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Integration by parts

  1. Mar 12, 2009 #1
    I found this passage in a book:
    http://img8.imageshack.us/img8/1452/75717730.jpg [Broken]
    where Φ(x) is the c.d.f. of the normal distribution.
    However, using integration by parts I also get this term which is not included in the passage:

    [tex] [ e^{itx} \phi(x) ]^{+ \infty}_{- \infty} [/tex]
    where i is the imaginary unit and [tex] \phi(x) [/tex] is the normal p.d.f.
    So,
    [tex] e^{itx} \phi(x) = \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right) [/tex]
    So, is this term:
    [tex] \left[ \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right) \right]^{+ \infty}_{- \infty} [/tex]
    equal to zero and why? It is not clearly evident to me
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 12, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, as x go to infinity,
    [tex]e^{-x^2}[/tex]
    goes to 0 very quickly. Quickly enough so that even
    [tex]e^{-x^2}e^{x}[/tex] goes to 0 as x goes to positive infinity.
     
  4. Mar 12, 2009 #3
    I kept doing the same fault, but I figured it out now, thanks
     
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