Dirac delta function integrated on a finite interval

In summary, the Dirac delta function is a mathematical construct that represents an idealized point mass or charge concentrated at a single point. When integrated over a finite interval that includes the location of the delta function, the integral evaluates to one, reflecting the total "weight" of the function at that point. If the interval does not include the point of concentration, the integral is zero. This property makes the delta function a useful tool in physics and engineering, particularly in systems involving impulse forces or point sources.
  • #1
Hill
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How to account for integration limits with Dirac delta function?
Which is correct:
$$\int_{-1}^1 \delta (x-x_0) \, dx =\begin{cases} 1, -1\leq x_0 \leq 1 \\ 0, \text { otherwise} \end{cases}$$
or
$$\int_{-1}^1 \delta (x-x_0) \, dx =\begin{cases} 1, -1< x_0 < 1 \\ 0, \text { otherwise} \end{cases}$$
?
 
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  • #2
OK, I know. The first one.
 
  • #3
Sorry, not so sure, again. Because, e.g., $$\int_{-1}^2 \delta (x-1) \, dx = 1$$ but if the first case above is correct, then $$\int_{-1}^2 \delta (x-1) \, dx = \int_{-1}^1 \delta (x-1) \, dx +\int_1^2 \delta (x-1) \, dx =2$$ and if the second case is correct, then $$\int_{-1}^2 \delta (x-1) \, dx = \int_{-1}^1 \delta (x-1) \, dx +\int_1^2 \delta (x-1) \, dx =0$$ So, how does it work?
 
  • #4
By symmetry, we must have:
$$\int_0^1 \delta(x) dx = \frac 1 2$$I use a rule of thumb that what holds for ##\delta (x)## must hold for the limit of appropriate Gaussians.
 
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  • #5
PeroK said:
By symmetry, we must have:
$$\int_0^1 \delta(x) dx = \frac 1 2$$I use a rule of thumb that what holds for ##\delta (x)## must hold for the limit of appropriate Gaussians.
I see. Then you disagree with the following example from the QFTftGA:

1700330922760.png


By your definition, it should rather be $$...=\begin{cases} 1, -1< x_0 < 1 \\ \frac 1 2, x_0=-1\\ \frac 1 2, x_0=1\\ 0, \text { otherwise} \end{cases}$$ Is it so?
 
  • #6
I'd say so. I can check my notes, but it seems a minor detail. I told you the maths was wild and woolly.

As I recall, Tobias Osborne gives a good definition of the functional derivative somewhere in his lectures.
 
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  • #7
PeroK said:
I told you the maths was wild and woolly.
Yes, you did, and I keep it in mind. :smile:
 
  • #8
PeroK said:
I'd say so. I can check my notes, but it seems a minor detail. I told you the maths was wild and woolly.

As I recall, Tobias Osborne gives a good definition of the functional derivative somewhere in his lectures.
This other example from the book is perhaps more "wild and woolly," but I might be wrong.

They use $$\frac {\partial} {\partial y}[f(y)+\epsilon\delta(y-x)]=f'+\epsilon\delta'(y-x)$$ Can we do this? Derivative of ##\delta()##?

A bit later, they integrate by parts:

1700340733766.png


taking ##\delta## outside. Is it OK?

P.S. There should be a typo: ##dy## is missing in the last integral.
 

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  • #9
Finally, I'll go ahead and trust that these manipulations with Dirac delta are fine. I like this derivation:

1700343564515.png


They do not explain why the term in square brackets on the third line vanishes, but I understand that it is so because ##t_i<t<t_f##, and this is because we vary the trajectory ##x(t)## everywhere except the end points.
 
  • #10
This is Chapter 1.3. I studied from this book in 2020 so I can't remember everything. I have, however, found my notes.

Interesting, I have no notes on this section! This suggests there was something I didn't like about it! I guess I must have decided that L&B had produced the E-L equations by some exotic mathematics that I didn't understand or didn't like or both. But, the net result was just the E-L equations. Which I was already completey familair with. So, I moved on. I guess I was hoping that their particular derivation of E-L wouldn't turn out to be important. In the worst case, I'd have to come back to this section and try to figure it out.

If I'm studying advanced material like this, with no tutor, then I have to be pragmatic.

Also, when I later watched Tobias Osborne's videos, his approach to functional derivatives made much more sense.
 
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  • #11
Here are two lecture notes (physics motivated, both under 6 pages) that helped me:
This one does go into the splitting bounds apart on page 5: http://jacobi.luc.edu/DiracDelta.pdf
This one is more pertaining to your HW thread (Higher order functional derivatives), and it may be more useful for you: https://www.physics.usu.edu/Wheeler/QFT2016/Notes/QFT09FunctionalDerivatives.pdf

However, you'll most likely need to go review a more mathematical oriented text on distributions if you want more generalizations and rigor.
 
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FAQ: Dirac delta function integrated on a finite interval

What is the Dirac delta function?

The Dirac delta function, denoted as δ(x), is a mathematical construct used to model an idealized point mass or point charge. It is not a function in the traditional sense but rather a distribution that has the property of being zero everywhere except at a single point, where it is infinitely high, and its integral over the entire real line is equal to one.

What happens when you integrate the Dirac delta function over a finite interval that includes its point of support?

When you integrate the Dirac delta function over a finite interval that includes its point of support (for example, the interval [a, b] where a < 0 < b), the result of the integral is 1. This is due to the sifting property of the delta function, which effectively "picks out" the value at the point where it is centered.

What is the result of integrating the Dirac delta function over a finite interval that does not include its point of support?

If you integrate the Dirac delta function over a finite interval that does not include its point of support (for example, the interval [a, b] where both a and b are either less than or greater than the point of support), the result of the integral is 0. This is because the delta function has no contribution outside its point of support.

How can the Dirac delta function be represented mathematically?

The Dirac delta function can be represented mathematically using the limit of a sequence of functions. One common representation is δ(x) = lim(ε→0) (1/(2ε)) for |x| < ε, and δ(x) = 0 for |x| ≥ ε. This representation shows how the delta function behaves like a peaked function that becomes infinitely narrow and tall while maintaining an area of one under the curve as ε approaches zero.

What are some applications of the Dirac delta function in physics and engineering?

The Dirac delta function is widely used in physics and engineering, particularly in the fields of signal processing, control systems, and quantum mechanics. It is used to model instantaneous impulses, point sources, and to simplify the representation of systems in differential equations. For example, in signal processing, it is used to represent an impulse response of a system, while in quantum mechanics, it is used to describe the position of a particle in space.

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