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How to obtain this?

  1. Apr 11, 2010 #1
    Hello,

    I am reading some material that using mathematics extensively, and I encountered with the following result:

    [tex]\frac{N}{\overline{\gamma}}\,\int_0^{\infty}\gamma\,\left[1-\mbox{e}^{-\gamma/\overline{\gamma}}\right]^{N-1}\,\mbox{e}^{-\gamma/\overline{\gamma}}\,d\gamma=\,\overline{\gamma}\sum_{k=1}^N\frac{1}{k}[/tex]

    How did they get there? I tried to use the binomial expansion and assemble the exponentials, but the result was totally different. Any hint will be highly appreciated.

    Thanks in advance
     
  2. jcsd
  3. Apr 11, 2010 #2

    CompuChip

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    Note that the integrand looks very much like a derivative of
    [tex]\left( 1 - e^{-\gamma / \overline{\gamma}} \right)^N[/tex]
    with respect to either gamma or gamma-bar.
    Maybe you can use that to your advantage.
     
  4. Apr 11, 2010 #3
    Yes, you are right. the term [tex]\left[1-\mbox{e}^{-\gamma/\overline{\gamma}}\right]^N[/tex] is the CDF of [tex]\gamma[/tex]. In the integral it is intended to find the statistical average (the expected value) of [tex]\gamma[/tex] which needs the PDF of [tex]\gamma[/tex] which is the derivative of the CDF with respect to [tex]\gamma[/tex]. But I am still stuck. Any further hint?

    Thanks in advance
     
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