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Pesky integral involving exponentials

  1. Dec 16, 2007 #1


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    I am trying to see how, in my textbook,

    [tex]\frac{1}{3}\int_{0}^{\infty}x^3 ln(1-e^{-x}) dx
    = \frac{1}{3}\left[ x^3 ln(1-e^{-x}) \right]_{0}^{\infty} - \frac{1}{3}\int_{0}^{\infty} \frac{x^3 dx}{e^x - 1}
    = - \frac{\pi^4}{45}[/tex]

    The integration by parts is clear enough, it's clear that first term of the middle expression evaluates at zero, and I recognise the standard integral that has been used to resolve the remaining integral... but how has exponent of e in the second integral suddenly gone from negative to being positive, without any change in the limits etc?

    I would have written:

    [tex]\frac{1}{3}\int_{0}^{\infty}x^3 ln(1-e^{-x}) dx
    = \frac{1}{3}\left[ x^3 ln(1-e^{-x}) \right]_{0}^{\infty} - \frac{1}{3}\int_{0}^{\infty} \frac{x^3 dx}{1 - e^{-x}} = \frac{1}{3}\int_{0}^{\infty} \frac{x^3 dx}{e^{-x} - 1}[/tex]

    I'm sure it's obvious, but I'm not seeing it just now.

    Last edited: Dec 16, 2007
  2. jcsd
  3. Dec 16, 2007 #2
    Hopefully this is what you are asking:

    if you think of integration by parts as being f g' = f g - [tex]\int g f[/tex], then they are letting ln (1-e^-x) be the f in this case.

    Taking the derivative gives [tex]1/1 - e^-x[/tex](-e^-x)(-1) by the chain rule. The two negatives cancel out to give

    [tex]1/1 - e^-x[/tex](e^-x).

    Bring e^-x = [tex]1/e^x[/tex] so your f' will be [tex]1/e^x (1 - e^-x)[/tex]

    Multiply through to get f' = [tex]1/e^x - 1[/tex]
  4. Dec 16, 2007 #3


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    Of course. I just wasn't handling the log properly when I differentiated it (!).

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