Integral Involving the Dirac Delta Function

  • #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
 

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  • #8
PeroK
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I know intuitively that the integral is 1, can you explain it to me mathematically?
By definition:
$$\int_{-\infty}^{+\infty}\delta(x) f(x) dx = f(0)$$
 
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  • #9
wrobel
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Actually it is not an integral it just a symbol. As well as ##\delta-##function is not a function in usual sense. These things are clarified in functional analysis
 
  • #10
vanhees71
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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.
 
  • #11
wrobel
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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})##.
 
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  • #12
vanhees71
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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"?
 
  • #13
wrobel
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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})##
 
  • #14
vanhees71
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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!
 

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