Integral involving Dirac Delta generalized function

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The discussion centers around evaluating integrals involving the Dirac delta function, highlighting its key property that the integral over its entire range equals one. Participants clarify that the confusion arises from a Chegg solution incorrectly applying the delta function instead of the Heaviside step function. The correct evaluation of the first integral is confirmed to be one, while the second integral evaluates to zero since the point of interest is outside the integration limits. There is a consensus that proper notation and showing work directly in posts would enhance understanding and communication. The thread emphasizes the importance of distinguishing between related functions like the Dirac delta and the Heaviside step function.
Legend101
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Homework Statement


Evaluate the integrals in the attached image

Homework Equations

The Attempt at a Solution

 

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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":

x &lt; 0 \Rightarrow \Theta(x) = 0

x = 0 \Rightarrow \Theta(x) = 1/2

x &gt; 0 \Rightarrow \Theta(x) = 1

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

\Theta(x) = \int_{-\infty}^x \delta(s) ds
 
  • #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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