Question

@dan815 how to integrate e^-x^2 a quick guide.

Answer 1

Wait, you're looking to calculate the mean and variance and stuff with this right? That implies an integration over all space. Otherwise it is not integrable.

Answer 2

i cud say k=x^2 and rewrite that series what else do u mean?

Answer 3

Dan, you noob. Use polar coordinates.

Answer 4

oh isnt that for positibee x^2

Answer 5

I'm retracting my medal. You don't deserve it.

Answer 6

:'(

Answer 7

HA!

Answer 8

To be clear this is what I'll show, how to integrate this. But we can always throw in constants or whatever and work it through, but I'll leave that to you. \[\Large \int\limits_{- \infty}^\infty e^{-x^2}dx\]

So we square it and then square root it to get the same value since we know it's positive. An integral just has dummy variables so we make it have y's for the other integral but it's really all the same. \[\Large \sqrt{\int\limits_{- \infty}^\infty e^{-x^2}dx \int\limits_{- \infty}^\infty e^{-y^2}dy }\]We can go ahead and combine them too: \[\Large \sqrt{\int\limits_{- \infty}^\infty \int\limits_{- \infty}^\infty e^{-(x^2+y^2)}dxdy }\] Go to polar we have: \[\Large \sqrt{\int\limits_{\theta = 0}^{\theta = 2\pi}\int\limits_{r=0}^{r=\infty} r e^{-r^2}rdr d \theta } \] You can see a u-sub right here so yeah the rest is easy