Why is the integral of the Dirac delta distribution equal to unity?

[SW VandeCarr]

Previously I posted a question on the Dirac delta function and was informed it was not a true function, but rather a distribution. However, I have to admit I still did not understand why its integral (neg inf to pos inf) is unity. I've thought about this and came up with the following:

Consider a normal distribution with a variance of 0. Since the integral of a normal distribution is constrained to be unity, the mean must go to infinity as the variance approaches zero. This is the Dirac delta. Is this reasonable or am I wandering in the wilderness?

 

[ice109]

you're confusing probability distribution and distribution.

http://en.wikipedia.org/wiki/Distribution_(mathematics )

they're probably related cause a lot of the words in that article are in probability articles but not like you think they are.

 

[mathman]

Consider a normal distribution with a variance of 0. Since the integral of a normal distribution is constrained to be unity, the mean must go to infinity as the variance approaches zero. This is the Dirac delta. Is this reasonable or am I wandering in the wilderness?

The mean doesn't have to go to infinity. It can stay at zero. In that case the density function has as its "limit" the Dirac delta function at 0. In other words the limit of the probability distribution is a Schwartz distribution.

 

[SW VandeCarr]

The mean doesn't have to go to infinity. It can stay at zero. In that case the density function has as its "limit" the Dirac delta function at 0. In other words the limit of the probability distribution is a Schwartz distribution.

I don't know much about general distribution theory, but I read that Schwartz "legitimized" the Dirac delta function by redefining it as a functional. In any case, I'm happy the mean can stay at zero. Your description of the Dirac delta as the 'limit' of the density function at 0 is very clear. Thanks mathman.

 

[Quantumjump]

Previously I posted a question on the Dirac delta function and was informed it was not a true function, but rather a distribution. However, I have to admit I still did not understand why its integral (neg inf to pos inf) is unity. I've thought about this and came up with the following:

Consider a normal distribution with a variance of 0. Since the integral of a normal distribution is constrained to be unity, the mean must go to infinity as the variance approaches zero. This is the Dirac delta. Is this reasonable or am I wandering in the wilderness?

Yes the idea is okay, but since you cannot consider the dirac-delta as a function, you cannot either interpret the integral in its usual way. Basically, the idea is that the delta can be seen as a limit of a sequence of functions (also called distribution). Typically, that function can be a gaussian (but they are other equally acceptable definitions, in terms of sine for example). A gaussian depends upon x and sigma. If you choose to tends sigma to zero, since the gaussian is normalized (that is, overal area=1), the width tends to zero, the central peak tends to infinity, while everwhere else it tends to zero. Now, if you integrate such a gaussian multiplied by any function, then apply the limit for sigma tending to zero, the value of the function can be considered as a constant upon the withd, so, if the with is centered around zero, you can put the f(0) out of the integral, and because the gaussian is normalize, the integral gives you 1, so it remains just f(0). Hence the definition, at least from a heuristic (physics) point of view.

 

[Tac-Tics]

For intuitions about the Dirac Delta, you should think of it as a bell curve with infinitesimal variance (spread). It's still got unit area under the curve, but all that area is concentrated in within an infinitesimal neighborhood of the mean.

EDIT. It seems you are thinking about this more or less correctly. And you're right, that the value at the mean is infinite (the inverse of an infinitesimal).