Here's the question you clicked on:
liliy
converge (to what number) or diverge: integral from -infinity to infinity of e^-x^2 dx
converges to 2sqrt(pi) i guess.
never mind, i see my mistake, thanks!
http://www.wolframalpha.com/input/?i=integrate+e%5E-x%5E2+from+-inf+to+%2B+inf
actually its converge to zero. you break it into two integrals and you equal -1/2 and 1/2
I thought so .. but wolf doesn't agree.
This is the error function, yo. http://en.wikipedia.org/wiki/Error_function
you cant plug these into wolfram
\[ e^{-x^2} < e^{-x} \] for all x except \[ 0< x < 1 \] By Comparison test, your integral is convergent.
p.s i have the answerkey and it says 0 sooo... lol
Taylor expansion and integration should give an even function that evaluates to \(0\), yes.
I'm mistaken, then. I remember seeing/working on a proof a while ago. Let me find it.
My memory is the opposite of perfect. :| And I'm prone to mistakes.
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.
Oh, right, it was easier than I thought. Switch to polar. :P http://mathworld.wolfram.com/GaussianIntegral.html
wowowoooooohhh!!! sooooooorrryy!! i left out an x... this is awkward.. lol the qestion should be: xe^-x^2
@eliassaab Yup, I was considering the fact that you obtain the \(\text{erf}\) component while integrating the gaussian.
That's just integration by parts, yo.
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 \(0\).
\[ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi} \]
@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 ) ;)