This obviously causes problems, so I was wondering is there another way to do this proof without resorting to an integral or is there an easy way of doing this integral?

P.S. Sorry if this thread should be in the advanced physics part. I only thought of that after I posted.

 

[George Jones]

Do you know how to calculate (via the standard trick)

$$\int_0^\infty e^{-a x^2} dx?$$

If you do, then using integration by parts a couple of times will reduce your integral to an integral of this form.

 

[AKG]

1. Do a change of variable just to neaten up the expression (i.e. absorb the constants into your variable with respect to which you're integrating so that you can take all constants out of the integral)
2. Use integration by parts
3. Fact:

$$\int _{-\infty} ^{\infty}e^{-x^2}dx = \sqrt{\pi }$$

 

[dextercioby]

Do you know how to calculate (via the standard trick)

$$\int_0^\infty e^{-a x^2} dx?$$

If you do, then using integration by parts a couple of times will reduce your integral to an integral of this form.

It's nicer if you treat "a" as a parameter and consider the second derivative of the result...