- #1
rwooduk
- 762
- 59
Say we have
[tex]\int_{0}^{\omega _{D}} \frac{\hbar \omega^{3}}{exp (\frac{\hbar\omega}{kT}) - 1} d\omega[/tex]
let
[tex]x = \frac{\hbar\omega}{kT}[/tex]
if we sub in we get
[tex](\frac{kT}{\hbar})^{3} (\frac{kT}{\hbar}) \int_{0}^{\omega _{D}} \frac{x^{3}}{exp (x) - 1} dx[/tex]
my question is how would the limits and integral change to get the limits in terms of x?
ANSWER:
[tex](\frac{kT}{\hbar})^{3} {kT} \int_{0}^{x _{D}} \frac{x^{3}}{exp (x) - 1} dx[/tex]
i.e. it removes a h-bar when the limits are changed, please could someone explain this?
thanks for any help
[tex]\int_{0}^{\omega _{D}} \frac{\hbar \omega^{3}}{exp (\frac{\hbar\omega}{kT}) - 1} d\omega[/tex]
let
[tex]x = \frac{\hbar\omega}{kT}[/tex]
if we sub in we get
[tex](\frac{kT}{\hbar})^{3} (\frac{kT}{\hbar}) \int_{0}^{\omega _{D}} \frac{x^{3}}{exp (x) - 1} dx[/tex]
my question is how would the limits and integral change to get the limits in terms of x?
ANSWER:
[tex](\frac{kT}{\hbar})^{3} {kT} \int_{0}^{x _{D}} \frac{x^{3}}{exp (x) - 1} dx[/tex]
i.e. it removes a h-bar when the limits are changed, please could someone explain this?
thanks for any help
Last edited: