Integral of Gaussian function

By Graham Lee (P1021) on Thursday, January 6, 2000 - 02:07 pm :

Can anyone prove that the definite integral between minus infinity and infinity of the Gaussian function is 1?

'Cos I can't.

Love,
GL.

By Alex Barnard (Agb21) on Thursday, January 6, 2000 - 02:27 pm :

If by the Gaussian function you mean something like e^-x²/2 then here is how:

Consider instead the function e^-x²/2-y²/2 on the x,y-plane. What is the integral of this over the whole plane? Well you can do the x and the y integrals separately and get that the answer is the square of the Gaussian integral you wanted - so if we can do this new question then we will be able to do the one you wanted.

Okay - the whole function is very rotationally symmetric about the origin so we should change variables to plane polar coordinates (r,t). The function becomes e^-r²/2 and the area-element dx.dy becomes r.dr.dt. So we have to integrate:

r.e^-r²/2 dr.dt from r=0..infinity and t=0..2p

Hopefully you will be able to do this integral and you'll see that the overall integral is 2p.

So the original Gaussian integral had value equal to either
+
Ö

2p

or
-
Ö

2p

. But it was the integral of an always positive function and so it must be the former.

AlexB.