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Question on changing limits of integral

  1. Feb 7, 2015 #1
    Say we have

    [tex]\int_{0}^{\omega _{D}} \frac{\hbar \omega^{3}}{exp (\frac{\hbar\omega}{kT}) - 1} d\omega[/tex]


    [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?


    [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: Feb 7, 2015
  2. jcsd
  3. Feb 7, 2015 #2


    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##.
  4. Feb 7, 2015 #3
    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
  5. Feb 7, 2015 #4


    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.
  6. Feb 9, 2015 #5
    ahhh, ok thank you! i thought changing the limit effected the term somehow :-/ many thanks!!!!!
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