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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

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.

No It is not zero.

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.

That's just integration by parts, yo.

\[
\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}
\]