Integral Involving the Dirac Delta Function

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
13 replies · 2K views
gabriellelee
Messages
21
Reaction score
1
Homework Statement
What is the integral of the following equation?
Relevant Equations
See below for the equation.
Screen Shot 2020-03-25 at 2.15.47 AM.png
 
Physics news on Phys.org
What are your limits of integral?
 
Abhishek11235 said:
What are your limits of integral?
-∞ to +∞
 
gabriellelee said:
-∞ to +∞
So what do you think?
 
Abhishek11235 said:
So what do you think?
1?
 
gabriellelee said:
1?
Yes
 
Abhishek11235 said:
Yes
I know intuitively that the integral is 1, can you explain it to me mathematically?
 
gabriellelee said:
I know intuitively that the integral is 1, can you explain it to me mathematically?
By definition:
$$\int_{-\infty}^{+\infty}\delta(x) f(x) dx = f(0)$$
 
  • Like
Likes   Reactions: PhDeezNutz and gabriellelee
Hm, is this an exercise in a physics course or a trick question in a mathematics course? In the latter case, I think, the answer is that the integral is undefined, because the exponential function is not a proper test function in the domain (of, e.g., Schwartz functions), the Dirac distribution is defined on.
 
There is no sense to speak about integrals here. It is nothing more than symbol in this context.
A space of test functions is a conditional question. In different problems this space is introduced differently this depends of problem we consider . Loosely speaking, the smaller space of test functions the bigger space of generalized functions.

For example, consider a subspace of ##\mathcal D'(\mathbb{R})## that consists of generalized functions with compact support. Such generalized functions are naturally defined on ##C^\infty(\mathbb{R})##.
 
Last edited:
  • Like
Likes   Reactions: jasonRF
I'm only a bit familiar with the formalism, but I don't think that ##\exp(\mathrm{i} x)## is a proper test function in any case. I'm a bit unsure, whether the Dirac ##\delta## function (which has compact support, namely only 1 point) is defined on the entire ##C^{\infty}(\mathbb{R})##. Isn't it usually either ##C_0^{\infty}(\mathbb{R})## or the "Schwartz space of quickly falling functions"?
 
What is the problem to define the Dirac δ function on ##C^\infty(\mathbb{R})##?
Just put
$$(\delta, \varphi):=\varphi(0),\quad \varphi\in C^\infty(\mathbb{R})$$
This is a continuous linear functional with respect the standard topology in ##C^\infty(\mathbb{R})##
 
I don't know. Maybe you can do that, but in quantum mechanics you often have confused students, because they think ##\exp(\mathrm{i} p x)## is a "eigenstate of the momentum operator", but it is not square integrable nor in the domain of the momentum operator as an essentially self-adjoint operator. Here you need the rigged-Hilbert space description!