Linear degree of freedom - Equipartition theorem

[steve233]

Homework Statement

Consider a classical 'degree of freedom' that is linear rather than quadratic: E = c|q| for some constant c. Derive the equipartition theorem using this energy and show that the average energy is Ebar = kT.

Homework Equations

[itex] Z = \sum e^{-\beta E(q)} = \sum e^{-\beta c|q|} [/itex]

[itex] Z = \frac{1}{\Delta q} \int_{-\infty}^{+\infty} e^{-\beta c |q|}dq [/itex]

The Attempt at a Solution

The question seems straight forward, but I'm having a hard time grasping it.

Using the second equation, If I carry out that integral I get:

[itex] \frac{1}{\Delta q} \frac {-1}{\beta c} \left [ e^{-\beta cq} \right ]_{-\infty}^{+\infty} = 0 [/itex]

Which doesn't help at all. I'm not sure if there is a trick to the integral or I have to use another method.

Any help will be much appreciated.

PS. This is coursework but not a homework question. I am just doing this question to study for a test.

 

[cragar]

it works if you go from zero to infinity on the bounds. and not -inf to +inf.
then use the formula <E>=-1/Z(dz/dB)