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## Main Question or Discussion Point

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

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