A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
converge (to what number) or diverge:
integral from infinity to infinity of e^x^2 dx
 2 years ago
converge (to what number) or diverge: integral from infinity to infinity of e^x^2 dx

This Question is Closed

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

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

experimentX
 2 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

liliy
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0I thought so .. but wolf doesn't agree.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1This is the error function, yo. http://en.wikipedia.org/wiki/Error_function

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

eliassaab
 2 years ago
Best 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.

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

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

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

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1My memory is the opposite of perfect. : And I'm prone to mistakes.

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1It 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.

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

liliy
 2 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

badreferences
 2 years ago
Best 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.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1That's just integration by parts, yo.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1But 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 \(0\).

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

liliy
 2 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 ) ;)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.