Question

Show the following:
NOTE: I would like a fair hint please, not the answer.

int_{0}^{\infty}x^2e^{-x^2}dx = \frac{1}{2} int_{0}^{\infty}e^{-x^2}dx

Answer 1

@dpaInc
@Mertsj
@Zarkon
@amistre64
@TuringTest
@SmoothMath
@lalaly
@AccessDenied
@aroub
@experimentX
@eliassaab
@FoolForMath

Answer 2

\[\int\limits_{0}^{\infty}x^2e^{-x^2}dx = \frac{1}{2} \int\limits_{0}^{\infty}e^{-x^2}dx\]

Answer 3

See the second post^

Answer 4

do u know the gamma function?

Answer 5

I bet integration by parts first ... then ... then ...

Answer 6

How? The e^{-x^2} is not integrable (not an elementary function)...

Answer 7

thats why i asked u if u know the gamma fnction

Answer 8

It's a definite function ... half normal distribution function has standard value sqrt(pi)/2

Answer 9

Well ... one way is Gamma function
the other way it changing to polar form ... quite a bit messy

Answer 10

Sorry I do not know gamma function @lalaly. Suppose I know the following:


- integration techniques (by parts, U-sub, trig sub)
- approximate integration
- improper integration
- and the pre-calculus stuff

Am I able to solve this? I have yet to hear of a gamma function


- no gamma functions

Answer 11

Oh .. sorry, it's *definite integration

Answer 12

ever done integration in polar coordinates??

Answer 13

Suppose I do not know that "polar form" you are talking about @experimentX

Answer 14

Then there is no other way?

Answer 15

ha ... i meant polar coordinate

Answer 16

Ok suppose I do not know polar coordinates.

Answer 17

Oh .. sorry,
dxdy = ??

Answer 18

No.

Answer 19

\[ dxdy = r drd\theta\]

Answer 20

No.

Answer 21

No it's a formula for unit incremental area like dxdy in polar coordinate form

Answer 22

x = rcos theta
y = rsin theta
dx dy = r dr d\theta

Answer 23

you should know these relations

Answer 24

What is the topic name?

Answer 25

Ah ... i really don't know .. just hit on google

Answer 26

Alright leave this question aside for now. I will talk to my instructor about it.

Answer 27

sure ...