Question

evaluate
integration from 0 to infinity e^-x^2

Answer 1

looks like error function(x)

Answer 2

\[\int\limits_{0}^{infinity} e ^{x ^{2}}\]

Answer 3

hold on
its -x

Answer 4

\[\int\limits_{0}^{infinity} e ^{-x ^{2}}\]

Answer 5

erf(inf)=1

Answer 6

use the substitution x^2=u and use integratoin by parts

Answer 7

it is to be done by gamma function
it says

Answer 8

no way you are going to get a nice closed form for this

Answer 9

help me evaluate

Answer 10

ok.. whenyou do the substitution i said, you'd get a gamma function
\[
x^2=t\quad dx={dt \over 2\sqrt{t}}={t^{-1/2}dt\over2}\\
I={1\over2}\int_0^\infty t^{-1/2}e^{-t}dt={\Gamma(1/2)\over2}
\]

Answer 11

i think you have to go to another dimension i.e. introduce \(y\)
honestly i forget, but you could google the proof,
the answer is well known \(\frac{\sqrt{\pi}}{2}\)

Answer 12

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

Answer 13

got it electro
right!!!!

Answer 14

wow much snappier!!! i have only seen it done this way
http://www.youtube.com/watch?v=fWOGfzC3IeY

Answer 15

that is from -infinity to + infinity

Answer 16

it is even, take half
now all we need is the proof that \(\Gamma(\frac{1}{2})=\sqrt{\pi}\)

Answer 17

well
i figured it out x) bt thanks