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

1. experimentX

converges to 2sqrt(pi) i guess.

2. anonymous

never mind, i see my mistake, thanks!

3. experimentX
4. anonymous

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

5. experimentX

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

6. anonymous

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

7. anonymous

you cant plug these into wolfram

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

No It is not zero.

12. anonymous

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

13. anonymous

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

14. anonymous

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.

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

That's just integration by parts, yo.

19. anonymous

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 $$0$$.

20. anonymous

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

21. anonymous

@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 ) ;)