Question

An Integral a Day:
day 3

\[\int_{-\infty}^\infty e^{ax^2}dx\]

Answer 1

this will only converge if a<0

Answer 2

double integral in polar coordinates?

Answer 3

oh, I don't know that...
is that like the CPV ?

Answer 4

not really sure ... values are usually in pi

Answer 5

and i would love to derive that.

Answer 6

I got up to\[\int_{-\infty}^{\infty}e^{-ax^2}dx=2\int_{0}^{\infty}e^{-ax^2}dx;a>0\]\[u=ax^2\implies x=\sqrt{\frac ua}\]\[du=2axdx\implies dx=\frac{du}{2\sqrt{au}}\]and the integral becomes\[\frac1{\sqrt a}\int_{0}^{\infty}u^{-1/2}e^{-u}du\]at which point I'm stuck.

Answer 7

apparently (according to wolfram) the integral above converges to \(\sqrt\pi\) but I have no idea how to show that

Answer 8

looks like gamma function

Answer 9

not something I've worked with, though I sure know about it

Answer 10

http://en.wikipedia.org/wiki/Gamma_function
it sure is a gamma function.

Answer 11

ah, and it specifically mentions the proof for the gamma function of argument 1/2 is \(sqrt\pi\)

Answer 12

\(\sqrt\pi\)*

Answer 13

1/a * integ 0 to inf e^-u t^(1-3/2) du

Answer 14

that is equal to 1/sqrt(a) * ( gamma(3/2) = 1/2 * sqrt(pi))
that is our answer.

Answer 15

yep :)
cool, first time I'm understanding it
so we have\[\frac1{\sqrt a}\Gamma(\frac12)\]

Answer 16

you made mistake .. gamma 3/2

Answer 17

must be 1-z

Answer 18

yes, my mistake
like I said, I'm new to the gamma function

Answer 19

i think it got something to do with real analysis.

Answer 20

check this out too ... it occurs frequently has some kinda relation with gamma function ... though i forgot.
http://en.wikipedia.org/wiki/Beta_function

Answer 21

wait, no, my exponent is negative
1-z=1-1/2=-1/2
so its gamma(1/2) isn't it?

Answer 22

that should read z-1=1/2-1=-1/2
dyslexia!

Answer 23

no ,,, you had -1/2 to begin with and that is 1-3/2

Answer 24

yes, but \[\Gamma(z)=\int_{0}^{\infty}e^{-u}u^{z-1}\]plug in z=1/2 and you get my integral

Answer 25

I'm sorry if I'm confused, I just want to understand...

Answer 26

now i am confused.

Answer 27

scroll down in the wiki article to "general principles"
it specifically mentions \[\Gamma(\frac12)=\sqrt\pi\]

Answer 28

1-z = -1/2

Answer 29

yeah i know that ... and gamma works like factorial.

Answer 30

but it is defined as t^(z-1)

Answer 31

not 1-z...

Answer 32

looks like ... i am nuts. you are right.

Answer 33

ok, sweet!
I finally learned something :)

Answer 34

thanks for hanging in there with me, I needed to think it through

Answer 35

lol ... it was the first thing written in wikipedia article.

Answer 36

momentary dyslexia
so frustrating

Answer 37

;)

Answer 38

and the answer was concluded with
\[ \sqrt{\frac{\pi}{a}}\]

Answer 39

...with \(a>0\)
The original problem has \(e^{ax^2}\)so that answer would be technically be\[\sqrt{\frac\pi {-a}};a<0\]but same thing of course

Answer 40

Turing, you were right, double integral in polar coordinates is the way I went. Also I made a mistake and missed a minus sign in the exponent writing it down.
\[ I = \int_{-\infty}^\infty e^{-ax^2}dx\]\[I^2 = \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-a(x^2+y^2)}dxdy\]then switch to polar:\[I^2 = \int_0^\infty \int_0^{2\pi}\rho e^{-a\rho^2}d\rho d\phi\]\[I^2 = 2\pi \int_0^\infty \rho e^{-a\rho^2}d\rho\]substitution as follows:
\[ \rho^2 =u ~~~~~~~~ du = 2\rho d\rho \] and our integral becomes:
\[I^2 = \pi \int_0^\infty e^{-a u} du = \frac{-\pi e^{-a u }}{a}|_0^\infty \] which evaluates to:
\[I^2 = lim ~~u\to \infty~~( \frac{-\pi e^{-au}}{a} + \frac{\pi}{a}) = \frac{pi}{a}\]\[and~~so~~I = \sqrt{\frac{\pi}{a}}\]