Question

I don't get why this is zero..i got \(\infty\)

\[\LARGE \int_{-\infty}^{\infty} xe^{-x^2} dx\]

Answer 1

Well, whats
\[\int\limits_{}^{}xe ^{-x^2}dx\]

Answer 2

Hint:
\[\int\limits_{}^{}xe^{-x^2} dx \text{ Let } u=-x^2 => du=-2 x dx => \frac{-1}{2} du =dx\]

Answer 3

hmm lemme type my solution..gonna be a lot of lateces...

Answer 4

\[\lim_{t \rightarrow \infty} \int_{0}^{t} xe^{-x^2}dx + \lim_{t \rightarrow -\infty} \int_{-t}^{0} xe^{-x^2} dx\]
\[\large -\frac{1}{2}\lim_{t \rightarrow \infty} [e^u] \stackrel{t}{0} - \frac{1}{2}\lim_{t \rightarrow -\infty} [e^u] \stackrel{0}{-t} \]

by the way...it would be fun to learn how to do that latex other than \stackrel that looks like a fraction without the fraction line

anyway

\[ [e^{\infty} - e^0] = \infty\]

\[e^0 - e^{-\infty}] = 1 - 0 = 1\]

\[\frac{-\infty}{2} - \frac{1}{2} = -\infty\]

right?

Answer 5

\[\int\limits_{}^{}xe^{-x^2} dx=\int\limits_{}^{}e^u \frac{-1}{2} du=\frac{-1}{2}e^u+C=\frac{-1}{2}e^{-x^2}+C\]

\[\int\limits_{-\infty}^{\infty} xe^{-x^2} dx=\int\limits_{-\infty}^{0}xe^{-x^2} dx+\int\limits_{0}^{\infty}xe^{-x^2} dx\]
\[\lim_{a \rightarrow -\infty}[\frac{-1}{2}e^{-x^2}]_{a}^0+\lim_{b \rightarrow \infty}[\frac{-1}{2}e^{-x^2}]_0^{b}\]

\[\lim_{a \rightarrow -\infty}[\frac{-1}{2}-\frac{-1}{2}e^{-a^2}]+\lim_{b \rightarrow \infty}[\frac{-1}{2}e^{-b^2}-\frac{-1}{2}]\]
But \[\lim_{u \rightarrow \infty}e^{-u}=0\]
So we have both parts converge :)
\[\frac{-1}{2}-\frac{-1}{2}\]

Answer 6

x^2 is positive no matter if x approaches negative infinity or positive infinity

Answer 7

So -x^2 is always negative if x approaches negative infinity or positive infinity

Answer 8

\[\lim_{u \rightarrow \pm \infty}e^{-u^2}=0\]

Answer 9

uhmm okay...i think i stopped getting it after

\[\lim_{a \rightarrow -\infty} [\frac{-1}{2} - \frac{1}{2}\] thingy

what does that mean??

Answer 10

uhmm @myininaya ?

Answer 11

Those are parenthesis

Answer 12

@myininaya just uses those to show that it's the limit of the whole thing, not just the part.

Answer 13

Did that explain you question @Igbasallote because I didn't understand your question?

Answer 14

what i dont get is where the second -1/2 sprouted from

Answer 15

Plugin in the upper and lower limits?

Answer 16

ohhhh thanks...ill go back to the solution again now

Answer 17

i get it...seems the probelm with what i did is that i didnt bring back to -x^2 *facepalm*

Answer 18

\[\frac{-1}{2}e^{-x^2}|_{a^2}^0=\frac{-1}{2}e^{-0^2}-\frac{-1}{2}e^{-a^2}=\frac{-1}{2}(1)+\frac{1}{2}e^{-a^2}=\frac{-1}{2}+\frac{1}{2}e^{-a^2}\]

Answer 19

We do the other part similarly

Answer 20

how do you do those limits o.O the one by the brackers

Answer 21

Are you still talking about pluggin' in the upper and lower limit?

Answer 22

no..the latex..how?

Answer 23

You know you can look at what I type by clicking it and going to tex commands

Answer 24

google chrome haha =)) it doesnt view there

Answer 25

And it will show you what I typed to get the prettiness

Answer 26

Like does it automatically minimize the window? And you can't open it?

Answer 27

It really is annoying. I have chrome. I know what you are talking about but you can go to your chrome thingy at the task bar below (or to the side if you are that type) and if you maximize the screen you will see what I wrote
OR
I can just tell you
lol