# Question on changing limits of integral

1. Feb 7, 2015

### rwooduk

Say we have

$$\int_{0}^{\omega _{D}} \frac{\hbar \omega^{3}}{exp (\frac{\hbar\omega}{kT}) - 1} d\omega$$

let

$$x = \frac{\hbar\omega}{kT}$$

if we sub in we get

$$(\frac{kT}{\hbar})^{3} (\frac{kT}{\hbar}) \int_{0}^{\omega _{D}} \frac{x^{3}}{exp (x) - 1} dx$$

my question is how would the limits and integral change to get the limits in terms of x?

$$(\frac{kT}{\hbar})^{3} {kT} \int_{0}^{x _{D}} \frac{x^{3}}{exp (x) - 1} dx$$

i.e. it removes a h-bar when the limits are changed, please could someone explain this?

thanks for any help

Last edited: Feb 7, 2015
2. Feb 7, 2015

### Staff: Mentor

Since you are replacing $\omega$ with x, you're going to need to calculate dx as well.
When $\omega$ = 0, it's pretty obvious that x = 0 as well. Use your substitution above to calculate x when $\omega = \omega_D$.

3. Feb 7, 2015

### rwooduk

Thanks for the reply, I already calculated dx/dw = h-bar/kT , thats where the second kT/h-bar comes from in the second equation. Not sure how what you are saying helps me determine how the new limit x on the integral changes what is in the equation. i.e. where the h-bar goes in the answer

4. Feb 7, 2015

### Staff: Mentor

You have made a mistake when you did your substitution. After your substitution you should have this:
$$\int \frac{\hbar(\frac{kT}{\hbar})^3x^3 \frac{kT}{\hbar}dx}{e^x - 1}$$
When simplified, this gives what you showed as the answer.

5. Feb 9, 2015

### rwooduk

ahhh, ok thank you! i thought changing the limit effected the term somehow :-/ many thanks!!!!!