How can we solve this temperature dependent integral in solid state physics?

[cscott]

Homework Statement

Trying to solve this integral

$$\int_0^T \frac{dT'}{T'}\frac{d}{dT'}U(T',V)$$

where the temperature dependent part of U is

$$\Sigma \frac{h\omega}{\exp(\beta\omega)-1}$$

The Attempt at a Solution

using x = hw/T I find that I need to integrate

$$\frac{x^3 \exp(x/k)}{(\exp(x/k)-1)^2 h\omega k}$$

with limits going to 0 -> inf and T -> hw/T

I just don't see how with these limits it will work out nicely (the result is used to get the pressure in the harmonic approximation) but I can't find my mistake.

 

[mark726]

Thank you for sharing your question with us. In order to solve this integral, we can use the substitution u = x/k, which will simplify the integral to:

\int_0^\infty \frac{u^3 \exp(u)}{(\exp(u)-1)^2 h\omega} du

This integral can be solved using integration by parts, with the first term being \frac{u^3}{\exp(u)-1} and the second term being \frac{3u^2}{(\exp(u)-1)^2}. After integrating and simplifying, the final result should be:

\frac{\pi^4}{15 h\omega k^4}

I hope this helps with your calculations. Good luck!