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liliy

  • 2 years ago

converge (to what number) or diverge: integral from -infinity to infinity of e^-x^2 dx

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  1. experimentX
    • 2 years ago
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    converges to 2sqrt(pi) i guess.

  2. liliy
    • 2 years ago
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    never mind, i see my mistake, thanks!

  3. experimentX
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=integrate+e%5E-x%5E2+from+-inf+to+%2B+inf

  4. liliy
    • 2 years ago
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    actually its converge to zero. you break it into two integrals and you equal -1/2 and 1/2

  5. experimentX
    • 2 years ago
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    I thought so .. but wolf doesn't agree.

  6. badreferences
    • 2 years ago
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    This is the error function, yo. http://en.wikipedia.org/wiki/Error_function

  7. liliy
    • 2 years ago
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    you cant plug these into wolfram

  8. eliassaab
    • 2 years ago
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    \[ e^{-x^2} < e^{-x} \] for all x except \[ 0< x < 1 \] By Comparison test, your integral is convergent.

  9. liliy
    • 2 years ago
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    p.s i have the answerkey and it says 0 sooo... lol

  10. badreferences
    • 2 years ago
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    Taylor expansion and integration should give an even function that evaluates to \(0\), yes.

  11. eliassaab
    • 2 years ago
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    No It is not zero.

  12. badreferences
    • 2 years ago
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    I'm mistaken, then. I remember seeing/working on a proof a while ago. Let me find it.

  13. badreferences
    • 2 years ago
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    My memory is the opposite of perfect. :| And I'm prone to mistakes.

  14. eliassaab
    • 2 years ago
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    It is a striclty poistive function, its integral cannot be zero. The integral of such a function is used in probabilty theory for the Normal Distribution.

  15. badreferences
    • 2 years ago
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    Oh, right, it was easier than I thought. Switch to polar. :P http://mathworld.wolfram.com/GaussianIntegral.html

  16. liliy
    • 2 years ago
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    wowowoooooohhh!!! sooooooorrryy!! i left out an x... this is awkward.. lol the qestion should be: xe^-x^2

  17. badreferences
    • 2 years ago
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    @eliassaab Yup, I was considering the fact that you obtain the \(\text{erf}\) component while integrating the gaussian.

  18. badreferences
    • 2 years ago
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    That's just integration by parts, yo.

  19. badreferences
    • 2 years ago
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    But you could simplify it. \(x\) is odd and \(e^{(-x^2)}\) is even around the same mid. Therefore, integrating from \(-\infty\to\infty\) should evaluate to \(0\).

  20. eliassaab
    • 2 years ago
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    \[ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi} \]

  21. liliy
    • 2 years ago
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    @eliassaab , i guess that would be the correct answer, but i wrote my question wrong and ment for it to be x e^-x^2 dx . but u get a medal anyway(well deserved ) ;)

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