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|dw:1353910548451:dw|

You need to use integration by parts

put u= -y^2/2 then.

Is it:\[\int\limits\limits_{-\infty}^{\infty}ke^{-\frac{y^2}{2}}dy=1\]?

Yes

u= -y^2/2
find dy =... ?

I used u=y^2

but maybe its easier with u=-y^2/2

Its not smart to have that negative there

Do you agree?

oh, sorry.... yeah, its easier with e^u
constants can be rearranged easily...

ohhhhh

Yes the answer is 1/ sqrt(2*pi)
I just so dont follow how they got that answer.

did u try that substitution and then integration by parts ?

I was getting weird answers. I have not really touched calculus for over a yr so i cld be messing up

\(\huge\int \frac{e^{-u}u^{-1/2}}{\sqrt2}du\)
when 2u=y^2
yes, its messy...

How would you continue from here?

Can you show me?

ok dont show me

hmmm I know the gamma distribution function

ohhhh I see it now lol

\( \Gamma(z) = \int_0^\infty e^{-t} t^{z-1} dt \)

Yesssssss

now , you'll get it easily.

ask if u don't get...

Thanks let me try to work it out then. Thanks :)))))

what?

\(\Gamma{1/2} = \sqrt{\pi}\)

you'll need it.

ohhh coool never knew that

I feel like i am missing the basics in math :(

http://en.wikipedia.org/wiki/Gamma_function