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anonymous
 4 years ago
converge (to what number) or diverge:
integral from infinity to infinity of e^x^2 dx
anonymous
 4 years ago
converge (to what number) or diverge: integral from infinity to infinity of e^x^2 dx

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experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0converges to 2sqrt(pi) i guess.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0never mind, i see my mistake, thanks!

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually its converge to zero. you break it into two integrals and you equal 1/2 and 1/2

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0I thought so .. but wolf doesn't agree.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is the error function, yo. http://en.wikipedia.org/wiki/Error_function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you cant plug these into wolfram

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0p.s i have the answerkey and it says 0 sooo... lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Taylor expansion and integration should give an even function that evaluates to \(0\), yes.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm mistaken, then. I remember seeing/working on a proof a while ago. Let me find it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My memory is the opposite of perfect. : And I'm prone to mistakes.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It 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
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, right, it was easier than I thought. Switch to polar. :P http://mathworld.wolfram.com/GaussianIntegral.html

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@eliassaab Yup, I was considering the fact that you obtain the \(\text{erf}\) component while integrating the gaussian.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's just integration by parts, yo.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \int_{\infty}^\infty e^{x^2} dx = \sqrt{\pi} \]

anonymous
 4 years ago
Best 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 ) ;)
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