Question

integrate from -infinity to infinity.
xe^((-x)^2)

Answer 1

i know that it turns out to be \[\frac{ -1 }{ 2 } e^{-x^{2}}\]

Answer 2

and im not sure if i need to split up the boundaries to -infinity to 0 and then 0 to infinity.

Answer 3

yep you need a splitting

Answer 4

i know when i split, the equation can't be improper, by splitting on 0 is that okay?

Answer 5

you can choose to split it up at 0 or at any constant number between -inf and inf

Answer 6

and , when i try to evaulate it from -infintiy to 0 i get -infinity, and if i evalulate it from 0 to infinity i get +infinity

Answer 7

would that be 0 or would i have to use l'hopital?

Answer 8

or am i wrong in my evaluation.

Answer 9

We need to evaluate this limit
\[\frac{-1}{2}e^{-x^2}|_{-\infty}^{0}\]
and also \[\frac{-1}{2}e^{-x^2}|_{0}^{\infty}\]
If both limits converge (say one to a and the other to b)
then the sum of these limits will converge to a+b
If one of diverges, then the sum diverges.

Answer 10

I don't even need l'hospital for this
x^2 is positive
-x^2 is negative
e^(really negative large numbers) approaches 0

Answer 11

but don't forget about the constant parts up there ...

Answer 12

the first evaluation is -infinity because its -1/2 -infinity, and the second one is 0 +1/2

Answer 13

They should both converge

Answer 14

first one is -1/2-0
second one is 0+1/2

Answer 15

is it because you're squaring the -infinity first before accounting the -x

Answer 16

squaring negative numbers is still positive

Answer 17

so like -(-infinity^2)

Answer 18

you mean -(-infinity)^2

Answer 19

i guess its order of operation LOL

Answer 20

Pretend f(x)=x^2
f(-x)=f(x)

Answer 21

f is an even function

Answer 22

Like pluggin in -1 and 1
both give the same f which is 1

Answer 23

\[g(x)=-x^2 \]

Answer 24

This is still an even function

Answer 25

pluggin in -1 or 1 still gives you -1

Answer 26

i understand now :) thank you for your help. im studying for an exam so if you wanna stick around i'll have more questions

Answer 27

It is actually really late here...But I might be on for like a few more minutes