anonymous
  • anonymous
converge (to what number) or diverge: integral from -infinity to infinity of e^-x^2 dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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experimentX
  • experimentX
converges to 2sqrt(pi) i guess.
anonymous
  • anonymous
never mind, i see my mistake, thanks!
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=integrate+e%5E-x%5E2+from+-inf+to+%2B+inf

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anonymous
  • anonymous
actually its converge to zero. you break it into two integrals and you equal -1/2 and 1/2
experimentX
  • experimentX
I thought so .. but wolf doesn't agree.
anonymous
  • anonymous
This is the error function, yo. http://en.wikipedia.org/wiki/Error_function
anonymous
  • anonymous
you cant plug these into wolfram
anonymous
  • anonymous
\[ e^{-x^2} < e^{-x} \] for all x except \[ 0< x < 1 \] By Comparison test, your integral is convergent.
anonymous
  • anonymous
p.s i have the answerkey and it says 0 sooo... lol
anonymous
  • anonymous
Taylor expansion and integration should give an even function that evaluates to \(0\), yes.
anonymous
  • anonymous
No It is not zero.
anonymous
  • anonymous
I'm mistaken, then. I remember seeing/working on a proof a while ago. Let me find it.
anonymous
  • anonymous
My memory is the opposite of perfect. :| And I'm prone to mistakes.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Oh, right, it was easier than I thought. Switch to polar. :P http://mathworld.wolfram.com/GaussianIntegral.html
anonymous
  • anonymous
wowowoooooohhh!!! sooooooorrryy!! i left out an x... this is awkward.. lol the qestion should be: xe^-x^2
anonymous
  • anonymous
@eliassaab Yup, I was considering the fact that you obtain the \(\text{erf}\) component while integrating the gaussian.
anonymous
  • anonymous
That's just integration by parts, yo.
anonymous
  • anonymous
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 \(0\).
anonymous
  • anonymous
\[ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi} \]
anonymous
  • anonymous
@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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