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In light of the not so well defined integral $\int_a^b \delta(x-a) dx$ and from David Z's comment at the end of this Math.SE post, consider the following equation, which I've come across in many sources on the discussion about cancellation of real and virtual corrections leading to infrared safety results: $$\int_0^1 dx \frac{dA(x)}{dx} + B = C = \text{finite}$$ which is then rewritten as $$\int_0^1 dx \left( \frac{dA(x)}{dx} + (B-C)\delta(1-x) \right) = \text{finite}.$$

So, here the integral $$\int_0^1 \delta(1-x)dx = 1$$ is used which is not the most sensible way to define the integral of the delta function with one of the end points coinciding with the zero of the delta argument.

If we demand $$ \int_{-1}^1 \delta(x)dx = 1 = \int_{-1}^0 \delta(x)dx + \int_0^1 \delta(x)dx = \theta(0) + (1- \theta(0)) $$ then $\int_0^1 \delta(x) dx$ is not well defined in so far as $\theta(0)$ is not so well defined. But if we want both contributions on the right hand side to yield the same result then we get $\theta(0) = 1/2$ which I understand to be the frequentist value taken for the integral. So, my question is

  1. What is the physical reasoning of allowing one to take $\int_0^1 \delta(1-x)dx = 1$ here in this particular set up?

For more context, see this page: http://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/QCD/DGLAP_2.html - the relevant equations are the third and fourth displays.

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You cannot have a satisfactory answer unless you think about the Dirac function in another way.

You see, the Dirac function is not a function, and the fact that we call it like that makes many mathematicians angry. It is rather a distribution. You can think about distributions as something like integration measures over the space.

Let us make an example. Usually, in integrals like $\int_0^1$, you don't specify wheter you have a closed interval, an open interval or a mixed one. This is because this choice does not matter for standard integrals. With this distribution, the detail is crucial.

A measure on the space is usually given in terms of a function that weighs the different areas, but in the case of Dirac distribution (let us use proper names) the definition is different: the distribution $\delta(x-x_0)$ acts on sets $A$ as $$ \delta(x-x_0)(A)=\left\{\begin{array}{cc}1&\text{if}\;x_0\in A\\0&\text{otherwise}\end{array}\right. $$ This distribution is assigned by assigning values on sets. In this case the question can be well posed: if you mean an integral over $(0,1]$ or $[0,1]$, the integral of the Dirac delta (or else, the measure of the interval) is $1$, else it is $0$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Feb 20 '17 at 12:00

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