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A change of variable

  • Thread starter Castilla
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  • #1
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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.
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
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[tex]\int_{0}^{+\infty}e^{-x^2}dx[/tex]
is just the half of
[tex]\int_{-\infty}^{+\infty}e^{-x^2}dx[/tex]

The second is usually done with Fubini's theorem and polar substitution.
 
  • #3
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Er... but the book from where I take the problem only deals with functions of one real variable... they don't use Fubini...
 
  • #4
VietDao29
Homework Helper
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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,
 
  • #5
240
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Yes, Viet Dao. Really thanks.
 

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