Delta function centered on the boundary of an integral Suppose I have a function $f : \mathbb{R} \rightarrow \mathbb{R}$ evaluated at some $k\in\mathbb{R}$. Often (espacially in physics contexts), one must represent $f(k)$ in the following way:
$$f(k) = \int_{-\infty}^{\infty} dx \delta(x - k)f(x).$$
My question concerns the bounds of this integral. For example, if $U$ is some open set containing $k$, it is fine to use the following representation instead:
$$f(k) = \int_U dx \delta(x - k)f(x).$$
But what if $k$ is in the boundary of the set I am integrating over; for example, if $C$ is an interval like $C = [k, k+1]$, can I write
$$f(k) = \int_C dx \delta(x - k)f(x)?$$
For regular functions, I know that the value of the function on the boundary of the integral does not affect the value of the integral, but I am not sure how this works for distributions.
 A: There is no unique way to define the integral if the delta function is located on the boundary of the interval, and the result will depend on context. Put in another way, the result is ambiguous from distribution theory alone and will take on different values depending on the model/representation of the delta function that is most useful in the case studied.
Some references assume that $\int_{0}^\infty \delta (x)\phi(x)dx=\phi(0)/2$ which is a valid choice, especially if the appearance of the delta function happens in the context of Fourier series (this demonstrated in the link). This is corroborated by the fact that Fourier series for a discontinuous function always end up assigning the value $\frac{f(0^+)+f(0^-)}{2}$ to the discontinuity. This definition works for most linear differential equations that appear in the field of quantum mechanics (as long as they are second order), but this should come as no surprise, given that they are intimately connected to the Fourier transform.
However, representations of the delta function can be found where the integral can have an arbitrary value in $(0,1)$. Consider the one-parameter family of indicator functions
$$\phi_a(x)=\frac{1}{(\kappa+1)a}\mathbb{I}_{(-a, \kappa a)}~~~~~ a,\kappa>0$$
Then it is easy to show they are normalized and $\int_0^\infty \phi_a(x)dx=\kappa/(\kappa+1)$. Also one can show directly by integrating against a test function that
$$\lim_{a\to 0}\phi_a(x)=\delta(x)$$
$$\delta(bx)=\frac{\delta(x)}{b}~~,~~ b>0$$
so this distribution has the desired distribution behavior but is asymmetric around the origin, yet scale invariant. What gives? The point of the standard delta function lore to be contested here is whether one considers the property $\delta(-x)=\delta(x)$ to be of fundamental importance. I personally think that this property is model dependent.
To further support the model dependence of the delta function, the discussion here shows various contexts in which the traditional assumptions (at least in physics) about the interactions of the delta function with discontinuities are violated. From a physicist's point of view, it seems reasonable to assume that whenever it appears in contexts where the model allows for a continuous response function, the traditional delta function formalism seems to hold very well. However, whenever it sources a discontinuous response, the solutions obtained by assuming a standard symmetric model may be ambiguous, hinting at the fact that our model of the response needs to be more detailed.
