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Homework Help: Help me understand the math behind this

  1. Jan 9, 2006 #1

    while solving for the average energy given by the following formula:

    [tex] \overline {E} = \frac {\int_{0}^{\infty} E e^\frac{-E}{kT}dE}{\int_{0}^{\infty} e^\frac{-E}{kT}dE} [/tex]

    where E bar is average energy, k is the Boltzmann's constant, and T is temperature

    I had to use integration by parts for the numerator.

    Integration by parts formula is [tex] \int u dv = uv - \int v du [/tex]

    So I made the following choices (and so did my textbook):

    [tex] u = E [/tex]

    then [tex] du = dE [/tex]

    [tex] dv = e^\frac{-E}{kT} [/tex]

    and so [tex] v = -kTe^\frac{-E}{kT} [/tex]

    Then I proceeded by applying the integration by parts formula, and the integral of the numerator would be:

    [tex] \int_{0}^{\infty} E e^\frac{-E}{kT}dE = -EkTe^\frac{-E}{kT} + kT \int_{0}^{\infty} e^\frac{-E}{kT}dE [/tex]

    but to my surprise, the book proceeded in the following manner:

    [tex] \int_{0}^{\infty} E e^\frac{-E}{kT}dE = kT \left[e^\frac{-E}{kT} \right]
    _{0}^{\infty} + kT \int_{0}^{\infty} e^\frac{-E}{kT}dE [/tex]

    That first term to the right of the equal sign threw me off....where did it come from?
    Last edited: Jan 9, 2006
  2. jcsd
  3. Jan 9, 2006 #2


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    Homework Helper


    [tex] dv=e^{-\frac{E}{kT}} \ dE [/tex]

    then you're right. It has to be some error in the book. Perhaps they meant

    [tex] kT \left[e^{-\frac{E}{kT}}\right]_{+\infty}^{0} [/tex]

    ,that is converting the minus before the whole term into an interchange of integration limits.

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