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

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experimentXBest ResponseYou've already chosen the best response.0
converges to 2sqrt(pi) i guess.
 one year ago

liliyBest ResponseYou've already chosen the best response.0
never mind, i see my mistake, thanks!
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=integrate+e%5Ex%5E2+from+inf+to+%2B+inf
 one year ago

liliyBest ResponseYou've already chosen the best response.0
actually its converge to zero. you break it into two integrals and you equal 1/2 and 1/2
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
I thought so .. but wolf doesn't agree.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
This is the error function, yo. http://en.wikipedia.org/wiki/Error_function
 one year ago

liliyBest ResponseYou've already chosen the best response.0
you cant plug these into wolfram
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
\[ e^{x^2} < e^{x} \] for all x except \[ 0< x < 1 \] By Comparison test, your integral is convergent.
 one year ago

liliyBest ResponseYou've already chosen the best response.0
p.s i have the answerkey and it says 0 sooo... lol
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
Taylor expansion and integration should give an even function that evaluates to \(0\), yes.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
I'm mistaken, then. I remember seeing/working on a proof a while ago. Let me find it.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
My memory is the opposite of perfect. : And I'm prone to mistakes.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
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.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
Oh, right, it was easier than I thought. Switch to polar. :P http://mathworld.wolfram.com/GaussianIntegral.html
 one year ago

liliyBest ResponseYou've already chosen the best response.0
wowowoooooohhh!!! sooooooorrryy!! i left out an x... this is awkward.. lol the qestion should be: xe^x^2
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
@eliassaab Yup, I was considering the fact that you obtain the \(\text{erf}\) component while integrating the gaussian.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
That's just integration by parts, yo.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
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 \(0\).
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
\[ \int_{\infty}^\infty e^{x^2} dx = \sqrt{\pi} \]
 one year ago

liliyBest ResponseYou've already chosen the best response.0
@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 ) ;)
 one year ago
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