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Homework Help: A change of variable

  1. Sep 27, 2005 #1
    I am trying to follow a proof of
    [tex] \int_{0}^{+\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2} [/tex] but the first impasse I find is that, "with the change of variable" [tex]\sqrt{n}x = y [/tex] they justify this equality:
    [tex] {\frac{1}{\sqrt{n}}\int_{0}^{+\infty}e^{-y^2}dy = \int_{0}^{+\infty}e^{-nx^2}dx [/tex].

    Maybe you can help me to see how they did it? Thanks.
  2. jcsd
  3. Sep 28, 2005 #2


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    is just the half of

    The second is usually done with Fubini's theorem and polar substitution.
  4. Sep 28, 2005 #3
    Er... but the book from where I take the problem only deals with functions of one real variable... they don't use Fubini...
  5. Sep 28, 2005 #4


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    [tex]\sqrt{n}x = y[/tex], so, you have: [tex]d(\sqrt{n}x) = \sqrt{n}dx = dy[/tex]
    Then, you have:
    [tex]\frac{1}{\sqrt{n}} \int_{0} ^ {+ \infty} e ^ {-y^2} dy = \frac{1}{\sqrt{n}} \int_{0} ^ {+\infty} e ^ {-nx ^ 2} (\sqrt{n} dx)[/tex]
    [tex]= \sqrt{n} \times \frac{1}{\sqrt{n}} \int_{0} ^ {+\infty} e ^ {-nx ^ 2} dx = \int_{0} ^ {+\infty} e ^ {-nx ^ 2} dx[/tex].
    Can you get it now?
    Viet Dao,
  6. Sep 28, 2005 #5
    Yes, Viet Dao. Really thanks.
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