Integral involving Dirac Delta generalized function

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Homework Help Overview

The discussion revolves around evaluating integrals involving the Dirac delta function. Participants are examining properties of the delta function and its implications for specific integrals, while also addressing potential misunderstandings related to the definitions and applications of the function.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the property of the Dirac delta function and its evaluation in integrals. There are conflicting interpretations regarding the results of specific integrals, particularly the first two integrals mentioned. Some participants express confusion over a Chegg solution that appears to misapply the delta function.

Discussion Status

There is ongoing exploration of the properties of the Dirac delta function, with some participants providing guidance on its correct application. Multiple interpretations of the integrals are being considered, and participants are questioning the validity of external sources while seeking clarity on their own reasoning.

Contextual Notes

Some participants note the importance of showing work directly in posts rather than relying on images, which may hinder understanding. There is also mention of homework rules regarding the effort shown in posts.

Legend101
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Homework Statement


Evaluate the integrals in the attached image

Homework Equations

The Attempt at a Solution

 

Attachments

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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.
 
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 .
 

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Using the rule you attached, you should get 1 in the first integral. What about the second?
 
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 .
 

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Where did you get that image? That is not how the Dirac Delta is defined...
 
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
 
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.
 
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 :)
 
  • #13
@Legend101, if you start another thread with no effort shown, it will be deleted.
 
  • #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.
 

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