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Need to help to solve an integral from quantum conductance

  1. Aug 7, 2006 #1
    Hi All,

    While going through quantum conductnace derivation, I got stuck by an integral.

    I = ((e*(h/2pi))/(m*pi)) Integral from 0 to infinity

    dt / ( exp (((((h/2pi)^2 * t)/(2*m)) - Ef) / kT)) + 1)

    The answer is
    I = (2e/h)*kT*ln(1 + exp(Ef/kT))

    I am stuggling to find the answer. Any help will be appreciated.

  2. jcsd
  3. Aug 7, 2006 #2


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    Homework Helper

    Cannot really get what you are asking here... :frown:
    Are you trying to evaluate:
    [tex]\mathop{\int} \limits_{0} ^ {\infty} \frac{dt}{e ^ { \frac{ \frac{ \left( \frac{h}{2 \pi} \right) ^ 2 t}{2m} - Ef}{kT} + 1}}[/tex]?
    If that's what you mean, then what does the part before it: I = ((e*(h/2pi))/(m*pi)) have to do here? Is it also an integral? Or a constant?
    Or do you mean:
    Are you trying to evaluate:
    [tex]I = \frac{e ^ {\frac{h}{2 \pi}}}{2m} \mathop{\int} \limits_{0} ^ {\infty} \frac{dt}{e ^ {\frac{\frac{ \left( \frac{h}{2 \pi} \right) ^ 2 t}{2m} - Ef}{kT} + 1}}[/tex]?
    Last edited: Aug 7, 2006
  4. Aug 7, 2006 #3
    I think [itex]e[/itex] is the electronic charge (hence the exp notation later on), it looks like he's trying to calculate a charge density (the integrand is the Fermi-Dirac distribution for average number of fermions in a given state).

    To evaluate

    [tex]\int_0^\infty \frac{1}{e^{\frac{ax-b}{c}} + 1}dx[/tex]

    where a,b,c > 0, just sub

    [tex]u = e^{\frac{ax-b}{c}}.[/tex]

    Then [itex]\frac{c}{au}du = dx[/itex], and when [itex]x=0[/itex] you get [itex]u = e^{-b/c}[/itex], so the integral is

    [tex]\int_{e^{-b/c}}^\infty \frac{c}{au(u+1)}du.[/tex]

    You can split the integrand up by partial fractions:

    [tex]\frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1},[/tex]

    and I think you should be able to finish from there (you'll need to simplify a bit after integrating to get to the form of the answer you posted above, but it's right) :smile:
    Last edited: Aug 7, 2006
  5. Aug 11, 2006 #4
    Thanks Data. I am able to get the answer after following your suggestions.
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