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

Question

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

experimentx · Answer

converges to 2sqrt(pi) i guess.

anonymous · Answer

never mind, i see my mistake, thanks!

experimentx · Answer

http://www.wolframalpha.com/input/?i=integrate+e%5E-x%5E2+from+-inf+to+%2B+inf

anonymous · Answer

actually its converge to zero. you break it into two integrals and you equal -1/2 and 1/2

experimentx · Answer

I thought so .. but wolf doesn't agree.

anonymous · Answer

This is the error function, yo. http://en.wikipedia.org/wiki/Error_function

anonymous · Answer

you cant plug these into wolfram

anonymous · Answer

$$
e^{-x^2} < e^{-x}
$$
for all x except 
$$ 0< x < 1
$$
By Comparison test, your integral is convergent.

anonymous · Answer

p.s i have the answerkey and it says 0 sooo... lol

anonymous · Answer

Taylor expansion and integration should give an even function that evaluates to $0$, yes.

anonymous · Answer

No It is not zero.

anonymous · Answer

I'm mistaken, then. I remember seeing/working on a proof a while ago. Let me find it.

anonymous · Answer

My memory is the opposite of perfect. :| And I'm prone to mistakes.

anonymous · Answer

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.

anonymous · Answer

Oh, right, it was easier than I thought. Switch to polar. :P http://mathworld.wolfram.com/GaussianIntegral.html

anonymous · Answer

wowowoooooohhh!!! sooooooorrryy!! i left out an x... this is awkward.. lol
the qestion should be: xe^-x^2

anonymous · Answer

@eliassaab Yup, I was considering the fact that you obtain the $\text{erf}$ component while integrating the gaussian.

anonymous · Answer

That's just integration by parts, yo.

anonymous · Answer

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 $0$.

anonymous · Answer

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

anonymous · Answer

@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 ) ;)