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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 +sqrt(2p) or -sqrt(2p). But it was the integral of an always positive function and so it must be the former.

AlexB.