Question on changing limits of integral

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  • #1
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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
 
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Answers and Replies

  • #2
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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]
Since you are replacing ##\omega## with x, you're going to need to calculate dx as well.
rwooduk said:
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?
When ##\omega## = 0, it's pretty obvious that x = 0 as well. Use your substitution above to calculate x when ##\omega = \omega_D##.
rwooduk said:
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
 
  • #3
747
55
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##.
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
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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
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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.
ahhh, ok thank you! i thought changing the limit effected the term somehow :-/ many thanks!!!!!
 

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