Integral involving Dirac Delta generalized function

In summary: I tried to include everything in the post.See my post in that thread. I tried to include everything in the post.
  • #1
Legend101
15
0
Member warned about posting with no effort

Homework Statement


Evaluate the integrals in the attached image

Homework Equations

The Attempt at a Solution

 

Attachments

  • image.jpg
    image.jpg
    23.3 KB · Views: 877
Physics news on Phys.org
  • #2
The dirac delta function has the great property that ##\int_{-\infty}^{\infty} \delta(x) dx = 1 ##.
Couple that with what you know about when it is not equal to zero, and these integrals are quick and easy.
In the future, please put a little more meat into you post, so we can better understand what you already know and what you need help fixing.
 
  • #3
RUber said:
The dirac delta function has the great property that ##\int_{-\infty}^{\infty} \delta(x) dx = 1 ##.
Couple that with what you know about when it is not equal to zero, and these integrals are quick and easy.
In the future, please put a little more meat into you post, so we can better understand what you already know and what you need help fixing.
thanks . I already know this property and many others (including the one in this new attached picture) . If I apply that rule (image) , i would get 1 in the first integral while many others told me 0 . I have problem with the first 2 integrals only .
 

Attachments

  • image.jpg
    image.jpg
    9.4 KB · Views: 633
  • #4
Using the rule you attached, you should get 1 in the first integral. What about the second?
 
  • #5
RUber said:
Using the rule you attached, you should get 1 in the first integral. What about the second?

The second one would be 0 since -0.5 is not in the interval of integration .
I found a solution to the first integral which totally confused me ( in the attached image) claiming that the answer should be 0 for 1st integral .
 

Attachments

  • image.jpg
    image.jpg
    28.8 KB · Views: 2,728
  • #6
Where did you get that image? That is not how the Dirac Delta is defined...
 
  • #7
Matterwave said:
Where did you get that image? That is not how the Dirac Delta is defined...
It is a Chegg solution . And it confused me .
I believe that the first integral should be 1 , i just need more proof
 
  • #8
Legend101 said:
It is a Chegg solution . And it confused me .
I believe that the first integral should be 1 , i just need more proof

I think the proof is the rule you quoted in post 3. The Chegg solution appears to be for a unit step function, not for a delta function. It's wrong.
 
  • #9
Using the fact that $$\int_a^b f(x)\delta(x-c)dx = f(c),~\text{iff}~~a<c<b$$ we can get immediately $$\int_0^5 \cos(2\pi t)\delta(t-2)dt=\left.\cos(2\pi t)\right|_{t=2}=\cos(4\pi)=1$$

The Chegg solution is clearly wrong. The function used there is not a dirac delta function.
 
  • #10
There is a related function, called the "Heaviside step function" or just the "step function":

[itex]x < 0 \Rightarrow \Theta(x) = 0[/itex]

[itex]x = 0 \Rightarrow \Theta(x) = 1/2[/itex]

[itex]x > 0 \Rightarrow \Theta(x) = 1[/itex]

The definition in that image seems correct for the step function, but not the delta function. The two are related by:

[itex]\Theta(x) = \int_{-\infty}^x \delta(s) ds[/itex]
 
  • #11
I agree with Dick and Matterwave. They were defining the Heavyside function which is related to the dirac delta by:
##H(x) = \int_{-\infty}^x \delta(t) dt## or ##\frac{d}{dx} H(x) = \delta(x) ##

Edit: Got scooped by stevendaryl...Exactly.
 
  • #12
Thank you guys for your contribution :)
I'll be posting a new topic regarding convolution integral :)
 
  • #14
Mark44 said:
@Legend101, if you start another thread with no effort shown, it will be deleted.
No worries . I showed my efforts in attached file
 
  • #15
Legend101 said:
No worries . I showed my efforts in attached file
You need to show what you have tried in the first post of the thread. And it would be better to include the work directly in the post, rather than an image of the work.
 
  • #16
Mark44 said:
You need to show what you have tried in the first post of the thread. And it would be better to include the work directly in the post, rather than an image of the work.
The work involves many integrals and mathematical notations . It would be hard to show all the work directly written on the post especially if someone is using a smartphone .
 
  • #17
Legend101 said:
The work involves many integrals and mathematical notations . It would be hard to show all the work directly written on the post especially if someone is using a smartphone .
See my post in that thread.
 

1. What is a Dirac Delta generalized function?

A Dirac Delta generalized function, also known as the Dirac delta function, is a mathematical concept used in physics and engineering to represent an idealized point of infinite density and zero width. It is often used to model point particles or to describe the distribution of point charges in an electric field.

2. What is the purpose of using Dirac Delta generalized functions in integrals?

Dirac Delta generalized functions are used in integrals to simplify calculations and provide a way to handle discontinuous or singular functions. They allow for the representation of a point source or impulse in a mathematical expression, which can be useful in various physical and engineering applications.

3. How is a Dirac Delta generalized function defined mathematically?

The Dirac Delta generalized function is defined as a distribution, rather than a traditional function, and is represented by the symbol δ(x). Its mathematical definition is given by the following properties:

  • δ(x) = 0 for all x ≠ 0
  • ∫ δ(x) dx = 1
  • ∫ f(x)δ(x) dx = f(0)

where f(x) is a continuous function.

4. Can Dirac Delta generalized functions be integrated like regular functions?

No, Dirac Delta generalized functions cannot be integrated in the traditional sense. They are not defined at a single point and do not have a definite value, making traditional integration techniques invalid. Instead, they are integrated using the concept of distribution, which involves using test functions to evaluate the integral.

5. What are some common applications of Dirac Delta generalized functions?

Dirac Delta generalized functions have various applications in physics and engineering. For example, they are used to model point sources of gravitational and electromagnetic fields, and to describe the behavior of particles in quantum mechanics. They are also used in signal processing to represent impulse signals and in control systems to model instantaneous changes in a system.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
31
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus
Replies
25
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
Back
Top