Question

Proving definite integral

Trying to prove that \[\frac{ 1 }{ \sqrt \pi } \int\limits_{-\infty}^{\infty} e^{-y^2} dy = 1\]

This actually stems from an electromagnetism question solving the delta function, however I've gotten this far and now am drawing a blank as to how to prove this? I 've used substitution to obtain this simplified form and am not sure where to go from here?

Answer 1

I have now reached the same point with a different delta funcion, obtaining \[ \int\limits_{- \infty}^{\infty} \frac{ 1 }{ \pi } \frac{ \sin y }{ y} dy = 1\]
I have tried to use integration by parts for this but I seem to be stuck at the evaluation part?

Any help would be greatly appreciated

Answer 2

substitute y^2=t and use the properties of even function prior to it...

Answer 3

siny/y is not integratable

Answer 4

I remember showing something similar to this using the Gamma Function.
Lemme keep trying, I think I've almost got it :U

Answer 5

@ankitshaw It definitely does work, http://www.wolframalpha.com/input/?i=%5Cint%5Climits_%7B-+%5Cinfty%7D%5E%7B%5Cinfty%7D+%5Cfrac%7B+1+%7D%7B+%5Cpi+%7D+%5Cfrac%7B+%5Csin+y+%7D%7B+y%7D+dy+

I'm just not sure now, hanks @zepdrix :-)

Answer 6

Grr I feel like I'm getting close :[
But there are still some holes in my steps...
Lemme show you what I was trying,

Answer 7

Starting with this,
\[\Large\rm \int\limits\limits_{-\infty}^{\infty} e^{-y^2} dy\]Making the substitution,\[\Large\rm u=y^2\]\[\Large\rm du=2y~dy\]And then you have to do some fancy jazz to get it in terms of u,
Oh let's uhhhh split the original integral in half, that might solve the problem I was running into,\[\Large\rm \int\limits_{-\infty}^{0} e^{-y^2}dy+\int\limits_{0}^{\infty} e^{-y^2}dy\]And then use rules of integration,\[\Large\rm -\int\limits\limits_{0}^{-\infty} e^{-y^2}dy+\int\limits\limits_{0}^{\infty} e^{-y^2}dy\]So after some fancy foot work you get it in terms of u,\[\Large\rm -\frac{1}{2}\int\limits_0^{\infty}e^{-u}u^{-1/2}~du+\frac{1}{2}\int\limits_0^{\infty}e^{-u}u^{-1/2}~du\]Hmm but then they end up canceling out :-(
Yah I made a boo boo somewhere in there...
Anyway, I'm trying to write it in this form:\[\Large\rm \Gamma(t)=\int\limits_0^{\infty}e^{-u}u^{t-1}~du\]So we would just change the -1/2 to 1/2-1 which let's us know we're dealing with Gamma of 1/2.\[\Large\rm \Gamma(1/2)\]Which is equal to sqrt(pi).
(We would have to go through a bit of a longer process if we needed to prove that though).

Answer 8

http://raghumahajan.wordpress.com/2010/07/26/integral-of-sinxx/

Answer 9

bah i dunno :c i gotta get some sleep right now

Answer 10

Try to see an easy proof using polar coordinates and double integration;
The details are posted in
http://en.wikipedia.org/wiki/Gaussian_integral

Answer 11

Yeah the methods described are right, but the polar one is what's used often. It's easier.

Answer 12

You can also use the Laplace Transform (or Fourier Transform if you have the time).