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Dirac delta   (5.8.02)

Herman Jaramillo

When I compute the integral of the Dirac delta function between 0 and infinity using Maple 7, I get "1". When I do it using Mathematica I get "1/2".

When I try to do it using distribution theory by choosing a test function which has bounded support on the positive real line and using the limit from the left and the right of the interval of integration (epsilon, infinity) as epsilon goes to 0 I get that the integral should be "0".

Does anyone understand this? Can somebody tell me where I can I find a formal proof using distribution theory whether this integral is 1,1/2 or zero?.

Fred Chapman   (6.8.02)

By definition, Dirac's delta distribution satisfies (in Maple notation)

                    infinity
                   /
                  |
                  |          f(x) Dirac(x) dx = f(0)
                  |
                 /
                   -infinity

In your case, f(x) is implicitly defined as the Heaviside function H(x), because

           infinity                  infinity
          /                         /
         |                         |
         |          Dirac(x) dx =  |          H(x) Dirac(x) dx
         |                         |
        /                         /
          0                         -infinity

where

                              { 0        x < 0
                       H(x) = {
                              { 1        0 < x

The value of your integral is therefore H(0). The problem is that the definition of H(0) is itself ambiguous! Some people define H(0) = 1 so that H(x) is right-continuous at x = 0; some people define H(0) = 0 so that H(x) is left-continuous at x = 0; some people define H(0) = 1/2, which is the average (arithmetic mean) of the right-sided and left-sided limits 1 and 0.

I recommend that you make an arbitrary choice for the value of H(0), defining it to be always 0 or always 1/2 or always 1 according to what makes the most sense in the context of your larger problem. Whatever value you assign to H(0) is also the value of your original integral. You can't avoid making an arbitrary choice, but you can apply your choice consistently throughout your work, which is the important thing here.

For a practical introduction to distribution theory, I recommend Bernard Friedman's books: "Principles and Techniques of Applied Mathematics" (Dover, 1990) and "Lectures on Applications-Oriented Mathematics" (Wiley, 1991). For a more in-depth theoretical treatment, I recommend A.H. Zemanian's "Distribution Theory and Transform Analysis: An Introduction to Generalized Functions, with Applications" (Dover, 1987).

Robert Israel   (6.8.02)

The answer is that the question is not well-defined.

The Dirac delta is a Borel probability measure on the real line with a unit mass at the origin. So you can integrate this measure over any Borel set, in particular an interval from 0 to infinity. But when you said "between 0 and infinity" you didn't specify whether the interval is [0,infinity) or (0,infinity), i.e. whether or not 0 is included in the set. The integral over [0,infinity) is 1, the integral over (0,infinity) is 0. Mathematica's answer of 1/2 is not the integral over any set (which doesn't mean it's necessarily wrong - you could interpret it as integrating a function H(x) that is 0 for x < 0, 1/2 for x = 0 and 1 for x > 0).

John Robert Kitchin   (6.8.02)

The integral of a dirac delta function over any interval that contains it is by definition 1 (it is zero in any interval that does not contain it). I don't know how you got 1/2 out of mathematica, my version correctly outputs 1.

Check out this website, it is one of many I found by googling dirac delta function:

http://www.chm.uri.edu/urichm/chm532/delta/node4.html

it also gives an example of deriving the delta function from the limit of a distribution.

Richard B. Kreckel   (7.8.02)

It is not defined. Both Maple and Mathematica define it in some way they consider meaningful or elegant. Mathematica obviously defines it to be symmetric.

You may "represent" the Dirac delta function as any ordinary sharply peaked function (such that their integral is normalized to unity) and take the limit where the peak gets infinitely sharp. Take step functions for instance, you may chose a step sitting just right of zero, or just left of zero or centered around zero. From the point of view of distribution theory it does not matter, that representation is at best a visualization anyway.

| Does anyone understand this?. Can somebody tell me ...

There is no such thing.

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