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Limits of integration

  1. Feb 25, 2008 #1
    [SOLVED] limits of integration

    1. The problem statement, all variables and given/known data
    I want to substitute x=t^2 in [itex]\int_{-\infty}^{\infty}{\exp(-t^4)} dt[/itex]. What are the new limits of integration? They are both infinity aren't they? But the integral is clearly not zero? Is the problem that the substitution rule only holds for finite limits of integration?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 25, 2008 #2
    The substitution introduces dx = 2t dt within the integral. Perhaps a polar coordinate form would be helpful.
     
  4. Feb 25, 2008 #3
    Of course, but [tex]2t = 2 \sqrt{x} [/tex]. Why is this not valid:

    [tex]\int_{-\infty}^{\infty}\exp(-t^4)dt = \int_{\infty}^{\infty}\exp(-x^2)dx/(2\sqrt{x}) [/tex]

    ?
     
  5. Feb 25, 2008 #4

    NateTG

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    Well, you need to use [itex]-\sqrt{-x}[/itex] where [itex]x<0[/itex].
     
  6. Feb 25, 2008 #5
    Maybe you should break the integral up into the part over t < 0 and t > 0. What does that turn into? I don't know the answer, just a suggestion.
     
  7. Feb 25, 2008 #6
    I don't care how to evaluate the specific integral. My question is about the substitution rule. No one has yet explained to me why this

    [tex]\int_{-\infty}^{\infty}\exp(-t^4)dt = \int_{\infty}^{\infty}\exp(-x^2)dx/(2\sqrt{x}) [/tex]

    is not a valid use of the substitution rule?
     
  8. Feb 25, 2008 #7
    I think it's just because t^2 = x doesn't necessarily imply that t = sqrt(x).
     
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